Question

A bag contains 4 white and 3 black balls. Four players (A,B,C, and D) agree to take turns drawing balls from this bag(without replacement) until a white ball is drawn. If the first player (A) draws a white ball, he gets $56. If he draws a black ball, he must pay $28 into the pot. the game continues until a white ball is drawn. Find the expected value to each of the 4 players, assuming each player initially pays in $14 to form the initial pot amount. Winner gets the pot.

Answer 1

In the first draw, probability of picking a white ball is 4/7
In the first draw, probability of picking a black ball is 3/7
We use a Random Variable, I_A, as:

The expected value to A is $ (56 * 4/7 - 28 * 3/7) = $20


If A succeeds in the 1st trial, no further trial is made.
Else, we move further
In the second draw, probability of picking a white ball is 4/6 [Since, the 1st black ball is taken out and not returned]
In the second draw, probability of picking a black ball is 2/6
We use a Random Variable, I_B, as:

The expected value to B is $ (56 * 4/6 - 28 * 2/6) = $28



If B succeeds in the 2nd trial, no further trial is made.
Else, we move further
In the third draw, probability of picking a white ball is 4/5 [Since, the 1st and 2nd black balls are taken out and not returned]
In the third draw, probability of picking a black ball is 1/5
We use a Random Variable, IC, as:

The expected value to C is $ (56 * 4/5 - 28 * 1/5) = $39.2

If C succeeds in the 3rd trial, no further trial is made.
Else, we move further
In the fourth draw, probability of picking a white ball is 4/4 [Since, the three black balls are taken out and not returned]
In the fourth draw, probability of picking a black ball is 0
We use a Random Variable, ID, as:

The expected value to D is $ (56 * 1) = $56






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