## Friday, July 11, 2014

### Unizor - Probability Examples - Dice Games

Craps

In the game of craps a player rolls two dice. Two numbers on top of these dice are summed up. The sums that are equal to 7 or 11 represent a win for some players.

What is the probability of having 7 or 11 as a sum of the numbers on two dice?

Solution

First of all, we have to establish the set of elementary events (or the sample space, as it is called) as the equally probable symmetrical results of this experiment. Obviously, all the pairs of numbers from 1 to 6 can be on top of two dice with equal chances. Therefore, our set contains 6·6=36 elementary events, each of which has the probability of 1/36.

Now we have to determine the number of those pairs of numbers from 1 to 6 that in sum equal to 7 or 11.

To get the sum of 7 on two dice we can have combinations {1,6}, {2,5}, {3,4}, {4,3}, {5,2} and {6,1} - 6 different combinations.

To get the sum of 11 on two dice we can have combinations {5,6} and {6,5} - 2 different combinations.

Therefore, there are 6+2=8 different elementary events (pairs of numbers from 1 to 6) that make up an event of having 7 or 11 as a sum of the numbers on two dice. Since the probability of each elementary event is 1/36, the probability of the event of having 7 or 11 equals to 8/36=2/9.

Yahtzee

This game is played with five dice and different combinations have different seniority.

What is the probability of having a combination of three dice having the same number on top and the other two dice having the same number on top, which is different from the first number?

Solution

The sample space in this case is a sequence of five numbers from 1 to 6 each. Therefore, there are 6^5 different elementary events, each having a probability of 1/6^5.

Now let's count the "good" elementary events - those that have three equal numbers and two other equal numbers among five numbers in a sequence.

First, count all the distributions of five dice into two groups of three and two dice. This can be done in C(5,3)=10 ways. Then pick a number for the triplet of dice, one of 6. Then we have to pick the number for the pair of dice out of remaining 5 numbers.

The result is a multiplication of the numbers of choices:

10·6·5 = 300

Therefore, the probability of the Full House in the game of Yahtzee equals to

300/6^5 ~= 0.03858 ~= 4%.

First Game of Chevalier de Mere

Chevalier de Mere lived in XVII century and was a gambler. Here is a dice game he suggested.

He rolls a dice four times. If number 1 comes up at least ones, he wins. If not, he loses.

What's the probability to win for him?

Solution

The sample space in this case is all sequences of four numbers from 1 to 6 each. There are 6^4 different statistically equivalent elementary events in this space, each having a probability of 1/6^4.

Now let's count the "good" elementary events - those that have at least one number 1 among four numbers in a sequence. We will count the sequences that do not contain the number 1 and then subtract it from the total number of elementary events. So, the total number of all sequences equals to 6^4, as indicated above. The count of sequences without number 1, similarly, is 5^4, since we can use only five digits from 2 to 6.

Therefore, the number of "good" elementary events equals to

6^4 − 5^4.

The probability of winning, therefore, equals to

(6^4−5^4)/6^4 = 1−(5/6)^4 ≅ 0.5177

This probability is greater than 50% and, therefore, chevalier de Mere more often won than lost this game.

Second Game

of Chevalier de Mere

After winning some money in the game of hitting at least once number 1 in four rolls of a dice and losing many companions because of that chevalier de Mere decided to invent another game.

He rolls two dice 24 times. If number 1 comes up simultaneously on both dice at least once in this series, he wins. If not, he loses.

Unfortunately for him, he has lost a lot in this game.

What's the probability to win for him?

Solution

The sample space in this case is all sequences of 24 pairs of numbers from 1 to 6. There are 36^24 different elementary events in this space, each having a probability of 1/36^24.

Now let's count the "good" elementary events - those that have at least one pair {1,1} among 24 pairs in a sequence. We will count the sequences that do not contain the pairs {1,1} and then subtract it from the total number of elementary events. So, the total number of all sequences of pairs equals to 36^24, as indicated above. The count of sequences without pairs {1,1}, similarly, is 35^24, since we can use any other pair of numbers from 1 to 6 but {1,1} on each of 24 positions.

Therefore, the number of "good" elementary events equals to

36^24 − 35^24.

The probability of winning equals to

(36^24−35^24)/36^24 = 1−(35/36)^24 ≅ 0.4914

Its less than 50% and, therefore, chevalier de Mere more often lost than won this game.

After losing a lot of money he requested a help of a famous mathematician Blaise Pascal, which was the beginning of the Theory of Probabilities as a mathematical subject.

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