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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.

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    Commented Apr 5 at 16:55
  • $\begingroup$ Thanks, i have edited my question. Hopefully it is more clear and aligned to the math SE guidelines. $\endgroup$
    – user147813
    Commented Apr 5 at 17:00

2 Answers 2

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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))
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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\$ $

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