Consider the following game with three balls (could do 10 balls, but 3 is much easier, and illustrates the point). Player Y puts 3 balls in a bag. He chooses n red balls and 3-n green balls with probability y_n for n = 0,1,2,3. Player X then takes two balls from the bag. He then guesses whether the next ball will be red or green. If he guesses red and is correct he wins $1, otherwise he loses $1. If he guesses green and is correct he wins $G, otherwise he loses $G. If he has seen m green balls so far, he guesses red with probability x_m, for m = 0,1,2. What y_n should Y choose, and what x_m should x choose in order not to be exploitable?
I'll leave out the details (maybe someone could check this?), but the profit for X is
(1+G)(x_0(-y_0+y_1/3)+(2/3)x_1(-y_1+y_2)+x_2(-y_2/3+y_3))+y_0+y_1/3-y_2/3-y_3
so we find that Y should choose y_0 = y_3 = 1/8 and y_1 = y_2 = 3/8. In other words, he should choose all green or all red a quarter of the time. Interestingly, this is independent of G.
We can also write this as
y_0(1-(1+G)x_0)+(y_1/3)(1+(1+G)(-2x_1+x_0))+(y_2/3)(-1+(1+G)(2x_1-x_2))+y_2(-1+(1+G)x_2)
so X should guess green with probability 1/(1+G) and red with probability G/(1+G), independent of the colour of the balls he takes out of the bag (x_0 = x_1 = x_2 = 1/(1+G)). If either X or Y follows this unexploitable strategy, their profit in the long run is zero.
Notice that if G is large, X should hardly ever guess green, but if he never guesses green, Y can exploit him by biassing the balls he puts in the bag towards green, and make a profit. It's just like poker. If someone pushes all in on the river a lot, you have to call them very infrequently in order to stop their bluffs being profitable.
Anyone fancy trying the general case of N balls in the bag?
My intuition says that
i) X should guess green with probability 1/(1+G) for all N>1.
ii) In the case of N=10 and G=1, even when X draws out 9 red balls, he should still therefore guess red and green with equal probability, because Y biasses his choice of balls so that 1 green and 9 red balls is more likely than 10 red balls by just the right (probably large) factor to make himself unexploitable.
iii) In fact, I bet the probabilities for Y are just those in Pascal's triangle
1
(11) / 2
(121) / 4
(1331) / 8
(14641) / 16
etc.
This may give a clue to an easy proof for general N.
So for N=10, Y chooses 9 red and 1 green ball 10 times as often as he chooses 10 red balls, so that the chance of drawing 9 green balls and leaving a red ball in the bag (1/10) is balanced by the fact that he puts these balls in the bag 10 times as often as he puts 10 green balls in the bag. y_10 = 1/1024, y_9 = 10/1024. Looks right to me.
