I’m not sure what your point is. If the prize is behind the 2nd contestants door then I don’t get a chance to switch. In the only games I get a chance to switch the prize has an equal chance of being behind the door I picked or behind the other door assuming that Monty is fair in choosing which door to reveal when the prize is behind the 3rd door. My simulation shows this clearly, it doesn’t matter whether or not you switch.
It seems that’s settled at last. 
But no one’s presented an answer to #58, the loosely-related bridge problem.
(Or is it that the problem is “old hat” to Dopers who play bridge, and incomprehensible to those who don’t?)
Assume 99 runs:
Never Switching:
33 times, player A picked the winner
33 times player B picked the winner
33 times neither picked the winner.
In the 33 times A picked the winner, B was eliminated, and A offered option to switch. If he never switches, then he wins 33 times.
In the 33 times B picked the winner, A was eliminated, and B offered option to switch. If he never switches, then he wins 33 times.
In the 33 times neither picked the winner:
16 times Player A was eliminated and B offered option to switch… If he never switches, B wins 0 times.
17 times PLayer B was eliminated and A offered option to switch. If he never switches, A wins 0 times.
So when no one ever switches, Player A wins 33, Player B wins 33, and 33 times neither player win.
Always Switching
33 times, player A picked the winner
33 times player B picked the winner
33 times neither picked the winner.
In the 33 times A picked the winner, B was eliminated, and A offered option to switch. If he always switches, then he wins 0 times.
In the 33 times B picked the winner, A was eliminated, and B offered option to switch. If he always switches, then he wins 0 times.
In the 33 times neither picked the winner:
16 times Player A was eliminated and B offered option to switch… If he always switches, B wins 16 times.
17 times Player B was eliminated and A offered option to switch. If he always switches, A wins 17 times.
So when players always switch, Player A wins 17, Player B wins 16, and 66 times neither player win.
There’s four possibilities:
-
1st contestant picks correct door (prob. =1/3), 2nd contestant picks incorrect door (prob. =1), Monty reveals 2nd contestant’s door (prob. = 1). Total probability of occurring = 1/3 x 1 x 1 = 1/3
-
1st contestant picks incorrect door (prob. =2/3), 2nd contestant picks correct door (prob. = 1/2), Monty reveals 1st contestant’s door (prob. =1). Total probability of occurring = 2/3 x 1/2 x 1 = 1/3
-
1st contestant picks incorrect door (prob. =2/3), 2nd contestant picks incorrect door (prob. =1/2), Monty reveals 1st contestant’s door (prob. = A). Total probability of occurring = 2/3 x 1/2 x A = 1/3*A
-
1st contestant picks incorrect door (prob. =2/3), 2nd contestant picks incorrect door (prob. =1/2), Monty reveals 2nd contestant’s door (prob. = 1- A). Total probability of occurring = 2/3 x 1/2 x (1-A) = 1/3*(1-A)
In 1) and 4) the 1st contestant can win and in 2) and 3) the 2nd contestant can win.
In 1) and 2) the winning choice is to stick and in 3) and 4) the winning choice is to change.
As 0≤A≤1, at best changing your choice will not improve your probability of winning and at worst it will ensure you don’t win. When A = 1/2, changing your choice will cause you to lose 2/3 of the time.
(continued from my post above)
** 50/50 switching split **
33 times, player A picked the winner
33 times player B picked the winner
33 times neither picked the winner.
In the 33 times A picked the winner, B was eliminated, and A offered option to switch. If he switches half the time, then he wins 17 times.
In the 33 times B picked the winner, A was eliminated, and B offered option to switch. If he always half the time, then he wins 16 times.
In the 33 times neither picked the winner:
16 times Player A was eliminated and B offered option to switch… If he switches half the time, B wins 8 times.
17 times Player B was eliminated and A offered option to switch. If he switches half the time, A wins 9 times.
So when players switch half the time, Player A wins 24, Player B wins 25, and 50 times neither player win.
Never switching is clearly best
Son of bitch! When I set up the simulation using Septimus’ method it’s clear you don’t switch. I’ll be damned if I’ve got enough energy left to figure out why tonight. I’m pretty sure it’s because half the time that prize is behind the third unselected door I don’t get to play.
Okay, let’s add one more wrinkle to the problem to help folks understand: the goats have names. Clarice and Boomer.
Here are six possibilities:
- You picked the right prize, Bob picked Clarice.
- You picked the right prize, Bob picked Boomer.
- Bob picked the right prize, you picked Clarice.
- Bob picked the right prize, you picked Boomer.
- You picked Clarice, Bob picked Boomer.
- You picked Boomer, Bob picked Clarice.
In scenario 1 and 2, Monty is gonna reveal Bob’s goat. Right? You’ll get a chance to switch.
In scenario 3 and 4, Monty is gonna reveal your goat. Right? You’ll be out of the game.
In scenario 5 and 6, Monty has to choose which goat to reveal. Let’s say that Monty is secretly in love with Clarice, and given the choice will always reveal her; that means you get eliminated in scenario 5, but stay in the game in scenario 6.*
Do you agree with me so far?
So now you’ve got three scenarios in which you stay in the game and get a choice: 1, 2, and 6 (or, see the note, 5a and 6a, which together are as likely as a 1 or a 2–feel free to ignore this note). In two of those scenarios where you stay in the game, you have the winning prize. In one of them, you don’t.
Now, you may object, who gives a shit about the goats’ names? They’re functionally equivalent, they just bleat and shit and mark you as a loser, right? Sure, that’s true–but they still comprise different scenarios, even if the scenarios are functionally similar.
- If that really bothers you, you can flip a coin for both scenario 5 and scenario 6, leading to 5a and 5b and 6a and 6b, each of which is half as likely as any other scenario, and in which the a subscenarios keep you in the game and the b subscenarios kick you out. Your odds will come out the same.
I didn’t see anyone post a truth table. I made one when trying to rewrite my old Monty Hall JavaScript program.
A:1 B:2, prize 1 : **no switch**
A:1 B:2, prize 2 : **no switch**
A:1 B:2, prize 3 : **switch**
A:1 B:3, prize 1 : **no switch**
A:1 B:3, prize 2 : **switch**
A:1 B:3, prize 3 : **no switch**
A:2 B:3, prize 1 : **switch**
A:2 B:3, prize 2 : **no switch**
A:2 B:3, prize 3 : **no switch**
A:2 B:1, prize 1 : **no switch**
A:2 B:1, prize 2 : **no switch**
A:2 B:1, prize 3 : **switch**
A:3 B:1, prize 1 : **no switch**
A:3 B:1, prize 2 : **switch**
A:3 B:1, prize 3 : **no switch**
A:3 B:2, prize 1 : **switch**
A:3 B:2, prize 2 : **no switch**
A:3 B:2, prize 3 : **no switch**
So, out of 18 possible scenarios, we have 6 where you would need to switch to win, and 12 where you should not switch.
Hence not switching is the better choice, giving you a 12/18 = 67% chance.
No, I think you’re misreading the situation. Keep in mind the question is only addressed to the second contestant. The first contestant has his door and loses; he isn’t given any options. So by specifying the question is being asked to the surviving contestant, you’ve already closed off some possibilities.
The problem with looking at it this way though is that if you are A, Monty’s reveal eliminates those 6 situations when B picks the right door.
BigT, I think you got the right answer for the wrong reason.
You don’t have the option to switch in this case: you’re out.
ETA: Or what Asympotically fat said.
Some Dopers are making this harder than it is. Like so many such problems, the “shortcut” is to “keep your eye on the ball” – the Information ball:
Initially the chance for curtain #3 was 33% and nothing has happened to change that.
… a more interesting problem … and of real-world relevance (if bridge-players live in the real world. :rolleyes: )
Thanks to whoever finally cleared things up for me upthread, the simplest explanation is:
There are three possible initial scenarios:
- You: car, Bob: nothing, third door: nothing - probability 1/3
- You: nothing, Bob: nothing, third door: car - probability 1/3
- You: nothing, Bob: car, third door: nothing - probability 1/3
When Monty opens a door that has nothing, he can either open yours or Bob’s door only. So the three scenarios now become four:
- You: car, Bob: nothing, third door: nothing, Monty HAS to open Bob’s door- probability still 1/3
2a. You: nothing, Bob: nothing, third door: car, Monty picks your door to open - probability is 1/6
2b. You: nothing, Bob: nothing, third door: car, Monty picks Bob’s door to open - probability is 1/6 - You: nothing, Bob: car, third door: nothing, Monty HAS to open your door - probability still 1/3
From the four above, 2a and 3 are removed, since you are eliminated in those scenarios, and the original question presupposes that you stay to the next round. So what is left is 1 and 2b. The chance of 1 happening is double that of 2b. In 1 you win. In 2b you lose. If you don’t switch, your chance of winning is double that of losing. Or 66.7%.
Again, thanks to zut for making this crystal clear.
Whilst that explanation is correct (though as I’ve tried to point out it does assume that Monty doesn’t favour either contestant when revealing the doors when both choose incorrectly), Big|T’s explanation isn’t correct even if it gets to the right answer.
To simplify this as much as I can:
The prize has equal chances of being behind the door you select and the 3rd unselected door. You get to play every time the prize is behind the door you selected, you only get to play half the time the prize is behind the 3rd unselected door.
Yes, that is correct.
But are you forgetting that Monty is always going to pick the losing door?
This is a tree diagram for when A=1/2 (where A is the probability that Monty reveals the 1st contestants door if both contestant initially choose the incorrect doors):
I thought this was the classic Monty Hall problem. It’s not. The key isn’t the other player. The key difference is that Monty, even though he knows where the prize is, has his hand forced by the rule that one player must learn that they’ve lost. And with three doors and one prize, at least one player must always lose.
So let’s go through the choices. Player one will always pick door one. Player two always door two.
Big prize behind door #1: Monty opens door two. He has no choice. Player one should stay put to win.
Big prize behind door #2: Monty opens door one. He has no choice. Player two should stay put to win
Big prize behind door #3: Now Monty has a choice. Both players have losing doors. Who does he screw over? Whoevers door he opens, the other should switch.
Ok, choices all out in the open, what are the odds?
One third of the time you’re eliminated from the game immediately. This in itself is a huge change from the Monty Hall problem btw.
One third of the time you’re given a choice and should stay put.
One half of one third of the time you’re also eliminated, this time by chance.
One half of one third of the time you should switch.
Let’s convert those thirds to 2/6 and sum up.
3/6 times you’re out of the game.
2/6 times you should stay put.
1/6 times you should switch.
It’s conclusive. I voted wrong. You’re twice as likely to win by staying put than switching .
Alrighty, fair enough.
Although, I wouldn’t call this a classic riddle so much as it is a probability problem.
The OP knew going in many would already be familiar with the Monty Hall Problem, so it was a trick question as well.