# Throwing The Dice

A common problem to solve for various university courses are gambling games. In these games we are usually interested in the probability that we actually win. One of these game consists in throwing dice and the higher number wins, now a more complicated and from a statistical point of view more interesting game is where each player is throwing its die twice and the person which has the overall higher sum wins. In the following I am going through an example to calculate the winning probability step by step and in the end I attach an excel file were the game can be simulated using various kind of dice.

Let us have two four-sided and fair dice. Dice A has the numbers 1,2,3,4 and Dice B has the numbers 1,2,3,4. Each die is assigned to one person, let’s say person 1 and person 2. Each person throws its die twice, where the appearing numbers are written down. The player with the highest Sum wins. How high is the probability that player 2 wins?

Solution

First calculate probabilities associated to each possible outcome of each player. Player one can get a sum of 2,3,4,5,6,7,8 as shown in the table below. Each outcome can be assigned with a certain probability. In order to know the probabilities that we get a sum of 4 or 5 or whatsoever we have to simply sum up the probabilities. Now as player 2 has exactly the same die the feasible sums and the probability tables are the same as for player 2. All we have to do now is to find out at which combinations player 2 wins and sum up the probabilities. Knowing this we can easily calculate the probability that player 2 wins. In order to do so we have simply to compute the probability that player 2 wins for each possible sum of player two. For example, how high is the probability that player 2 wins when the sum is 2? The answer is zero. Why? If player 2 has a sum of two it is impossible to win as the lowest what player 1 can get is also a two and it would be a draw. Lets make another example. How high is the probability that player two wins when the sum of the die is 6? Well in order to win player 1 has to have something strictly smaller, which is 5,4,3,2. Now the probability that player 1 has a sum of 2,3,4 or 5 is simply the sum of the probabilities taken from the table above and is 10/16 (1/16 + 2/16 + 3/16 + 4/16). The table below shows the corresponding winning probabilities given the sum. Now we know the probabilities that player 2 wins given the sum of player 1. In order to know the overall winning probability, meaning that we do not know that player two has, we calculated the expected value over all possible sum that player 2 can achieve. 1/16 * 0+2/16 *1/16+3/16 *3/16+4/16 *6/16+3/16 * 10/16 + 2/16 * 13/16 + 1/16 *15/16 = 0.4140625 We see that the probability that player 2 wins in 0.4140625.

The computation file:

Throwing The Dice

1. sk.np says: