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Two people want to play a game in which the expected amount of money that each of them wins is equal to zero. After having chosen a number x, the game is played as follows: Player 1 rolls a fair die, independently, three times.

  • If none of the three rolls results in 6, then Player 1 pays one dollar to Player 2.
  • If exactly one of the rolls results in 6, then Player 2 pays one dollar to Player 1.
  • If exactly two rolls result in 6, then Player 2 pays two dollars to Player 1.
  • If all three rolls result in 6, then Player 2 pays x dollars to Player 1.

Determine the value of x.


So I determined the probability of each roll outcome but I am not sure how to figure out what the value of x is with this information.

P(no sixes) = 125/216

P(1 six) = 75/216

P(2 sixes) = 15/216

P(3 sixes) = 1/216

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The different cases $$ \text{payoff} = \begin{cases} -1 \,\,\text{No. of heads}=0\\ 1 \,\,\text{No. of heads}=1\\ 2 \,\,\text{No. of heads}=2\\ x \,\,\text{No. of heads}=3\\ \end{cases} $$ thus the expectation is $$ E(\text{payoff}) = P(H=0)\cdot (-1) + P(H=1)\cdot 1 + P(H=2)\cdot 2 + P(H=3)\cdot x = 0 $$ so you need to rearrange to get $x$.

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After a theoretical 216 games, player 1 will be down \$125, player 2 will be down \$(75+2*15+x)=\$105+x. Therefore $x=20$.

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