Dice Roll Probabilities

Question

I'm trying to figure out the probabilities for the following casino game:

You and the dealer each roll a pair of dice and the person with the highest individual die roll wins. If its a tie, you win.

First, what is the probability you win?

Second, given that you've won, whats the probability that the game resulted in a tie?

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Here are my thoughts:

For the first - the expected value of your die roll is 4.25; which means you will win unless the dealer rolls a 5 or 6, giving you a 2/3 chance of winning.

For the second I'm not sure how to think about it.

Answer 1

Suppose $2$ dice are rolled, say by you. We will find the probability that the "larger" result is $i$, for $i=1$ to $6$.

The probability that the larger is $1$ is $\frac{1}{36}$. The probability that they are both $\le 2$ is $\frac{4}{36}$, so the probability the larger is $2$ is $\frac{4-1}{36}$. The probability they are both $\le 3$ is $\frac{9}{36}$, so the probability the larger is $3$ is $\frac{9-4}{36}$. And so on.

So the probability the larger is $i$, for $i=1$ to $6$, is $$\frac{1}{36},\quad\frac{3}{36},\quad \frac{5}{36}, \quad\frac{7}{36},\quad\frac{9}{36}, \quad\frac{11}{36}.\tag{1}$$

Now suppose that you and the dealer each toss two dice. From (1), the probability the higher of your tosses is equal to the higher of the dealer's tosses (that is, the probability $t$ of a tie) is given by $$t=\frac{1}{36^2}\left(1^2+3^2+5^2+7^2+9^2+11^2\right).$$

Let $p$ be the probability you win. Then $p-t$ is the probability your number is bigger than the dealer's. By symmetry this is the same as the probability the dealer's number is bigger than yours. Thus $p+p-t=1$, and therefore $p=\frac{1+t}{2}$. We know $t$, so we know $p$.

As to the conditional probability, by the usual conditional probability calculation, or by inspection, the probability you won through a tie, given that you won, is $\frac{t}{p}$.

Remark: Presumably the real game is that if there is a tie the casino wins. A casino that lets customers win with probability $\gt \frac{1}{2}$ will not be long in business.

Answer 2

Denote $a,b$ are the results of my dices with $6 \ge a>b \ge 1$ We consider $6$ cases:

1. If the largest result of my dice is $1$, or $a=1$ then $b=1$. Hence, the dealer's results must less than or equal to $1$. We obtain $1$ way to determine the dealer's results such that the condition is satisfied. Thus, there are $1 \times 1$ ways that I won when $1$ is my largest result.
2. If $a=2$ then $b=1$ or $b=2$. Now we need to determine the dealer's results so that I could win. This means, the largest result of the dealer is $2$ or $1$. There are $2$ ways that the dealer gets the same result in two dices ($11$ and $22$). We consider the case that the result in two dices is different. We get $\frac{(2-1) \times 2}{2}=1$ ways. This means the dealer has $1+2=3$ ways to get the result. Therefore, there are $2 \times 3$ ways so that I won when my largest result is $2$
3. If $a=3$ then $b=1,2$ or $3$. Similarly, there are $3$ ways so that the dealer can get the same result in 2 dices ($11,22$ and $33$) and $\frac{(3-1) \times 3}{2}=3$ ways in order to get the different results in 2 dices. This means the dealer has $3+3=6$ ways to get the result. Hence, there are $6 \times 3$ ways so that I won when my largest result is $3$.
4. We obtain the formula $n \times \left(n+ \frac{(n-1)n}{2} \right)=\frac{n^2(n+1)}{2}$, which is the number of ways that I won when my largest result is $n$.
5. In conclusion, the probability to win is $$\dfrac{\sum_{n=1}^6 \frac{n^2(n+1)}{2}}{\left[ 6+ \frac{(6-1)6}{2} \right]^2}= \dfrac{\sum_{n=1}^6 \frac{n^2(n+1)}{2}}{21^2}$$