[Math] Dice Roll Probabilities

probability

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.

Best Answer

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.

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Probability – Most Unfair Set of Three Nontransitive Dice

It turns out that you can do very slightly better than the dice in your example. The probability has to be a multiple of $1/36$, your example has $20/36$, and the best possible value is $21/36=7/12$.

From my answer to the question that anon linked to, we know that the probability of one $6$-sided die beating another while having a lower median can be at most $3/4-1/12=2/3=24/36$. Wikipedia gives an example of a cycle of four dice in which each die beats the next one with this probability.

You're looking for a cycle of three dice in which each die beats the next one with the same probability. Without loss of generality we can assume that the dice carry $18$ different numbers. Then there are $18!/6!^3=17153136$ different sets of dice, so it's well within the reach of our electronic friends to try them all out. Here's code that does that (rather inefficiently). The result is that the maximal probability for such a cycle is $21/36=7/12$ and there are eight different sets of dice that attain that probability. Here they are, with consecutive stretches of numbers collapsed into one as in your example:

0 3 3 3 6 6
2 2 2 5 5 5
1 4 4 4 4 4

0 0 3 3 3 6
2 2 2 2 2 5
1 1 1 4 4 4

0 3 5 5 5 5
2 4 4 4 4 7
1 1 1 6 6 6

0 3 5 7 7 7
2 2 4 4 6 9
1 1 1 8 8 8

0 3 3 5 5 5
2 2 2 4 4 7
1 1 1 6 6 6

0 3 5 5 7 7
2 2 2 4 6 9
1 1 1 8 8 8

0 3 3 3 3 5
2 2 2 2 4 7
1 1 1 6 6 6

0 0 3 6 6 6
2 2 2 5 5 5
1 1 1 4 7 7

There are two sets with $7$ different numbers, four sets with $8$ and two with $10$, and your near-optimal example has $9$, whereas the mean number of different numbers after collapsing, averaged over all sets with $18$ different numbers, is $18-17\cdot(5/17)=13$, so consecutive stretches of numbers and thus high collapsibility are a striking feature of these sets.

P.S.: I just realized that there's a symmetry in that you can invert the orders of both the dice and the numbers and get another set with the same probability. The first two sets are related by this symmetry, as are the third and seventh, the fourth and sixth and the fifth and eighth. Thus there are only four really different sets of dice, with $7$, $8$, $10$ and $8$ different numbers, respectively.

[Math] Probability of winning dice roll-off with a re-roll

I gather that only the last roll of player 2 "counts."

If player 2 is losing after her first roll, she will certainly roll again. If player 2 is winning after her first roll, she will certainly not roll again. If player 2 ties after her first roll, she will roll again if she has a 1 or 2 or 3 (since there is a greater chance she'll tie or win on her second roll) and she will not roll again if she has a 4, 5 or 6 (since there is a greater chance she'll lose on her second roll).

One can enumerate all these results in a decision tree and simply count the wins and losses.

Player 1: ${\bf 1}$

Player 2:

${\bf 1}$ followed by ${\bf 1}$ or $\{ {\bf 2}-{\bf 6} \}$: P[tie] = 1/6; P[win] = 5/6
$\{ {\bf 2}-{\bf 6} \}$ P[win] = 1

So if Player 1 rolls a ${\bf 1}$, the Player 2 has the following probabilities:

Tie: $1/6$

Win: $5/6$

Continue through other cases of player 1 rolling ${\bf 2}$ through ${\bf 6}$, and be sure to weight each result by the probability of the rolls.

Then add up the results!

Best Answer

Related Solutions

Probability – Most Unfair Set of Three Nontransitive Dice

[Math] Probability of winning dice roll-off with a re-roll

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