Two white and two black balls, kept in two bins, are arranged in four different ways as given below. In each arrangement, a bin is to be chosen randomly and only one ball is needed to be picked randomly from the chosen bin. Which one out of the following arrangements has the highest probability of getting a White ball picked?
Arrangement 1: 1st bin (one white and one back ball),
2nd bin (one white and one back ball)
Arrangement 2: 1st bin (two white balls),
2nd bin (two back balls)
Arrangement 3: 1st bin (one white ball),
2nd bin (one white and two back balls)
Arrangement 4: 1st bin (one back ball),
2nd bin (two white balls and one back ball)
Answers
Answer:
Assume that we have a white balls and b blacks balls. We can choose between two things: to play or not to play. In the first case our profit is 0. Now assume that we choose to play. With probability aa+b we gain one dollar and we are left with a−1 balls and b black balls. Next, with probability ba+b we lose one dollar and we are left with a white balls and b−1 black balls.
This gives the following reccurence formula for Pa,b, the best expected profit we can get for a white balls and b black balls:
Pa,b=max{0,aa+b(1+Pa−1,b)+ba+b(−1+Pa,b−1)}
Now, if we have a white balls and no black balls, then obviously the best we can do is to win a dollars. On the other hand, if there is no white balls, then we should not play at all. I.e., Pa,0=a,P0,b=0.
Using these formulas, we obtain:
P0,0=P0,1=P0,2=P0,3=0,
P1,0=1,P2,0=2,
P1,1=max{0,12(1+P0,1)+12(−1+P1,0)}=12,
P2,1=max{0,23(1+P1,1)+13(−1+P2,0)}=43,
P1,2=max{0,13(1+P0,2)+23(−1+P1,1)}=0,
P1,3=max{0,14(1+P0,3)+34(−1+P1,2)}=0,
P2,2=max{0,12(1+P1,2)+12(−1+P2,1)}=23,
P2,3=max{0,25(1+P1,3)+35(−1+P2,2)}=15,
i.e., on average we can gain 1/5 dollars in the inital game. And the strategy is as follows. Look at the numbers above. Assume that we are left with a white balls and b black balls. If Pa,b>0, draw a ball, otherwise stop.