A B C Pr 1 P F . 1/3 * 1/2 win 2 P . F 1/3 * 1/2 win 3 F P . 1/3 * 1/2 other contestant wins 4 . P F 1/3 * 1/2 lose 5 F . P 1/3 * 1/2 other contestant wins 6 . F P 1/3 * 1/2 lose3 and 5 can be ruled out, as the other Contestant didn't in fact pick the prize, but he might have done. That leaves equal probabilities of winning and losing if you don't switch. See also NotMontyHallSimulation

Compare with MontyHallProblem: You pick A; prize is in P; host shows you H. As before, the probabilities given are a-priori. Given that the prize is in A, the chance of you picking B and the host showing you C are 1 in 3.

A B C Pr 1 P H . 1/3 * 1/2 win 2 P . H 1/3 * 1/2 win 3 . P H 1/3 lose 4 . H P 1/3 loseThe probability of winning if you don't switch is now only 1/3, while if you do switch it's 2/3. Because the host knows where the prize is, there is no randomness in which door he picks when you don't pick the prize: he must pick the only other door which doesn't have the prize. The two problems look the same, the only difference is in whether the door that is shown to you is picked at random or not. That is why it is difficult. -- AndrewMcGuinness

AndrewMcGuinness wrote:

Exactly. If you're telling the problem as a "brain-teaser", you need to state that the Host is

I may be getting this wrong, but from the description:

Here's another NotTheMontyHallProblem that may help explain why people make the wrong choice: Suppose you pick one door out of three, knowing that there is one prize behind one of the doors. Then the host (the "Monty Hall"-like guy) has free choice to open another door or do nothing. We can assume (or suspect) that he knows which door holds the prize. If he knows that you selected a losing door, he'd just do nothing, letting you lose. But if he knows that you selected the winning door (and that he loses if you win), then he'd be highly motivated to convince you to change your choice. So he opens a losing door and "gives you another chance to chose." In this scenario, if he's offering to let you change doors

The solution to the first problem on this page is wrong. The table analysis admits that cases 3 and 5 are eliminated. Doing so results in the remaining probabilities not adding up to one. Cases 4 and 6 have probabilities 1/3 * 1. This is because when the first contestant picks a non-prize, the second contestant *HAS NO CHOICE* but to pick the other non-prize. To not do so violates the conditions of the problem. Suggesting the second contestant can pick the prize is along the lines of adding another case where there is more or less than one prize. After the first guess, the first contestant *KNOWS* that at least one door could be opened to reveal a non-prize. Whether that door is opened by a knowing host or an unknowing contestant is irrelevant, because doing so adds no new information. The true NotTheMontyHallProblem would be the following: Contestant 1 picks a door, it is opened to reveal a non-prize. The second contestant *then* picks a door from the remaining two. Now the second contestant has an even chance of winning. -- JohnVriezen (I now realize that the above is incorrect. -- JohnVriezen

The solution to the first problem is not wrong: I said the probabilities were as calculated in advance. They are correct. Once you have seen the second contestant has been unlucky, you then eliminate 3 and 5, and you are left with two equal probabilities. In short, 1/3 of the time you will switch and win, 1/3 of the time you will switch and lose, and 1/3 of the time you won't get the chance to switch because the second contestant will have got the prize. In other words, in the 2/3 of cases where the second contestant doesn't happen to get the prize, you have a 50/50 chance whether you switch or not. The same applies if the second contestant is a host who doesn't actually know where the prize is, which the overly vague definition on the MontyHallProblem page doesn't rule out. -- AndrewMcGuinness

I stand corrected. -- JohnVriezen

A bit of explanation to explain the difference between the two cases: In the case where the host (or whoever opens a door second)

Case You 2nd Probability Win if switch A W L1 1/6 No B W L2 1/6 No C L1 W 1/6 N/A D L1 L2 1/6 Yes E L2 W 1/6 N/A F L2 L1 1/6 YesThus, 1/3 of the time you win if you switch, 1/3 of the time you lose if you switch, 1/3 of the time the other guy wins and it doesn't matter. Simple enough. The error that people make, in assuming that this equivalent to the MontyHallProblem, is assuming that in the case where Monty knows where the winner is and avoids it, that this 1/3 is eliminated and is distributed equally among the other two choices, leaving 1/2 possibility of winning if you siwtch. Not so; any instance where Monty MIGHT have picked the winner is forced into Monty picking the other loser, leaving the winner behind the last door - in other words, the 1/3 probability that Monty picks the winner is COMPLETELY given to the case where the winner is the last one remaining, and you win if you switch. The table is as follows, changes from the above table are in

Case You 2nd Probability Win if switch A W L1 1/6 No B W L2 1/6 No C L1 WAdding them up, gives 2/3 chance of winning if you switch, 1/3 chance if you don't.0N/A D L1 L21/3Yes E L2 W0N/A F L2 L11/3Yes

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