Question

A game is to be played by first flipping a fair coin, and then drawing balls without replacement from a bin. If you flip heads you get to draw one ball, and if you flip tails you get two draw two balls. The bin contains 7 red balls, and 4 white balls. Let W = 4^x - 2 be your win amount, where X represents the number of white balls drawn. What is your expected win amount on any given play?

Answer 1

When we flip a coin possible outcomes are head and tail. So,

P(H) = P(T) = 0.5

Total number of balls: 7+4 =11

Now if head occur then you get ot draw one ball. That is X can take values 0, 1.

There are 7 red balls out of 11. So

P(X=0 | H) = 7 /11 = 0.6364

P(X=1 |H) = 4/11 = 0.3636

P(X=2 |H) = 0

Now if head occur then you get ot draw two balls. That is X can take values 0, 1, 2.

There are 7 red balls and 4 white balls out of 11. So

P(X=0 | T) = C(7,2) / C(11,2) = 42 / 110 = 0.3818

P(X=1 | T) = [ C(4,1) * C(7,1)] / C(11,2) = 28/55 = 0.5091

P(X=2 | T) = C(4,2) / C(11,2) = 28/55 = 0.1091

By the law of total probability we have

P(X=0) = P(X=0|H)P(H) + P(X=0|T)P(T) = 0.6364 *0.5 + 0.3818 *0.5 = 0.5091

P(X=1) = P(X=1|H)P(H) + P(X=1|T)P(T) = 0.3636*0.5 + 0.5091*0.5 = 0.43635

P(X=2) = P(X=2|H)P(H) + P(X=2|T)P(T) = 0*0.5 + 0.1091*0.5 = 0.05455

Now following table shows the calculations for expected vaue of W:

X	P(X=x)	W=(4^X)-2	P(W=w)=P(X=x)	WP(W=w)
0	0.5091	-1	0.5091	-0.5091
1	0.43635	2	0.43635	0.8727
2	0.05455	14	0.05455	0.7637
Total	1			1.1273

That is:

Hence, your expected win amount on any given play is $1.13.