## monty hall problem

this is a fun problem, especially for those that have not seen it before.

imagine you are a contestant on the game show, Lets Make a Deal, with host, monty hall. in the game you are playing, you are presented with three doors. behind one of the doors is a prize and behind the other two are goats (non-prizes, unless you really like goats). you, the contestant, are then to select one of the doors. after you have made your choice, monty hall then opens one of the other doors revealing a goat and asks you whether you would like to keep the door you originally selected or if you would like to change to the other door. what should you do?

when i taught this to fourth grade students, we played this game a number of times for them to get a feel for how it works. they got the hang of it very quickly. after a while, i asked what strategy did they adopt to win. most knew that the game did not really begin until monty hall revealed a door, but almost everyone thought the game was an even chance of winning or losing.

to help them a little bit, we kept a tally of who switched doors and won, who switched doors and lost, who stayed doors and won, and who stayed doors and lost. they had expected all the numbers to come about equal, but what they found was that amongst people that switched doors, more of them won than lost, and those that stayed, more of them lost than won.

at this point, they were starting to catch on but there were still a few students that were not convinced that after removing a door the odds were not even. so i changed the game a little. this time, instead of three doors, there were 1000 doors. after picking a door, monty hall would remove 998 of doors which did not contain a prize. this left only two doors. at this point, the students knew that switching was the better strategy. they explained, that when picking the first door, there was a 1/1000 chance of selecting the door with the prize, and a 999/1000 chance it was in one of the other doors. when we removed 998 doors, the odds did not change to 50%, instead, the one door remaining assumed the 999/1000 chance of containing the prize.

after this change in the game, all the students understood the original game. they correctly found that there was a 2/3 chance the prize was in the other door.

to do this more precisely, we set up a few things. let $\Omega = \{ (w_1, w_2) \mid w_i \in \{1,2,3\}, w_1 \neq w_2 \}$ where $w_1$ is the door the prize is behind and $w_2$ is the door monty hall opens. let $A_i = \{ (w_1, w_2) \mid w_1 = i \}$ be the event the prize is behind door $i$. the door monty hall picks depends on the door we pick, so let $D = \{ (w_1, w_2) \mid w_2 = 3\}$ be the door monty hall opens, and let $E = \{ (w_1, w_2) \mid w_2 = 1\}$ represent the doors monty hall cannot open. we will need set $E$ later.

we want to calculate $P(\\{ \textrm{prize is in door 2} \\}\mid\\{\textrm{we pick door 1 and monty hall reveals door 3\\}).$