Probability Function - Betting

Question

The game is played as follows: Three fair dice are rolled. You, as the bettor, place a \$1 bet on the occurrence of one of the integers 1, 2, 3, 4, 5 or 6. Suppose you bet on the occurrence of a 5. If at least one 5 occurs on the three dice, you win back your stake of \$1 and in addition \$1 for each 5 that occurs, unless all three die show a 5, when you win \$5 along with your stake. If no 5s occur then you lose your stake of \$1. (The same rules apply to any of the 6 numbers.) Let V be the net amount you win in one play of this game (loss = negative gain).

How to find the probability function of V?

Workings: To any of these $1, 2, 3,4,5,6$ the probability to occur is: $\frac {1}{6}$

I need to find out the following $P(X=3), P(X=2), P(X=1), P(X=0)$ where $X$ is equal to number of 5s.

$P(X=0)=\frac {5}{6}\frac {5}{6}\frac {5}{6}=\frac {125}{216}$ loss

$P(X=1)=\frac {1}{6}\frac {5}{6}\frac {5}{6}+\frac {5}{6}\frac {1}{6}\frac {5}{6}+\frac {5}{6}\frac {5}{6}\frac {1}{6}=\frac {25}{72}$ wins 1$

$P(X=2)=\frac {1}{6}\frac {1}{6}\frac {5}{6}+\frac {1}{6}\frac {5}{6}\frac {1}{6}+\frac {5}{6}\frac {5}{6}\frac {1}{6}=\frac {5}{72}$ wins 1$

$P(X=3)=\frac {1}{6}\frac {1}{6}\frac {1}{6}=\frac {1}{216}$ wins $5

Out of the data above how should I write the V function now?

Answer 1

According to the rules you stated, a gambler

• loses $\$1$ if the number he or she bets on does not occur
• wins $\$1$ if the number he or she bets on occurs exactly once
• wins $\$2$ if the number he or she bets on occurs exactly twice
• wins $\$5$ if the number he or she bets on occurs exactly three times

Hence, \begin{align*} P(V = -1) & = P(X = 0)\\ P(V = 1) & = P(X = 1)\\ P(V = 2) & = P(X = 2)\\ P(V = 5) & = P(X = 3) \end{align*} With the exception of the typographical error I mentioned in the comments, you correctly calculated $P(X = 0)$, $P(X = 1)$, $P(X = 2)$, and $P(X = 3)$.

Another way to calculate the probability that the number the gambler bets on occurs in exactly $k$ of the $3$ rolls is to use the binomial distribution. The probability that the number the gambler bets on occurs exactly $k$ times in $3$ rolls is $$P(X = k) = \binom{3}{k}\left(\frac{1}{6}\right)^k\left(\frac{5}{6}\right)^{3 - k}$$ where $\binom{3}{k}$ represents the number of ways exactly $k$ of the dice show the number the gambler bets on, $1/6$ is the probability that number appears, and $5/6$ is the probability that number does not appear.