Murray, re: "And thank you to all who still dont understand the correct solution for inducing a warm glow of smug self-satisfaction in those who do." I am much amused, thank you.It seems like a number of converts have come over to the light side here... I'm curious to know if anyone, ever, sees the right answer right away. I certainly didn't.
Jon and Andres, thank you for making tables. You both made the same mistake; you just COUNTED the possibilities when you have to WEIGH them. Remember that having a certain number of possibilities doesn't necessarily mean that each of those possibilities is equally probable; in this case, they are not.
I will demonstrate this by editing Andres' table so that the probability of each outcome is weighed (Jon, I don't dare mess with your table). You remember, of course, that the probability of a complex outcome is the product of the probabilities of the events that compose it.
I will assume that the player is some silly person who believes that it doesn't matter if he switches or not, so the player's second choice is random.
And I will code each outcome: A means a switch resulting in a win, B means a switch resulting in a loss, C means a stay resulting in a win, and D means a stay resulting in a loss.
PRIZE IS IN CUP 1 (1/3)
-----------------
Choose 1(1/3), lift 2(1/2), change to 3(1/2): P=1/36,B
Choose 1(1/3), lift 3(1/2), change to 2(1/2): P=1/36,B
Choose 2(1/3), lift 3(1), change to 1(1/2) : P=1/18,A
Choose 3(1/3), lift 2(1), change to 1(1/2) : P=1/18,A
PRIZE IS IN CUP 2(1/3)
-----------------
Choose 1(1/3), lift 3(1), change to 2(1/2) : P=1/18,A
Choose 2(1/3), lift 1(1/2), change to 3(1/2) : P=1/36,B
Choose 2(1/3), lift 3(1/2), change to 1(1/2) : P=1/36,B
Choose 3(1/3), lift 1(1), change to 2(1/2) : P=1/18,A
PRIZE IS IN CUP 3(1/3)
-----------------
Choose 1(1/3), lift 2(1), change to 3(1/2) : P=1/18,A
Choose 2(1/3), lift 1(1), change to 3(1/2) : P=1/18,A
Choose 3(1/3), lift 2(1/2), change to 1(1/2) : P=1/36,B
Choose 3(1/3), lift 1(1/2), change to 2(1/2) : P=1/36,B
============================================
PRIZE IS IN CUP 1(1/3)
-----------------
Choose 1(1/3), lift 2(1/2), stay on 1(1/2) : P=1/36,C
Choose 1(1/3), lift 3(1/2), stay on 1(1/2) : P=1/36,C
Choose 2(1/3), lift 3(1), stay on 2(1/2) : P=1/18,D
Choose 3(1/3), lift 2(1), stay on 3(1/2) : P=1/18,D
PRIZE IS IN CUP 2(1/3)
-----------------
Choose 1(1/3), lift 3(1), stay on 1(1/2) : P=1/18,D
Choose 2(1/3), lift 1(1/2), stay on 2(1/2) : P=1/36,C
Choose 2(1/3), lift 3(1/2), stay on 2(1/2) : P=1/36,C
Choose 3(1/3), lift 1(1), stay on 3(1/2) : P=1/18,D
PRIZE IS IN CUP 3(1/3)
-----------------
Choose 1(1/3), lift 2(1), stay on 1(1/2) : P=1/18,D
Choose 2(1/3), lift 1(1), stay on 2(1/2) : P=1/18,D
Choose 3(1/3), lift 2(1/2), stay on 3(1/2) : P=1/36,C
Choose 3(1/3), lift 1(1/2), stay on 3(1/2) : P=1/36,C
(This is the same in essence as Andy's fixing of the table, but with everything spelled out.)
OK, so if you add up the probabilities of the all the A scenarios, and all the B scenarios, and so on, you get:
Probability that player will switch and win by doing so:
6/18 or 33%
Probability that player will switch and lose by doing so:
6/36 or 17%
Probability that player will stay and win by doing so:
6/18 or 33%
Probability that player will stay and lose by doing so:
6/36 or 17%
I think I can now say, with deep sincerity: QED. I have proved my case.
Marion, secretly hoping there'll be more holdouts so I'll have to learn another way to explain it