Question

A friend proposes a unique game of chance to you. He swears the game is straightforward but you're not sure. He proposes, you flip a coin and then roll a dice and your payout is the amount showing on the dice. However if you flip heads, you roll a 6 sided dice, and if you flip tails, you roll a 10 sided dice. What is the expected payout from this game?

Answer 1

X: Payout

X =1 ; If you flip heads and the '1' showing on the 6 sided dice or If you flip tail and '1' showing on the 10 sided dice

P(X=1) = (1/2 x 1/6) + (1/2) x (1/10) = 1/12+1/20 =8/60=2/15

X=2 ; If you flip heads and the '2' showing on the 6 sided dice or If you flip tail and '2' showing on the 10 sided dice

P(X=2) = (1/2 x 1/6) + (1/2) x (1/10) = 1/12+1/20 =8/60=2/15

Similarly

P(X=3)=P(X=4)=P(X=5)=P(X=6) =2/15

X=7 ; If you flip tail and '7' showing on the 10 sided dice

P(X=7) = (1/2)x(1/10) =1/20

X=8 ; If you flip tail and '8' showing on the 10 sided dice

P(X=8) = (1/2)x(1/10) =1/20

P(X=9)=P(X=10) =1/20;

Therefore the Probability distribution of X : payout is per below table

X	p(X=x)
1	2/15=0.1333
2	2/15=0.1333
3	2/15=0.1333
4	2/15=0.1333
5	2/15=0.1333
6	2/15=0.1333
7	1/20=0.05
8	1/20=0.05
9	1/20=0.05
10	1/20=0.05

expected payout from this game = E(X)

X	p	xP(x)
1	0.1333	0.1333
2	0.1333	0.2667
3	0.1333	0.4
4	0.1333	0.5333
5	0.1333	0.6667
6	0.1333	0.8
7	0.05	0.35
8	0.05	0.4
9	0.05	0.45
10	0.05	0.5
	Total	4.5

Expected payout from this game = 4.5