# In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins / loses.

Very nice solution

Given a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six.
P (getting a six) $p= \large\frac{1}{6}$$\rightarrow q = 1 - p =$$\large\frac{5}{6}$
There are three possibilities here:
1. He gets a six in the first throw and he will receive Rs. 1. The required probability is $\large\frac{1}{6}$
2. He gets some other number on the first throw and throws again and gets a six on the second throw.
Since he loses a Rupee for not getting a six on the first throw, but wins back the Rupee on the second throw, his net gain/loss = 0.
Probability of this occuring is $\large\frac{5}{6}$$\times\large \frac{1}{6} = \frac{5}{36} 3. Does not get a six on the first two throws, but on the third throw. The required probability is \large\frac{5}{6}$$\times \large\frac{5}{6}$ $\times\large\frac{1}{6} = \frac{25}{216}$
The amount he will receive/lose is Rs. ( - 1 -1 + 1) = -Rs. 1.
Expected value of the amount =
$$\;E(X)=1 \times \large \frac{1}{6}$$$$+\;0\; \times \large\frac{5}{36}$$$$+(-1)\times\large\frac{25}{216}\;$$
=$$\large\;\frac{36+0-25}{216}=\;\frac{11}{54}$$

answered Jun 22, 2013

If man does no get 6 in all 3 trials the ln total money= -1-1-1 = -3 If gets 1 in 1st trail then money = 1 If in 2nd trail then money = -1+1 = 0 If in 3rd trail then money = -1-1+1 = -1 X ( random variable) = -3,-1,0,1. P(x=-3) = 5/6×5/6×5/6 = 125/216 P(x=-1) = 5/6×5/6×1/6 P(x=0) = 5/6×1/6 P(x=1) = 1/6 Mean = 1×1/6+0×5/36+-1×25/216+-3×125/216 = -364/216