# expected value of a two dice rolling game [closed]

I am stuck on how to proceed with this question: you toss two six-sided fair dice. if the sum is 7, you win a dollar. if the sum is even, you lose a dollar. otherwise, you roll again. The question seeks the expected payoff? I know the probability for obtaining a sum = 7 is 6/36 and the probability for sum = even is 18/36. 1-6/36-18/36=12/36 is the probability of rolling again (the sum is neither a 7 nor an even number). I also know that i have to write this as a geometric series (infinite sum). Any tips/insights would be much appreciate it.

• Thanks, i have edited my question. Hopefully it is more clear and aligned to the math SE guidelines. Commented Apr 5 at 17:00

Let $$E$$ denoted the expected payoff per roll.

Considering only the rolls that either give us a dollar or we lose a dollar, there are $$6+18=24$$ possibilities. $$\frac{6}{24}=\frac{1}{4}$$ of those win us a dollar. $$\frac{18}{24}=\frac{3}{4}$$ lose us a dollar. Thus, $$E=\frac{1}{4}(+1\)+\frac{3}{4}(-1\)=-0.5\$$

If you're interested, here's a simply python program to verify this result:

import random
n=100000
T=0
def result():
d_1=random.randint(1,6)
d_2=random.randint(1,6)
if d_1+d_2==7:
return int(1)
if (d_1+d_2)%2==0:
return int(-1)
else:
return result() #Roll again

for i in range(1,n):
T=T+result()

print("Average value after 100000 games is "+str(T/n))


Denoting with $$X$$ the payoff you can compute its expected value as:$$\mathbb{E}[X] = \frac{1}{6}\cdot 1\ + \frac{1}{2} \cdot (-1\) + \frac{1}{3}\mathbb{E}[X] \rightarrow \frac{2}{3} \mathbb{E}[X] = -\frac{1}{3} \.$$ Therefore $$\mathbb{E}[X] = -0.5\$$