George Rebane

Fun time.  We all know the original MH problem – three closed doors behind one of which is a car (C) and goats behind the other two.  You pick one door, MH knows the C-door and opens one of the other doors with a goat.  You now get to stay with the original door or switch to the remaining unopened door.  A lot of clever people made fools of themselves saying switching didn’t matter, since the probability of picking the C-door from the remaining two closed doors was 50-50.  We now know that switching ups your probability of winning the car from 1/3 to 2/3.

So I was thinking of an interesting variation of the MH problem.  You now have N > 3 doors with C behind one of them.  You pick one, and then MH opens M < N-2 doors with goats staring at you.  What’s the probability of winning C by switching to one of the doors unopened by MH, and what is the probability of winning C by staying with your original pick?

Notationally P(C|N,M) is the probability of winning C when switching given N and M.  And P(C|N) is the probability of winning, given N, when you stick with your original pick – i.e. you don’t switch.  Hint: P(C|N) is the same probability as C not being behind one of the remaining doors that MH can open.  Formally we write P(C|N) = P(¬C|N,M).  The minus sign with the hangy-down part is the symbol for the logical NOT.  So, the probability of C not being behind one of the N-1 doors that you did not select is the same as its being behind the door that you did select.  Capice?

Finally, if the value of C is $VC = $100K in a game where N = 100 and M = 80, and you switch your door pick, then the expected value of the game is the maximum you should pay to play.  The expected game value $VG = P(C|N,M)*$VC = 0.0521*$100K = $5,210.  Most people would not go that high.  What amount would you be willing to pay for play?

Extra credit question – if C is valued at $100K, what is the maximum amount you should be willing to pay to play the game?  So here is the solution.