Brad DeLong has solved the
balls in the hat game, and in his comments section provides
the solution, which is a psychologically horrible one: you basically have
to run your losses and cut your winnings.
Keep playing until you get to (0,1) or (1,2) or (1,3) or (2,4) or (3,5) or
(3,6). At those points the game is no longer worthwhile.
If you recall, there are 4 white balls and 6 black balls in the hat, which
means the initial state is (4,6). You get $1 for every white ball you pick out,
and lose $1 for every black ball you pick out. You can stop picking out balls
at any time, and you want to play the game so as to have a positive expected
The interesting thing about this solution is that there’s no chance of getting
the best-case scenario (picking out four white balls in a row and then stoppping,
for a profit of $4) but there is a chance of getting the worst-case scenario
(picking out six black balls in a row and then recouping some of your losses
with the remaining white balls, for a loss of $2). In fact, if the first ball
out of the hat is white, then you stop then and there: you cash in your $1 and
And I think I’m right in saying that’s the only way you actually walk
away with a profit in this game. If the first ball out of the hat is black,
then your best-case scenario is to break even.
So your maximum upside is $1, your maximum downside is $2, and most of the
time the game consists of a desperate struggle to get back up to $0 after picking
out a black ball initially. You can take your +$0.07 EV, I’m not playing this
The best case scenario would be 4$ profit and then you stop… worst case is losing 2…
So you don’t actually need to stop when you are up by 1$ but it would be the smart decision.