Monty Hall Problem-Probability ParadoxThe Monty Hall problemVariation of the Monty Hall Problem.Modified monty hall problemThe Monty Hall Problem: 2 contrasting answers.Variation of Monty Hall problemMonty Hall Variation (100 doors)Does the Monty Hall Paradox hold true if the Game Host can open either door?Alternative analysis of the Monty Hall problemAnother Monty Hall problemHow to model and solve the variant of Monty Hall problem in which the host opens a door randomly?Monty Hall problem: If contestant knows the door with goat…
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Monty Hall Problem-Probability Paradox
The Monty Hall problemVariation of the Monty Hall Problem.Modified monty hall problemThe Monty Hall Problem: 2 contrasting answers.Variation of Monty Hall problemMonty Hall Variation (100 doors)Does the Monty Hall Paradox hold true if the Game Host can open either door?Alternative analysis of the Monty Hall problemAnother Monty Hall problemHow to model and solve the variant of Monty Hall problem in which the host opens a door randomly?Monty Hall problem: If contestant knows the door with goat…
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I just learned about the Monty Hall Problem and it seemed pretty much amazing to me.I am just a bit confused with it.
So,according to the problem we are on a game show, and we are given the choice of three doors: Behind one of them is a car and behind the others are goats. We start by picking one of the doors. After our selection, the host, who knows what's behind the doors reaveals one of the other two doors that has a goat. Now we are asked if we want to change our mind or stay with our initial pick.
According to Probabilities, if we don't swap and keep our first selection we get $(1/3)$ $33.3% $ chance of wining the car since the elimination of the door that was opened by the host doesn't affect the probability of our door having the car which remains $33.3%$ as it is at the initial problem.
On the other hand, if we switch the door with the other one left , then we get $(2/3)$ $66.6%$ chances of wining the car since only two doors are remaining and the fact that the host revealed a goat in one of the unchosen doors changed nothing about the initial probability of our door having the car.
Till this point it makes pretty much sense to me.
But what about assuming we have a man in the audience that makes a different initial choice in his head. For instance, let's say that the contestant picked door number $1$ and he picked door number $2$. If door number $3$ is the door that is being revealed by the host (has a goat) then both doors $1$ and $2$ remain in the game. For the contestant door number $2$ has $66.6%$ chances of having the car when for the man in the audience door number $1$ has $66.6%$ chances of having the car. Isn't that weird? From two different perspectives we get two different probabilities about the same unopened doors. How's that possible?
probability bayes-theorem monty-hall
asked Apr 20 at 13:17
MJ13MJ13
536
$endgroup$
add a comment |
$begingroup$
I just learned about the Monty Hall Problem and it seemed pretty much amazing to me.I am just a bit confused with it.
So,according to the problem we are on a game show, and we are given the choice of three doors: Behind one of them is a car and behind the others are goats. We start by picking one of the doors. After our selection, the host, who knows what's behind the doors reaveals one of the other two doors that has a goat. Now we are asked if we want to change our mind or stay with our initial pick.
According to Probabilities, if we don't swap and keep our first selection we get $(1/3)$ $33.3% $ chance of wining the car since the elimination of the door that was opened by the host doesn't affect the probability of our door having the car which remains $33.3%$ as it is at the initial problem.
On the other hand, if we switch the door with the other one left , then we get $(2/3)$ $66.6%$ chances of wining the car since only two doors are remaining and the fact that the host revealed a goat in one of the unchosen doors changed nothing about the initial probability of our door having the car.
Till this point it makes pretty much sense to me.
But what about assuming we have a man in the audience that makes a different initial choice in his head. For instance, let's say that the contestant picked door number $1$ and he picked door number $2$. If door number $3$ is the door that is being revealed by the host (has a goat) then both doors $1$ and $2$ remain in the game. For the contestant door number $2$ has $66.6%$ chances of having the car when for the man in the audience door number $1$ has $66.6%$ chances of having the car. Isn't that weird? From two different perspectives we get two different probabilities about the same unopened doors. How's that possible?
probability bayes-theorem monty-hall
asked Apr 20 at 13:17
MJ13MJ13
536
$endgroup$
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The main Monte Hall answer seems to be math.stackexchange.com/questions/96826/the-monty-hall-problem
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– Henry
Apr 20 at 23:58
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It is important to note that Monty's behavior is extremely important to this calculation. Even for the classic problem, if Monty cannot be depended on to always open a door with a goat, the probability calculation changes. When he can be depended on, it is only the contestant, not the audience member, who can depend on Monty's behavior, as lulu elaborates.
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– Paul Sinclair
Apr 21 at 0:56
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Note that it also happens that the observer choses the door that is going to be revealed empty. So even though he concludes that he should stick with his choce in the case you describe, that does not mean he will win 2/3 of attempts.
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 6:42
add a comment |
$begingroup$
I just learned about the Monty Hall Problem and it seemed pretty much amazing to me.I am just a bit confused with it.
So,according to the problem we are on a game show, and we are given the choice of three doors: Behind one of them is a car and behind the others are goats. We start by picking one of the doors. After our selection, the host, who knows what's behind the doors reaveals one of the other two doors that has a goat. Now we are asked if we want to change our mind or stay with our initial pick.
According to Probabilities, if we don't swap and keep our first selection we get $(1/3)$ $33.3% $ chance of wining the car since the elimination of the door that was opened by the host doesn't affect the probability of our door having the car which remains $33.3%$ as it is at the initial problem.
On the other hand, if we switch the door with the other one left , then we get $(2/3)$ $66.6%$ chances of wining the car since only two doors are remaining and the fact that the host revealed a goat in one of the unchosen doors changed nothing about the initial probability of our door having the car.
Till this point it makes pretty much sense to me.
But what about assuming we have a man in the audience that makes a different initial choice in his head. For instance, let's say that the contestant picked door number $1$ and he picked door number $2$. If door number $3$ is the door that is being revealed by the host (has a goat) then both doors $1$ and $2$ remain in the game. For the contestant door number $2$ has $66.6%$ chances of having the car when for the man in the audience door number $1$ has $66.6%$ chances of having the car. Isn't that weird? From two different perspectives we get two different probabilities about the same unopened doors. How's that possible?
probability bayes-theorem monty-hall
asked Apr 20 at 13:17
MJ13MJ13
536
$endgroup$
I just learned about the Monty Hall Problem and it seemed pretty much amazing to me.I am just a bit confused with it.
So,according to the problem we are on a game show, and we are given the choice of three doors: Behind one of them is a car and behind the others are goats. We start by picking one of the doors. After our selection, the host, who knows what's behind the doors reaveals one of the other two doors that has a goat. Now we are asked if we want to change our mind or stay with our initial pick.
According to Probabilities, if we don't swap and keep our first selection we get $(1/3)$ $33.3% $ chance of wining the car since the elimination of the door that was opened by the host doesn't affect the probability of our door having the car which remains $33.3%$ as it is at the initial problem.
On the other hand, if we switch the door with the other one left , then we get $(2/3)$ $66.6%$ chances of wining the car since only two doors are remaining and the fact that the host revealed a goat in one of the unchosen doors changed nothing about the initial probability of our door having the car.
Till this point it makes pretty much sense to me.
But what about assuming we have a man in the audience that makes a different initial choice in his head. For instance, let's say that the contestant picked door number $1$ and he picked door number $2$. If door number $3$ is the door that is being revealed by the host (has a goat) then both doors $1$ and $2$ remain in the game. For the contestant door number $2$ has $66.6%$ chances of having the car when for the man in the audience door number $1$ has $66.6%$ chances of having the car. Isn't that weird? From two different perspectives we get two different probabilities about the same unopened doors. How's that possible?
probability bayes-theorem monty-hall
probability bayes-theorem monty-hall
asked Apr 20 at 13:17
MJ13MJ13
536
asked Apr 20 at 13:17
MJ13MJ13
536
edited Apr 20 at 20:06
MJ13
asked Apr 20 at 13:17
MJ13MJ13
536
asked Apr 20 at 13:17
MJ13MJ13
536
asked Apr 20 at 13:17
MJ13MJ13
536
536
$begingroup$
The main Monte Hall answer seems to be math.stackexchange.com/questions/96826/the-monty-hall-problem
$endgroup$
– Henry
Apr 20 at 23:58
$begingroup$
It is important to note that Monty's behavior is extremely important to this calculation. Even for the classic problem, if Monty cannot be depended on to always open a door with a goat, the probability calculation changes. When he can be depended on, it is only the contestant, not the audience member, who can depend on Monty's behavior, as lulu elaborates.
$endgroup$
– Paul Sinclair
Apr 21 at 0:56
$begingroup$
Note that it also happens that the observer choses the door that is going to be revealed empty. So even though he concludes that he should stick with his choce in the case you describe, that does not mean he will win 2/3 of attempts.
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 6:42
add a comment |
$begingroup$
The main Monte Hall answer seems to be math.stackexchange.com/questions/96826/the-monty-hall-problem
$endgroup$
– Henry
Apr 20 at 23:58
$begingroup$
It is important to note that Monty's behavior is extremely important to this calculation. Even for the classic problem, if Monty cannot be depended on to always open a door with a goat, the probability calculation changes. When he can be depended on, it is only the contestant, not the audience member, who can depend on Monty's behavior, as lulu elaborates.
$endgroup$
– Paul Sinclair
Apr 21 at 0:56
$begingroup$
Note that it also happens that the observer choses the door that is going to be revealed empty. So even though he concludes that he should stick with his choce in the case you describe, that does not mean he will win 2/3 of attempts.
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 6:42
$begingroup$
The main Monte Hall answer seems to be math.stackexchange.com/questions/96826/the-monty-hall-problem
$endgroup$
– Henry
Apr 20 at 23:58
$begingroup$
The main Monte Hall answer seems to be math.stackexchange.com/questions/96826/the-monty-hall-problem
$endgroup$
– Henry
Apr 20 at 23:58
$begingroup$
It is important to note that Monty's behavior is extremely important to this calculation. Even for the classic problem, if Monty cannot be depended on to always open a door with a goat, the probability calculation changes. When he can be depended on, it is only the contestant, not the audience member, who can depend on Monty's behavior, as lulu elaborates.
$endgroup$
– Paul Sinclair
Apr 21 at 0:56
$begingroup$
It is important to note that Monty's behavior is extremely important to this calculation. Even for the classic problem, if Monty cannot be depended on to always open a door with a goat, the probability calculation changes. When he can be depended on, it is only the contestant, not the audience member, who can depend on Monty's behavior, as lulu elaborates.
$endgroup$
– Paul Sinclair
Apr 21 at 0:56
$begingroup$
Note that it also happens that the observer choses the door that is going to be revealed empty. So even though he concludes that he should stick with his choce in the case you describe, that does not mean he will win 2/3 of attempts.
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 6:42
$begingroup$
Note that it also happens that the observer choses the door that is going to be revealed empty. So even though he concludes that he should stick with his choce in the case you describe, that does not mean he will win 2/3 of attempts.
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 6:42
add a comment |
7 Answers
7
active
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The big difference between the two scenarios is that Monty deliberately avoids choosing the door the contestant chose, but he isn't even aware of what the man in the audience chose so that doesn't factor into his decision.
Let's do the math explicitly. Suppose (as you say) the contestant chooses $1$, the audience member chooses $2$, and Monty opens $3$. The audience member knows that Monty was choosing between doors $2$ and $3$, and could not open door $1$. Thus the audience member knows, as does the contestant, that Monty was forced to avoid door $1$ but elected to avoid $2$ as well. That deliberate selection is evidence that the prize is behind door $2$, so the audience member reasons in exactly the same way the contestant does, and concludes that he should stick with his original mental vote.
As another, similar, variant of the problem, suppose that Monty himself has no idea where the prize is, so just chooses between the unselected doors uniformly at random (possibly exposing the prize, in which case the game ends). In that case, convince yourself that there is no value to switching.
answered Apr 20 at 13:37
lulululu
44.1k25182
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Yeah but if the contestant and the man in the audience have a conversation, they would both claim that they have a selection with $66.6%$ chances of winning the car. However, these two different choices concern two unopened doors in the same problem.
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– MJ13
Apr 20 at 14:00
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No they would not. Their situations are not comparable, for the reason I mentioned, so they would not have done the same calculation.
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– lulu
Apr 20 at 14:01
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Nope. I agree with the $frac 23$ computation for door $2$. I should add: "evidence" is a word that comes up a lot in taking about Bayes' Theorem, which this is an instance of. Bayes tells us how we should adjust our estimate of probabilities as we get new information. Here we see that Monty could have opened door $2$ but elected not to, which is useful information. It increases our estimate for the door $2$ probability, but certainly doesn't settle the question of where the prize is.
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– lulu
Apr 20 at 14:41
2
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Well, the contestant (and any observers) need to know what rules Monty is following. Here, in the standard version of the problem, we assume that Monty will always open one of the worthless doors not selected by the contestant and that, if both of the unselected doors are worthless, he chooses between them uniformly at random.
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– lulu
Apr 20 at 14:59
1
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To be clear: if the audience member has no idea which door the contestant chose, so all he knows is that Monty opened $3$, then the audience member has no way to distinguish doors $1,2$. Thus the audience member sees equal probability between those two. That's not a contradiction! The contestant has better information so of course the contestant's estimate of the probability is going to reflect that extra knowledge.
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– lulu
Apr 20 at 20:45
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show 10 more comments
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lulu's answer is perfectly correct, but I wanted to show an alternative (frequentist) way of arriving at the same result. For very small problems, like Monty Hall, it is often possible to just list out the entire sample space:
- Car, Goat, Goat.
- Goat, Car, Goat.
- Goat, Goat, Car.
Each of these is equally probable, so they all have 1/3 probability. So far, so simple.
Suppose as before that the contestant chooses door #1, the audience member chooses door #2, and Monty opens door #3. Then, translating our cases into outcomes:
- Monty opens either door #2 or door #3 (1/3 probability). Assuming he picks one at random:
Monty opens door #2 (1/6 probability).- Monty opens door #3 (1/6 probability).
- Monty opens door #3 (1/3 probability).
Monty opens door #2 (1/3 probability).
I have struck out the cases where Monty opens door #2, since we assumed that those cases don't happen. That eliminates cases (3) and (1.1), leaving us with cases (1.2) and (2). Based on the probabilities shown above, case (2) is twice as probable as case (1.2), meaning that the car is twice as likely to be behind door #2 as door #1. These probabilities are "objective" (frequentist), so both the contestant and the audience member arrive at the same numbers by the same reasoning process.
However, we added an assumption. We assumed that Monty chooses a door to open at random in case (1). If (for example) Monty always picks the door on the right (with the greater number), then it might appear that this logic no longer goes through. That is not the case. If Monty always picks the door on the right, then the doors are no longer symmetrical, and we now have to account for the cases where the contestant and audience member select different doors. When this is done, we recover the same result.
answered Apr 20 at 15:35
KevinKevin
1,688722
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(Corrected for incorrect initial solution.)
It is not correct that opening one door does not change the probability of what's behind the other doors. Probability calculations require an assumption of a particular probability distribution. Your probability of 1/3 before any doors are opened is based on assuming a random distribution of one car behind three doors. If you assume the host was always going to open a door with a goat, then the chance that your initial choice was right is still 1/3, which means you have a 2/3 chance of being correct if you switch since there is now only one door left.
Don't confuse the audience member's choice with his probability for that choice. Just like you, the audience member would have assigned 1/3 probability to each door before opening, including the one he picked, but would change his probability calculation at the same time as you and would end up with the same answer as you. After all, if the host had picked the audience member's door to open, the audience member's probability for that door would fall to 0, and so would yours.
Actually, it gets worse than this. Buried in the Monty Hall problem is the assumption that the host was always going to open one of the doors. However, there is nothing in the statement of the problem that says this. It just describes what HAPPENED without describing what the host PLANNED to happen. It is possible that the host only would have opened another door if the car was behind your first pick, and otherwise would have just opened the door you just picked and says, "Congratulations on your new goat." In this case, switching gives you 0% chance of getting the car, and not switching gives you 100%. In the real world, this is actually what a conman would do, and this type of behavior is also the basis for certain magic tricks where the audience member picks out a concealed card by seemingly random guesses.
Bottom line: You cannot actually solve the Monty Hall problem without specifying the planned behavior of the host. Most presentations of the problem only specify what the host does for the show you attend without guaranteeing that the host would always open another door. This all comes back to the principle that probability calculations are based upon a particular assumed probability distribution, and that assumption changes based on the information you have.
answered Apr 20 at 14:08
magster2magster2
213
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Good point, I’ve described always switch as “not x and not whatever the host uncovers”, but that requires the host to actually uncover every time. If the host only does so when switching makes you loose, then it’s a totally different ball game.
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– jmoreno
Apr 20 at 19:47
add a comment |
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That's the initial trick. You are tricked into believing here that the 3rd curtain/door doesn't matter, since it's seemingly "eliminated" from the equation. This is an illusion and is meant to fool you into thinking the 3rd curtain doesn't matter. You HAVE to consider the 3rd curtain as well. Why? Because you made your pick BEFORE the 3rd curtain got eliminated. It would only be a true 50/50 chance if this 3rd curtain was eliminated BEFORE you made your pick and not after.
To see this more easily, you just have to exaggerate the numbers to the extreme. Suppose you had 1000 doors and only one door had the prize. The chances of picking the right door would be extremely small, right? Ok, you pick a door and afterwards, the host eliminates 998 doors, leaving the door you picked and a second door. Now,you tell me if you think it likely that your initial pick would just HAPPEN to be the right door chosen out of 1000 doors. It sure as hell wouldn't be 50%.
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
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If you can actually understand that the probability of the car being behind the original door you picked remains the same before and after Monty opens other doors, then three doors is just as convincing as $1000,$ because $frac13neqfrac12.$ On the other hand, if you mistakenly think that the opening of other doors changes the probability of the door you originally picked, how will additional doors change that misconception? If you make that mistake, more open doors just seems like more ways for the probability of your original door to change.
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– David K
Apr 21 at 14:17
add a comment |
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On the game show hosted by real-life Monty Hall, after a contestant picked a door, one of four things would happen (so far as I could tell, all four have occurred on the actual show):
Monty can reveal that the contestant chose the right door and wins.
Monty can reveal that the contestant chose the wrong door and loses.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the right door to the other wrong door.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the remaining wrong door to the right door.
If the contestant starts by picking the right door, only #1 and #3 can occur. If the contestant starts by picking the wrong door, only #2 and #4 can occur. The combined probability of #1 and #3 will thus be 1/3 and the combined probability of #2 and #4 will be 2/3. The "paradox" assumes that probabilities of #1 and #2 are zero, implying that #4 is twice as great as #3, but that version of the game doesn't coincide with the actual game show hosted by Monty Hall.
From the standpoint of an audience member who picks an arbitrary door unbeknownst to Monty Hall or the contestant, the same outcomes are possible, though unless the audience member's choice matches that of the contestant, the probabilities of the outcomes may be different. Those probabilities are affected upon the host's motivation, but in different ways. If the host only offers a player a chance to switch if the player was initially correct, then an audience member's guess that doesn't match the contestant's will never be right when the player is given a chance to switch. On the flip side, if the host only offers a player a chance to switch if the initial guess was wrong, then any audience member guess that matches neither the door picked by the contestant nor revealed empty by the host must contain the prize.
For the "classic" form where #1 and #2 never occur, an audience member whose choice doesn't match the contestant will have a 50% chance of having it revealed to be empty. If that doesn't occur, the audience member will have a 2/3 chance of being right with the initial guess.
answered Apr 20 at 23:28
supercatsupercat
46837
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66.6% is not the chance of the contenstant finding a car behind the second door, it is the chance of finding a car if he changes his choice. The same is true for the audience.
So, a better question is, what is the probability of the participant/audience winning a car between sticking to his initial choice or changing his choice. It is worth mentioning that, the probability of a car behind each door doesn't change. Finding a car behind the second door or the first door is still 33.3%.
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
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Basically, you must be careful what conditional likelihood/probability you are computing.
If the contestant and bystander both pick a door uniformly randomly, and happen to pick different doors, which we shall label 1 and 2 respectively, then there is only a $2/3$ likelihood that the host can pick a goat door that is not either picked door, which must happen if both contestant and bystander can invoke the so-called Monty-Hall argument. In this case, the likelihood that the prize door is door 1 is $1/2$, and same for door 2, so when both of them switch doors they are equally likely to get the prize. However, the Monty-Hall argument is not refuted, because each of them individually did have original likelihood $1/3$ of getting the prize! With likelihood $1/3$ the host would have been unable of picking a goat door that was different from both contestant and bystander choices, and in this case the bystander obviously cannot invoke the Monty-Hall argument, because the host opens the bystander's chosen door.
answered Apr 21 at 6:17
user21820user21820
40.5k544163
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7 Answers
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7 Answers
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The big difference between the two scenarios is that Monty deliberately avoids choosing the door the contestant chose, but he isn't even aware of what the man in the audience chose so that doesn't factor into his decision.
Let's do the math explicitly. Suppose (as you say) the contestant chooses $1$, the audience member chooses $2$, and Monty opens $3$. The audience member knows that Monty was choosing between doors $2$ and $3$, and could not open door $1$. Thus the audience member knows, as does the contestant, that Monty was forced to avoid door $1$ but elected to avoid $2$ as well. That deliberate selection is evidence that the prize is behind door $2$, so the audience member reasons in exactly the same way the contestant does, and concludes that he should stick with his original mental vote.
As another, similar, variant of the problem, suppose that Monty himself has no idea where the prize is, so just chooses between the unselected doors uniformly at random (possibly exposing the prize, in which case the game ends). In that case, convince yourself that there is no value to switching.
answered Apr 20 at 13:37
lulululu
44.1k25182
$endgroup$
$begingroup$
Yeah but if the contestant and the man in the audience have a conversation, they would both claim that they have a selection with $66.6%$ chances of winning the car. However, these two different choices concern two unopened doors in the same problem.
$endgroup$
– MJ13
Apr 20 at 14:00
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No they would not. Their situations are not comparable, for the reason I mentioned, so they would not have done the same calculation.
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– lulu
Apr 20 at 14:01
1
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Nope. I agree with the $frac 23$ computation for door $2$. I should add: "evidence" is a word that comes up a lot in taking about Bayes' Theorem, which this is an instance of. Bayes tells us how we should adjust our estimate of probabilities as we get new information. Here we see that Monty could have opened door $2$ but elected not to, which is useful information. It increases our estimate for the door $2$ probability, but certainly doesn't settle the question of where the prize is.
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– lulu
Apr 20 at 14:41
2
$begingroup$
Well, the contestant (and any observers) need to know what rules Monty is following. Here, in the standard version of the problem, we assume that Monty will always open one of the worthless doors not selected by the contestant and that, if both of the unselected doors are worthless, he chooses between them uniformly at random.
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– lulu
Apr 20 at 14:59
1
$begingroup$
To be clear: if the audience member has no idea which door the contestant chose, so all he knows is that Monty opened $3$, then the audience member has no way to distinguish doors $1,2$. Thus the audience member sees equal probability between those two. That's not a contradiction! The contestant has better information so of course the contestant's estimate of the probability is going to reflect that extra knowledge.
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– lulu
Apr 20 at 20:45
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show 10 more comments
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The big difference between the two scenarios is that Monty deliberately avoids choosing the door the contestant chose, but he isn't even aware of what the man in the audience chose so that doesn't factor into his decision.
Let's do the math explicitly. Suppose (as you say) the contestant chooses $1$, the audience member chooses $2$, and Monty opens $3$. The audience member knows that Monty was choosing between doors $2$ and $3$, and could not open door $1$. Thus the audience member knows, as does the contestant, that Monty was forced to avoid door $1$ but elected to avoid $2$ as well. That deliberate selection is evidence that the prize is behind door $2$, so the audience member reasons in exactly the same way the contestant does, and concludes that he should stick with his original mental vote.
As another, similar, variant of the problem, suppose that Monty himself has no idea where the prize is, so just chooses between the unselected doors uniformly at random (possibly exposing the prize, in which case the game ends). In that case, convince yourself that there is no value to switching.
answered Apr 20 at 13:37
lulululu
44.1k25182
$endgroup$
$begingroup$
Yeah but if the contestant and the man in the audience have a conversation, they would both claim that they have a selection with $66.6%$ chances of winning the car. However, these two different choices concern two unopened doors in the same problem.
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– MJ13
Apr 20 at 14:00
$begingroup$
No they would not. Their situations are not comparable, for the reason I mentioned, so they would not have done the same calculation.
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– lulu
Apr 20 at 14:01
1
$begingroup$
Nope. I agree with the $frac 23$ computation for door $2$. I should add: "evidence" is a word that comes up a lot in taking about Bayes' Theorem, which this is an instance of. Bayes tells us how we should adjust our estimate of probabilities as we get new information. Here we see that Monty could have opened door $2$ but elected not to, which is useful information. It increases our estimate for the door $2$ probability, but certainly doesn't settle the question of where the prize is.
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– lulu
Apr 20 at 14:41
2
$begingroup$
Well, the contestant (and any observers) need to know what rules Monty is following. Here, in the standard version of the problem, we assume that Monty will always open one of the worthless doors not selected by the contestant and that, if both of the unselected doors are worthless, he chooses between them uniformly at random.
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– lulu
Apr 20 at 14:59
1
$begingroup$
To be clear: if the audience member has no idea which door the contestant chose, so all he knows is that Monty opened $3$, then the audience member has no way to distinguish doors $1,2$. Thus the audience member sees equal probability between those two. That's not a contradiction! The contestant has better information so of course the contestant's estimate of the probability is going to reflect that extra knowledge.
$endgroup$
– lulu
Apr 20 at 20:45
|
show 10 more comments
$begingroup$
The big difference between the two scenarios is that Monty deliberately avoids choosing the door the contestant chose, but he isn't even aware of what the man in the audience chose so that doesn't factor into his decision.
Let's do the math explicitly. Suppose (as you say) the contestant chooses $1$, the audience member chooses $2$, and Monty opens $3$. The audience member knows that Monty was choosing between doors $2$ and $3$, and could not open door $1$. Thus the audience member knows, as does the contestant, that Monty was forced to avoid door $1$ but elected to avoid $2$ as well. That deliberate selection is evidence that the prize is behind door $2$, so the audience member reasons in exactly the same way the contestant does, and concludes that he should stick with his original mental vote.
As another, similar, variant of the problem, suppose that Monty himself has no idea where the prize is, so just chooses between the unselected doors uniformly at random (possibly exposing the prize, in which case the game ends). In that case, convince yourself that there is no value to switching.
answered Apr 20 at 13:37
lulululu
44.1k25182
$endgroup$
The big difference between the two scenarios is that Monty deliberately avoids choosing the door the contestant chose, but he isn't even aware of what the man in the audience chose so that doesn't factor into his decision.
Let's do the math explicitly. Suppose (as you say) the contestant chooses $1$, the audience member chooses $2$, and Monty opens $3$. The audience member knows that Monty was choosing between doors $2$ and $3$, and could not open door $1$. Thus the audience member knows, as does the contestant, that Monty was forced to avoid door $1$ but elected to avoid $2$ as well. That deliberate selection is evidence that the prize is behind door $2$, so the audience member reasons in exactly the same way the contestant does, and concludes that he should stick with his original mental vote.
As another, similar, variant of the problem, suppose that Monty himself has no idea where the prize is, so just chooses between the unselected doors uniformly at random (possibly exposing the prize, in which case the game ends). In that case, convince yourself that there is no value to switching.
answered Apr 20 at 13:37
lulululu
44.1k25182
edited Apr 20 at 14:28
answered Apr 20 at 13:37
lulululu
44.1k25182
answered Apr 20 at 13:37
lulululu
44.1k25182
answered Apr 20 at 13:37
lulululu
44.1k25182
44.1k25182
$begingroup$
Yeah but if the contestant and the man in the audience have a conversation, they would both claim that they have a selection with $66.6%$ chances of winning the car. However, these two different choices concern two unopened doors in the same problem.
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– MJ13
Apr 20 at 14:00
$begingroup$
No they would not. Their situations are not comparable, for the reason I mentioned, so they would not have done the same calculation.
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– lulu
Apr 20 at 14:01
1
$begingroup$
Nope. I agree with the $frac 23$ computation for door $2$. I should add: "evidence" is a word that comes up a lot in taking about Bayes' Theorem, which this is an instance of. Bayes tells us how we should adjust our estimate of probabilities as we get new information. Here we see that Monty could have opened door $2$ but elected not to, which is useful information. It increases our estimate for the door $2$ probability, but certainly doesn't settle the question of where the prize is.
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– lulu
Apr 20 at 14:41
2
$begingroup$
Well, the contestant (and any observers) need to know what rules Monty is following. Here, in the standard version of the problem, we assume that Monty will always open one of the worthless doors not selected by the contestant and that, if both of the unselected doors are worthless, he chooses between them uniformly at random.
$endgroup$
– lulu
Apr 20 at 14:59
1
$begingroup$
To be clear: if the audience member has no idea which door the contestant chose, so all he knows is that Monty opened $3$, then the audience member has no way to distinguish doors $1,2$. Thus the audience member sees equal probability between those two. That's not a contradiction! The contestant has better information so of course the contestant's estimate of the probability is going to reflect that extra knowledge.
$endgroup$
– lulu
Apr 20 at 20:45
|
show 10 more comments
$begingroup$
Yeah but if the contestant and the man in the audience have a conversation, they would both claim that they have a selection with $66.6%$ chances of winning the car. However, these two different choices concern two unopened doors in the same problem.
$endgroup$
– MJ13
Apr 20 at 14:00
$begingroup$
No they would not. Their situations are not comparable, for the reason I mentioned, so they would not have done the same calculation.
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– lulu
Apr 20 at 14:01
1
$begingroup$
Nope. I agree with the $frac 23$ computation for door $2$. I should add: "evidence" is a word that comes up a lot in taking about Bayes' Theorem, which this is an instance of. Bayes tells us how we should adjust our estimate of probabilities as we get new information. Here we see that Monty could have opened door $2$ but elected not to, which is useful information. It increases our estimate for the door $2$ probability, but certainly doesn't settle the question of where the prize is.
$endgroup$
– lulu
Apr 20 at 14:41
2
$begingroup$
Well, the contestant (and any observers) need to know what rules Monty is following. Here, in the standard version of the problem, we assume that Monty will always open one of the worthless doors not selected by the contestant and that, if both of the unselected doors are worthless, he chooses between them uniformly at random.
$endgroup$
– lulu
Apr 20 at 14:59
1
$begingroup$
To be clear: if the audience member has no idea which door the contestant chose, so all he knows is that Monty opened $3$, then the audience member has no way to distinguish doors $1,2$. Thus the audience member sees equal probability between those two. That's not a contradiction! The contestant has better information so of course the contestant's estimate of the probability is going to reflect that extra knowledge.
$endgroup$
– lulu
Apr 20 at 20:45
$begingroup$
Yeah but if the contestant and the man in the audience have a conversation, they would both claim that they have a selection with $66.6%$ chances of winning the car. However, these two different choices concern two unopened doors in the same problem.
$endgroup$
– MJ13
Apr 20 at 14:00
$begingroup$
Yeah but if the contestant and the man in the audience have a conversation, they would both claim that they have a selection with $66.6%$ chances of winning the car. However, these two different choices concern two unopened doors in the same problem.
$endgroup$
– MJ13
Apr 20 at 14:00
$begingroup$
No they would not. Their situations are not comparable, for the reason I mentioned, so they would not have done the same calculation.
$endgroup$
– lulu
Apr 20 at 14:01
$begingroup$
No they would not. Their situations are not comparable, for the reason I mentioned, so they would not have done the same calculation.
$endgroup$
– lulu
Apr 20 at 14:01
1
1
$begingroup$
Nope. I agree with the $frac 23$ computation for door $2$. I should add: "evidence" is a word that comes up a lot in taking about Bayes' Theorem, which this is an instance of. Bayes tells us how we should adjust our estimate of probabilities as we get new information. Here we see that Monty could have opened door $2$ but elected not to, which is useful information. It increases our estimate for the door $2$ probability, but certainly doesn't settle the question of where the prize is.
$endgroup$
– lulu
Apr 20 at 14:41
$begingroup$
Nope. I agree with the $frac 23$ computation for door $2$. I should add: "evidence" is a word that comes up a lot in taking about Bayes' Theorem, which this is an instance of. Bayes tells us how we should adjust our estimate of probabilities as we get new information. Here we see that Monty could have opened door $2$ but elected not to, which is useful information. It increases our estimate for the door $2$ probability, but certainly doesn't settle the question of where the prize is.
$endgroup$
– lulu
Apr 20 at 14:41
2
2
$begingroup$
Well, the contestant (and any observers) need to know what rules Monty is following. Here, in the standard version of the problem, we assume that Monty will always open one of the worthless doors not selected by the contestant and that, if both of the unselected doors are worthless, he chooses between them uniformly at random.
$endgroup$
– lulu
Apr 20 at 14:59
$begingroup$
Well, the contestant (and any observers) need to know what rules Monty is following. Here, in the standard version of the problem, we assume that Monty will always open one of the worthless doors not selected by the contestant and that, if both of the unselected doors are worthless, he chooses between them uniformly at random.
$endgroup$
– lulu
Apr 20 at 14:59
1
1
$begingroup$
To be clear: if the audience member has no idea which door the contestant chose, so all he knows is that Monty opened $3$, then the audience member has no way to distinguish doors $1,2$. Thus the audience member sees equal probability between those two. That's not a contradiction! The contestant has better information so of course the contestant's estimate of the probability is going to reflect that extra knowledge.
$endgroup$
– lulu
Apr 20 at 20:45
$begingroup$
To be clear: if the audience member has no idea which door the contestant chose, so all he knows is that Monty opened $3$, then the audience member has no way to distinguish doors $1,2$. Thus the audience member sees equal probability between those two. That's not a contradiction! The contestant has better information so of course the contestant's estimate of the probability is going to reflect that extra knowledge.
$endgroup$
– lulu
Apr 20 at 20:45
|
show 10 more comments
$begingroup$
lulu's answer is perfectly correct, but I wanted to show an alternative (frequentist) way of arriving at the same result. For very small problems, like Monty Hall, it is often possible to just list out the entire sample space:
- Car, Goat, Goat.
- Goat, Car, Goat.
- Goat, Goat, Car.
Each of these is equally probable, so they all have 1/3 probability. So far, so simple.
Suppose as before that the contestant chooses door #1, the audience member chooses door #2, and Monty opens door #3. Then, translating our cases into outcomes:
- Monty opens either door #2 or door #3 (1/3 probability). Assuming he picks one at random:
Monty opens door #2 (1/6 probability).- Monty opens door #3 (1/6 probability).
- Monty opens door #3 (1/3 probability).
Monty opens door #2 (1/3 probability).
I have struck out the cases where Monty opens door #2, since we assumed that those cases don't happen. That eliminates cases (3) and (1.1), leaving us with cases (1.2) and (2). Based on the probabilities shown above, case (2) is twice as probable as case (1.2), meaning that the car is twice as likely to be behind door #2 as door #1. These probabilities are "objective" (frequentist), so both the contestant and the audience member arrive at the same numbers by the same reasoning process.
However, we added an assumption. We assumed that Monty chooses a door to open at random in case (1). If (for example) Monty always picks the door on the right (with the greater number), then it might appear that this logic no longer goes through. That is not the case. If Monty always picks the door on the right, then the doors are no longer symmetrical, and we now have to account for the cases where the contestant and audience member select different doors. When this is done, we recover the same result.
answered Apr 20 at 15:35
KevinKevin
1,688722
$endgroup$
add a comment |
$begingroup$
lulu's answer is perfectly correct, but I wanted to show an alternative (frequentist) way of arriving at the same result. For very small problems, like Monty Hall, it is often possible to just list out the entire sample space:
- Car, Goat, Goat.
- Goat, Car, Goat.
- Goat, Goat, Car.
Each of these is equally probable, so they all have 1/3 probability. So far, so simple.
Suppose as before that the contestant chooses door #1, the audience member chooses door #2, and Monty opens door #3. Then, translating our cases into outcomes:
- Monty opens either door #2 or door #3 (1/3 probability). Assuming he picks one at random:
Monty opens door #2 (1/6 probability).- Monty opens door #3 (1/6 probability).
- Monty opens door #3 (1/3 probability).
Monty opens door #2 (1/3 probability).
I have struck out the cases where Monty opens door #2, since we assumed that those cases don't happen. That eliminates cases (3) and (1.1), leaving us with cases (1.2) and (2). Based on the probabilities shown above, case (2) is twice as probable as case (1.2), meaning that the car is twice as likely to be behind door #2 as door #1. These probabilities are "objective" (frequentist), so both the contestant and the audience member arrive at the same numbers by the same reasoning process.
However, we added an assumption. We assumed that Monty chooses a door to open at random in case (1). If (for example) Monty always picks the door on the right (with the greater number), then it might appear that this logic no longer goes through. That is not the case. If Monty always picks the door on the right, then the doors are no longer symmetrical, and we now have to account for the cases where the contestant and audience member select different doors. When this is done, we recover the same result.
answered Apr 20 at 15:35
KevinKevin
1,688722
$endgroup$
add a comment |
$begingroup$
lulu's answer is perfectly correct, but I wanted to show an alternative (frequentist) way of arriving at the same result. For very small problems, like Monty Hall, it is often possible to just list out the entire sample space:
- Car, Goat, Goat.
- Goat, Car, Goat.
- Goat, Goat, Car.
Each of these is equally probable, so they all have 1/3 probability. So far, so simple.
Suppose as before that the contestant chooses door #1, the audience member chooses door #2, and Monty opens door #3. Then, translating our cases into outcomes:
- Monty opens either door #2 or door #3 (1/3 probability). Assuming he picks one at random:
Monty opens door #2 (1/6 probability).- Monty opens door #3 (1/6 probability).
- Monty opens door #3 (1/3 probability).
Monty opens door #2 (1/3 probability).
I have struck out the cases where Monty opens door #2, since we assumed that those cases don't happen. That eliminates cases (3) and (1.1), leaving us with cases (1.2) and (2). Based on the probabilities shown above, case (2) is twice as probable as case (1.2), meaning that the car is twice as likely to be behind door #2 as door #1. These probabilities are "objective" (frequentist), so both the contestant and the audience member arrive at the same numbers by the same reasoning process.
However, we added an assumption. We assumed that Monty chooses a door to open at random in case (1). If (for example) Monty always picks the door on the right (with the greater number), then it might appear that this logic no longer goes through. That is not the case. If Monty always picks the door on the right, then the doors are no longer symmetrical, and we now have to account for the cases where the contestant and audience member select different doors. When this is done, we recover the same result.
answered Apr 20 at 15:35
KevinKevin
1,688722
$endgroup$
lulu's answer is perfectly correct, but I wanted to show an alternative (frequentist) way of arriving at the same result. For very small problems, like Monty Hall, it is often possible to just list out the entire sample space:
- Car, Goat, Goat.
- Goat, Car, Goat.
- Goat, Goat, Car.
Each of these is equally probable, so they all have 1/3 probability. So far, so simple.
Suppose as before that the contestant chooses door #1, the audience member chooses door #2, and Monty opens door #3. Then, translating our cases into outcomes:
- Monty opens either door #2 or door #3 (1/3 probability). Assuming he picks one at random:
Monty opens door #2 (1/6 probability).- Monty opens door #3 (1/6 probability).
- Monty opens door #3 (1/3 probability).
Monty opens door #2 (1/3 probability).
I have struck out the cases where Monty opens door #2, since we assumed that those cases don't happen. That eliminates cases (3) and (1.1), leaving us with cases (1.2) and (2). Based on the probabilities shown above, case (2) is twice as probable as case (1.2), meaning that the car is twice as likely to be behind door #2 as door #1. These probabilities are "objective" (frequentist), so both the contestant and the audience member arrive at the same numbers by the same reasoning process.
However, we added an assumption. We assumed that Monty chooses a door to open at random in case (1). If (for example) Monty always picks the door on the right (with the greater number), then it might appear that this logic no longer goes through. That is not the case. If Monty always picks the door on the right, then the doors are no longer symmetrical, and we now have to account for the cases where the contestant and audience member select different doors. When this is done, we recover the same result.
answered Apr 20 at 15:35
KevinKevin
1,688722
answered Apr 20 at 15:35
KevinKevin
1,688722
answered Apr 20 at 15:35
KevinKevin
1,688722
answered Apr 20 at 15:35
KevinKevin
1,688722
1,688722
add a comment |
add a comment |
$begingroup$
(Corrected for incorrect initial solution.)
It is not correct that opening one door does not change the probability of what's behind the other doors. Probability calculations require an assumption of a particular probability distribution. Your probability of 1/3 before any doors are opened is based on assuming a random distribution of one car behind three doors. If you assume the host was always going to open a door with a goat, then the chance that your initial choice was right is still 1/3, which means you have a 2/3 chance of being correct if you switch since there is now only one door left.
Don't confuse the audience member's choice with his probability for that choice. Just like you, the audience member would have assigned 1/3 probability to each door before opening, including the one he picked, but would change his probability calculation at the same time as you and would end up with the same answer as you. After all, if the host had picked the audience member's door to open, the audience member's probability for that door would fall to 0, and so would yours.
Actually, it gets worse than this. Buried in the Monty Hall problem is the assumption that the host was always going to open one of the doors. However, there is nothing in the statement of the problem that says this. It just describes what HAPPENED without describing what the host PLANNED to happen. It is possible that the host only would have opened another door if the car was behind your first pick, and otherwise would have just opened the door you just picked and says, "Congratulations on your new goat." In this case, switching gives you 0% chance of getting the car, and not switching gives you 100%. In the real world, this is actually what a conman would do, and this type of behavior is also the basis for certain magic tricks where the audience member picks out a concealed card by seemingly random guesses.
Bottom line: You cannot actually solve the Monty Hall problem without specifying the planned behavior of the host. Most presentations of the problem only specify what the host does for the show you attend without guaranteeing that the host would always open another door. This all comes back to the principle that probability calculations are based upon a particular assumed probability distribution, and that assumption changes based on the information you have.
answered Apr 20 at 14:08
magster2magster2
213
$endgroup$
$begingroup$
Good point, I’ve described always switch as “not x and not whatever the host uncovers”, but that requires the host to actually uncover every time. If the host only does so when switching makes you loose, then it’s a totally different ball game.
$endgroup$
– jmoreno
Apr 20 at 19:47
add a comment |
$begingroup$
(Corrected for incorrect initial solution.)
It is not correct that opening one door does not change the probability of what's behind the other doors. Probability calculations require an assumption of a particular probability distribution. Your probability of 1/3 before any doors are opened is based on assuming a random distribution of one car behind three doors. If you assume the host was always going to open a door with a goat, then the chance that your initial choice was right is still 1/3, which means you have a 2/3 chance of being correct if you switch since there is now only one door left.
Don't confuse the audience member's choice with his probability for that choice. Just like you, the audience member would have assigned 1/3 probability to each door before opening, including the one he picked, but would change his probability calculation at the same time as you and would end up with the same answer as you. After all, if the host had picked the audience member's door to open, the audience member's probability for that door would fall to 0, and so would yours.
Actually, it gets worse than this. Buried in the Monty Hall problem is the assumption that the host was always going to open one of the doors. However, there is nothing in the statement of the problem that says this. It just describes what HAPPENED without describing what the host PLANNED to happen. It is possible that the host only would have opened another door if the car was behind your first pick, and otherwise would have just opened the door you just picked and says, "Congratulations on your new goat." In this case, switching gives you 0% chance of getting the car, and not switching gives you 100%. In the real world, this is actually what a conman would do, and this type of behavior is also the basis for certain magic tricks where the audience member picks out a concealed card by seemingly random guesses.
Bottom line: You cannot actually solve the Monty Hall problem without specifying the planned behavior of the host. Most presentations of the problem only specify what the host does for the show you attend without guaranteeing that the host would always open another door. This all comes back to the principle that probability calculations are based upon a particular assumed probability distribution, and that assumption changes based on the information you have.
answered Apr 20 at 14:08
magster2magster2
213
$endgroup$
$begingroup$
Good point, I’ve described always switch as “not x and not whatever the host uncovers”, but that requires the host to actually uncover every time. If the host only does so when switching makes you loose, then it’s a totally different ball game.
$endgroup$
– jmoreno
Apr 20 at 19:47
add a comment |
$begingroup$
(Corrected for incorrect initial solution.)
It is not correct that opening one door does not change the probability of what's behind the other doors. Probability calculations require an assumption of a particular probability distribution. Your probability of 1/3 before any doors are opened is based on assuming a random distribution of one car behind three doors. If you assume the host was always going to open a door with a goat, then the chance that your initial choice was right is still 1/3, which means you have a 2/3 chance of being correct if you switch since there is now only one door left.
Don't confuse the audience member's choice with his probability for that choice. Just like you, the audience member would have assigned 1/3 probability to each door before opening, including the one he picked, but would change his probability calculation at the same time as you and would end up with the same answer as you. After all, if the host had picked the audience member's door to open, the audience member's probability for that door would fall to 0, and so would yours.
Actually, it gets worse than this. Buried in the Monty Hall problem is the assumption that the host was always going to open one of the doors. However, there is nothing in the statement of the problem that says this. It just describes what HAPPENED without describing what the host PLANNED to happen. It is possible that the host only would have opened another door if the car was behind your first pick, and otherwise would have just opened the door you just picked and says, "Congratulations on your new goat." In this case, switching gives you 0% chance of getting the car, and not switching gives you 100%. In the real world, this is actually what a conman would do, and this type of behavior is also the basis for certain magic tricks where the audience member picks out a concealed card by seemingly random guesses.
Bottom line: You cannot actually solve the Monty Hall problem without specifying the planned behavior of the host. Most presentations of the problem only specify what the host does for the show you attend without guaranteeing that the host would always open another door. This all comes back to the principle that probability calculations are based upon a particular assumed probability distribution, and that assumption changes based on the information you have.
answered Apr 20 at 14:08
magster2magster2
213
$endgroup$
(Corrected for incorrect initial solution.)
It is not correct that opening one door does not change the probability of what's behind the other doors. Probability calculations require an assumption of a particular probability distribution. Your probability of 1/3 before any doors are opened is based on assuming a random distribution of one car behind three doors. If you assume the host was always going to open a door with a goat, then the chance that your initial choice was right is still 1/3, which means you have a 2/3 chance of being correct if you switch since there is now only one door left.
Don't confuse the audience member's choice with his probability for that choice. Just like you, the audience member would have assigned 1/3 probability to each door before opening, including the one he picked, but would change his probability calculation at the same time as you and would end up with the same answer as you. After all, if the host had picked the audience member's door to open, the audience member's probability for that door would fall to 0, and so would yours.
Actually, it gets worse than this. Buried in the Monty Hall problem is the assumption that the host was always going to open one of the doors. However, there is nothing in the statement of the problem that says this. It just describes what HAPPENED without describing what the host PLANNED to happen. It is possible that the host only would have opened another door if the car was behind your first pick, and otherwise would have just opened the door you just picked and says, "Congratulations on your new goat." In this case, switching gives you 0% chance of getting the car, and not switching gives you 100%. In the real world, this is actually what a conman would do, and this type of behavior is also the basis for certain magic tricks where the audience member picks out a concealed card by seemingly random guesses.
Bottom line: You cannot actually solve the Monty Hall problem without specifying the planned behavior of the host. Most presentations of the problem only specify what the host does for the show you attend without guaranteeing that the host would always open another door. This all comes back to the principle that probability calculations are based upon a particular assumed probability distribution, and that assumption changes based on the information you have.
answered Apr 20 at 14:08
magster2magster2
213
edited Apr 20 at 14:36
answered Apr 20 at 14:08
magster2magster2
213
answered Apr 20 at 14:08
magster2magster2
213
answered Apr 20 at 14:08
magster2magster2
213
213
$begingroup$
Good point, I’ve described always switch as “not x and not whatever the host uncovers”, but that requires the host to actually uncover every time. If the host only does so when switching makes you loose, then it’s a totally different ball game.
$endgroup$
– jmoreno
Apr 20 at 19:47
add a comment |
$begingroup$
Good point, I’ve described always switch as “not x and not whatever the host uncovers”, but that requires the host to actually uncover every time. If the host only does so when switching makes you loose, then it’s a totally different ball game.
$endgroup$
– jmoreno
Apr 20 at 19:47
$begingroup$
Good point, I’ve described always switch as “not x and not whatever the host uncovers”, but that requires the host to actually uncover every time. If the host only does so when switching makes you loose, then it’s a totally different ball game.
$endgroup$
– jmoreno
Apr 20 at 19:47
$begingroup$
Good point, I’ve described always switch as “not x and not whatever the host uncovers”, but that requires the host to actually uncover every time. If the host only does so when switching makes you loose, then it’s a totally different ball game.
$endgroup$
– jmoreno
Apr 20 at 19:47
add a comment |
$begingroup$
That's the initial trick. You are tricked into believing here that the 3rd curtain/door doesn't matter, since it's seemingly "eliminated" from the equation. This is an illusion and is meant to fool you into thinking the 3rd curtain doesn't matter. You HAVE to consider the 3rd curtain as well. Why? Because you made your pick BEFORE the 3rd curtain got eliminated. It would only be a true 50/50 chance if this 3rd curtain was eliminated BEFORE you made your pick and not after.
To see this more easily, you just have to exaggerate the numbers to the extreme. Suppose you had 1000 doors and only one door had the prize. The chances of picking the right door would be extremely small, right? Ok, you pick a door and afterwards, the host eliminates 998 doors, leaving the door you picked and a second door. Now,you tell me if you think it likely that your initial pick would just HAPPEN to be the right door chosen out of 1000 doors. It sure as hell wouldn't be 50%.
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
$endgroup$
$begingroup$
If you can actually understand that the probability of the car being behind the original door you picked remains the same before and after Monty opens other doors, then three doors is just as convincing as $1000,$ because $frac13neqfrac12.$ On the other hand, if you mistakenly think that the opening of other doors changes the probability of the door you originally picked, how will additional doors change that misconception? If you make that mistake, more open doors just seems like more ways for the probability of your original door to change.
$endgroup$
– David K
Apr 21 at 14:17
add a comment |
$begingroup$
That's the initial trick. You are tricked into believing here that the 3rd curtain/door doesn't matter, since it's seemingly "eliminated" from the equation. This is an illusion and is meant to fool you into thinking the 3rd curtain doesn't matter. You HAVE to consider the 3rd curtain as well. Why? Because you made your pick BEFORE the 3rd curtain got eliminated. It would only be a true 50/50 chance if this 3rd curtain was eliminated BEFORE you made your pick and not after.
To see this more easily, you just have to exaggerate the numbers to the extreme. Suppose you had 1000 doors and only one door had the prize. The chances of picking the right door would be extremely small, right? Ok, you pick a door and afterwards, the host eliminates 998 doors, leaving the door you picked and a second door. Now,you tell me if you think it likely that your initial pick would just HAPPEN to be the right door chosen out of 1000 doors. It sure as hell wouldn't be 50%.
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
$endgroup$
$begingroup$
If you can actually understand that the probability of the car being behind the original door you picked remains the same before and after Monty opens other doors, then three doors is just as convincing as $1000,$ because $frac13neqfrac12.$ On the other hand, if you mistakenly think that the opening of other doors changes the probability of the door you originally picked, how will additional doors change that misconception? If you make that mistake, more open doors just seems like more ways for the probability of your original door to change.
$endgroup$
– David K
Apr 21 at 14:17
add a comment |
$begingroup$
That's the initial trick. You are tricked into believing here that the 3rd curtain/door doesn't matter, since it's seemingly "eliminated" from the equation. This is an illusion and is meant to fool you into thinking the 3rd curtain doesn't matter. You HAVE to consider the 3rd curtain as well. Why? Because you made your pick BEFORE the 3rd curtain got eliminated. It would only be a true 50/50 chance if this 3rd curtain was eliminated BEFORE you made your pick and not after.
To see this more easily, you just have to exaggerate the numbers to the extreme. Suppose you had 1000 doors and only one door had the prize. The chances of picking the right door would be extremely small, right? Ok, you pick a door and afterwards, the host eliminates 998 doors, leaving the door you picked and a second door. Now,you tell me if you think it likely that your initial pick would just HAPPEN to be the right door chosen out of 1000 doors. It sure as hell wouldn't be 50%.
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
$endgroup$
That's the initial trick. You are tricked into believing here that the 3rd curtain/door doesn't matter, since it's seemingly "eliminated" from the equation. This is an illusion and is meant to fool you into thinking the 3rd curtain doesn't matter. You HAVE to consider the 3rd curtain as well. Why? Because you made your pick BEFORE the 3rd curtain got eliminated. It would only be a true 50/50 chance if this 3rd curtain was eliminated BEFORE you made your pick and not after.
To see this more easily, you just have to exaggerate the numbers to the extreme. Suppose you had 1000 doors and only one door had the prize. The chances of picking the right door would be extremely small, right? Ok, you pick a door and afterwards, the host eliminates 998 doors, leaving the door you picked and a second door. Now,you tell me if you think it likely that your initial pick would just HAPPEN to be the right door chosen out of 1000 doors. It sure as hell wouldn't be 50%.
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
answered Apr 20 at 20:48
Shahid KhatibShahid Khatib
91
91
$begingroup$
If you can actually understand that the probability of the car being behind the original door you picked remains the same before and after Monty opens other doors, then three doors is just as convincing as $1000,$ because $frac13neqfrac12.$ On the other hand, if you mistakenly think that the opening of other doors changes the probability of the door you originally picked, how will additional doors change that misconception? If you make that mistake, more open doors just seems like more ways for the probability of your original door to change.
$endgroup$
– David K
Apr 21 at 14:17
add a comment |
$begingroup$
If you can actually understand that the probability of the car being behind the original door you picked remains the same before and after Monty opens other doors, then three doors is just as convincing as $1000,$ because $frac13neqfrac12.$ On the other hand, if you mistakenly think that the opening of other doors changes the probability of the door you originally picked, how will additional doors change that misconception? If you make that mistake, more open doors just seems like more ways for the probability of your original door to change.
$endgroup$
– David K
Apr 21 at 14:17
$begingroup$
If you can actually understand that the probability of the car being behind the original door you picked remains the same before and after Monty opens other doors, then three doors is just as convincing as $1000,$ because $frac13neqfrac12.$ On the other hand, if you mistakenly think that the opening of other doors changes the probability of the door you originally picked, how will additional doors change that misconception? If you make that mistake, more open doors just seems like more ways for the probability of your original door to change.
$endgroup$
– David K
Apr 21 at 14:17
$begingroup$
If you can actually understand that the probability of the car being behind the original door you picked remains the same before and after Monty opens other doors, then three doors is just as convincing as $1000,$ because $frac13neqfrac12.$ On the other hand, if you mistakenly think that the opening of other doors changes the probability of the door you originally picked, how will additional doors change that misconception? If you make that mistake, more open doors just seems like more ways for the probability of your original door to change.
$endgroup$
– David K
Apr 21 at 14:17
add a comment |
$begingroup$
On the game show hosted by real-life Monty Hall, after a contestant picked a door, one of four things would happen (so far as I could tell, all four have occurred on the actual show):
Monty can reveal that the contestant chose the right door and wins.
Monty can reveal that the contestant chose the wrong door and loses.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the right door to the other wrong door.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the remaining wrong door to the right door.
If the contestant starts by picking the right door, only #1 and #3 can occur. If the contestant starts by picking the wrong door, only #2 and #4 can occur. The combined probability of #1 and #3 will thus be 1/3 and the combined probability of #2 and #4 will be 2/3. The "paradox" assumes that probabilities of #1 and #2 are zero, implying that #4 is twice as great as #3, but that version of the game doesn't coincide with the actual game show hosted by Monty Hall.
From the standpoint of an audience member who picks an arbitrary door unbeknownst to Monty Hall or the contestant, the same outcomes are possible, though unless the audience member's choice matches that of the contestant, the probabilities of the outcomes may be different. Those probabilities are affected upon the host's motivation, but in different ways. If the host only offers a player a chance to switch if the player was initially correct, then an audience member's guess that doesn't match the contestant's will never be right when the player is given a chance to switch. On the flip side, if the host only offers a player a chance to switch if the initial guess was wrong, then any audience member guess that matches neither the door picked by the contestant nor revealed empty by the host must contain the prize.
For the "classic" form where #1 and #2 never occur, an audience member whose choice doesn't match the contestant will have a 50% chance of having it revealed to be empty. If that doesn't occur, the audience member will have a 2/3 chance of being right with the initial guess.
answered Apr 20 at 23:28
supercatsupercat
46837
$endgroup$
add a comment |
$begingroup$
On the game show hosted by real-life Monty Hall, after a contestant picked a door, one of four things would happen (so far as I could tell, all four have occurred on the actual show):
Monty can reveal that the contestant chose the right door and wins.
Monty can reveal that the contestant chose the wrong door and loses.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the right door to the other wrong door.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the remaining wrong door to the right door.
If the contestant starts by picking the right door, only #1 and #3 can occur. If the contestant starts by picking the wrong door, only #2 and #4 can occur. The combined probability of #1 and #3 will thus be 1/3 and the combined probability of #2 and #4 will be 2/3. The "paradox" assumes that probabilities of #1 and #2 are zero, implying that #4 is twice as great as #3, but that version of the game doesn't coincide with the actual game show hosted by Monty Hall.
From the standpoint of an audience member who picks an arbitrary door unbeknownst to Monty Hall or the contestant, the same outcomes are possible, though unless the audience member's choice matches that of the contestant, the probabilities of the outcomes may be different. Those probabilities are affected upon the host's motivation, but in different ways. If the host only offers a player a chance to switch if the player was initially correct, then an audience member's guess that doesn't match the contestant's will never be right when the player is given a chance to switch. On the flip side, if the host only offers a player a chance to switch if the initial guess was wrong, then any audience member guess that matches neither the door picked by the contestant nor revealed empty by the host must contain the prize.
For the "classic" form where #1 and #2 never occur, an audience member whose choice doesn't match the contestant will have a 50% chance of having it revealed to be empty. If that doesn't occur, the audience member will have a 2/3 chance of being right with the initial guess.
answered Apr 20 at 23:28
supercatsupercat
46837
$endgroup$
add a comment |
$begingroup$
On the game show hosted by real-life Monty Hall, after a contestant picked a door, one of four things would happen (so far as I could tell, all four have occurred on the actual show):
Monty can reveal that the contestant chose the right door and wins.
Monty can reveal that the contestant chose the wrong door and loses.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the right door to the other wrong door.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the remaining wrong door to the right door.
If the contestant starts by picking the right door, only #1 and #3 can occur. If the contestant starts by picking the wrong door, only #2 and #4 can occur. The combined probability of #1 and #3 will thus be 1/3 and the combined probability of #2 and #4 will be 2/3. The "paradox" assumes that probabilities of #1 and #2 are zero, implying that #4 is twice as great as #3, but that version of the game doesn't coincide with the actual game show hosted by Monty Hall.
From the standpoint of an audience member who picks an arbitrary door unbeknownst to Monty Hall or the contestant, the same outcomes are possible, though unless the audience member's choice matches that of the contestant, the probabilities of the outcomes may be different. Those probabilities are affected upon the host's motivation, but in different ways. If the host only offers a player a chance to switch if the player was initially correct, then an audience member's guess that doesn't match the contestant's will never be right when the player is given a chance to switch. On the flip side, if the host only offers a player a chance to switch if the initial guess was wrong, then any audience member guess that matches neither the door picked by the contestant nor revealed empty by the host must contain the prize.
For the "classic" form where #1 and #2 never occur, an audience member whose choice doesn't match the contestant will have a 50% chance of having it revealed to be empty. If that doesn't occur, the audience member will have a 2/3 chance of being right with the initial guess.
answered Apr 20 at 23:28
supercatsupercat
46837
$endgroup$
On the game show hosted by real-life Monty Hall, after a contestant picked a door, one of four things would happen (so far as I could tell, all four have occurred on the actual show):
Monty can reveal that the contestant chose the right door and wins.
Monty can reveal that the contestant chose the wrong door and loses.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the right door to the other wrong door.
Monty can open a door that does not contain the prize and offers the contestant a chance to switch from the remaining wrong door to the right door.
If the contestant starts by picking the right door, only #1 and #3 can occur. If the contestant starts by picking the wrong door, only #2 and #4 can occur. The combined probability of #1 and #3 will thus be 1/3 and the combined probability of #2 and #4 will be 2/3. The "paradox" assumes that probabilities of #1 and #2 are zero, implying that #4 is twice as great as #3, but that version of the game doesn't coincide with the actual game show hosted by Monty Hall.
From the standpoint of an audience member who picks an arbitrary door unbeknownst to Monty Hall or the contestant, the same outcomes are possible, though unless the audience member's choice matches that of the contestant, the probabilities of the outcomes may be different. Those probabilities are affected upon the host's motivation, but in different ways. If the host only offers a player a chance to switch if the player was initially correct, then an audience member's guess that doesn't match the contestant's will never be right when the player is given a chance to switch. On the flip side, if the host only offers a player a chance to switch if the initial guess was wrong, then any audience member guess that matches neither the door picked by the contestant nor revealed empty by the host must contain the prize.
For the "classic" form where #1 and #2 never occur, an audience member whose choice doesn't match the contestant will have a 50% chance of having it revealed to be empty. If that doesn't occur, the audience member will have a 2/3 chance of being right with the initial guess.
answered Apr 20 at 23:28
supercatsupercat
46837
answered Apr 20 at 23:28
supercatsupercat
46837
answered Apr 20 at 23:28
supercatsupercat
46837
answered Apr 20 at 23:28
supercatsupercat
46837
46837
add a comment |
add a comment |
$begingroup$
66.6% is not the chance of the contenstant finding a car behind the second door, it is the chance of finding a car if he changes his choice. The same is true for the audience.
So, a better question is, what is the probability of the participant/audience winning a car between sticking to his initial choice or changing his choice. It is worth mentioning that, the probability of a car behind each door doesn't change. Finding a car behind the second door or the first door is still 33.3%.
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
$endgroup$
add a comment |
$begingroup$
66.6% is not the chance of the contenstant finding a car behind the second door, it is the chance of finding a car if he changes his choice. The same is true for the audience.
So, a better question is, what is the probability of the participant/audience winning a car between sticking to his initial choice or changing his choice. It is worth mentioning that, the probability of a car behind each door doesn't change. Finding a car behind the second door or the first door is still 33.3%.
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
$endgroup$
add a comment |
$begingroup$
66.6% is not the chance of the contenstant finding a car behind the second door, it is the chance of finding a car if he changes his choice. The same is true for the audience.
So, a better question is, what is the probability of the participant/audience winning a car between sticking to his initial choice or changing his choice. It is worth mentioning that, the probability of a car behind each door doesn't change. Finding a car behind the second door or the first door is still 33.3%.
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
$endgroup$
66.6% is not the chance of the contenstant finding a car behind the second door, it is the chance of finding a car if he changes his choice. The same is true for the audience.
So, a better question is, what is the probability of the participant/audience winning a car between sticking to his initial choice or changing his choice. It is worth mentioning that, the probability of a car behind each door doesn't change. Finding a car behind the second door or the first door is still 33.3%.
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
answered Apr 21 at 3:39
AbhijitAbhijit
2,380819
2,380819
add a comment |
add a comment |
$begingroup$
Basically, you must be careful what conditional likelihood/probability you are computing.
If the contestant and bystander both pick a door uniformly randomly, and happen to pick different doors, which we shall label 1 and 2 respectively, then there is only a $2/3$ likelihood that the host can pick a goat door that is not either picked door, which must happen if both contestant and bystander can invoke the so-called Monty-Hall argument. In this case, the likelihood that the prize door is door 1 is $1/2$, and same for door 2, so when both of them switch doors they are equally likely to get the prize. However, the Monty-Hall argument is not refuted, because each of them individually did have original likelihood $1/3$ of getting the prize! With likelihood $1/3$ the host would have been unable of picking a goat door that was different from both contestant and bystander choices, and in this case the bystander obviously cannot invoke the Monty-Hall argument, because the host opens the bystander's chosen door.
answered Apr 21 at 6:17
user21820user21820
40.5k544163
$endgroup$
add a comment |
$begingroup$
Basically, you must be careful what conditional likelihood/probability you are computing.
If the contestant and bystander both pick a door uniformly randomly, and happen to pick different doors, which we shall label 1 and 2 respectively, then there is only a $2/3$ likelihood that the host can pick a goat door that is not either picked door, which must happen if both contestant and bystander can invoke the so-called Monty-Hall argument. In this case, the likelihood that the prize door is door 1 is $1/2$, and same for door 2, so when both of them switch doors they are equally likely to get the prize. However, the Monty-Hall argument is not refuted, because each of them individually did have original likelihood $1/3$ of getting the prize! With likelihood $1/3$ the host would have been unable of picking a goat door that was different from both contestant and bystander choices, and in this case the bystander obviously cannot invoke the Monty-Hall argument, because the host opens the bystander's chosen door.
answered Apr 21 at 6:17
user21820user21820
40.5k544163
$endgroup$
add a comment |
$begingroup$
Basically, you must be careful what conditional likelihood/probability you are computing.
If the contestant and bystander both pick a door uniformly randomly, and happen to pick different doors, which we shall label 1 and 2 respectively, then there is only a $2/3$ likelihood that the host can pick a goat door that is not either picked door, which must happen if both contestant and bystander can invoke the so-called Monty-Hall argument. In this case, the likelihood that the prize door is door 1 is $1/2$, and same for door 2, so when both of them switch doors they are equally likely to get the prize. However, the Monty-Hall argument is not refuted, because each of them individually did have original likelihood $1/3$ of getting the prize! With likelihood $1/3$ the host would have been unable of picking a goat door that was different from both contestant and bystander choices, and in this case the bystander obviously cannot invoke the Monty-Hall argument, because the host opens the bystander's chosen door.
answered Apr 21 at 6:17
user21820user21820
40.5k544163
$endgroup$
Basically, you must be careful what conditional likelihood/probability you are computing.
If the contestant and bystander both pick a door uniformly randomly, and happen to pick different doors, which we shall label 1 and 2 respectively, then there is only a $2/3$ likelihood that the host can pick a goat door that is not either picked door, which must happen if both contestant and bystander can invoke the so-called Monty-Hall argument. In this case, the likelihood that the prize door is door 1 is $1/2$, and same for door 2, so when both of them switch doors they are equally likely to get the prize. However, the Monty-Hall argument is not refuted, because each of them individually did have original likelihood $1/3$ of getting the prize! With likelihood $1/3$ the host would have been unable of picking a goat door that was different from both contestant and bystander choices, and in this case the bystander obviously cannot invoke the Monty-Hall argument, because the host opens the bystander's chosen door.
answered Apr 21 at 6:17
user21820user21820
40.5k544163
answered Apr 21 at 6:17
user21820user21820
40.5k544163
answered Apr 21 at 6:17
user21820user21820
40.5k544163
answered Apr 21 at 6:17
user21820user21820
40.5k544163
40.5k544163
add a comment |
add a comment |
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$begingroup$
The main Monte Hall answer seems to be math.stackexchange.com/questions/96826/the-monty-hall-problem
$endgroup$
– Henry
Apr 20 at 23:58
$begingroup$
It is important to note that Monty's behavior is extremely important to this calculation. Even for the classic problem, if Monty cannot be depended on to always open a door with a goat, the probability calculation changes. When he can be depended on, it is only the contestant, not the audience member, who can depend on Monty's behavior, as lulu elaborates.
$endgroup$
– Paul Sinclair
Apr 21 at 0:56
$begingroup$
Note that it also happens that the observer choses the door that is going to be revealed empty. So even though he concludes that he should stick with his choce in the case you describe, that does not mean he will win 2/3 of attempts.
$endgroup$
– Jean-Claude Arbaut
Apr 21 at 6:42