Probability 5. (Paradox of Chevalier de Méré) Chevalier de Méré mentions that wh

Question

Probability

5. (Paradox of Chevalier de Méré) Chevalier de Méré mentions that when rolling three fair non-distinguishable dice there are 6 different possibilities for obtaining either 11 or 12 the sun. Thus he concludes that both events (sum equals 11 or sum equals 12) should be equally likely. But experiments show that this is not the case. Why he was wrong and what are the right probabilities for both events? 1 P 6. A deck of 52 cards is randomly distributed to 4 players. Find the probability that each of the four players has exactly one ace and the probability that one player has all 4 aces. 7. In an urn are balls labeled from 0 to 6. Choose four balls with replacement. Find the probability that the sum of the chosen numbers equals 6.

Explanation / Answer

Sum of 12 can be achieved in following ways
6,5,1---Total cases = 3! = 6
6,4,2---Total cases = 3!= 6
6,3,3---Total cases = 3!/2! = 3
5,5,2---Total cases = 3!/2! = 3
5,4,3---Total cases = 3! = 6
4,4,4---Total cases = 3!/3! = 1

Total cases = 25

Probability = 25 * (1/6 * 1/6 * 1/6) = 25/216

Sum of 11 can be acheived in following ways

6,4,1 Total cases= 3! = 6

1,5,5 Total cases = 3!/2! = 3

5,4,2 Total cases = 3! = 6

3,3,5 Total cases =3!/2!=3

4,3,4 Total cases =3!/2!=3

6,3,2 Total cases =3!=6

Probability of a sum of 11: 27/216 = 12.5%

Probability of getting sum 11 is more than 12 both have 6 possible set but purmutation of numbers gives us different probability for both the cases.

When three different numbers form the partition, such as 12 = 6 + 2 + 4, there are 3! different ways of permuting these numbers. So this would count toward six outcomes in the sample space. When two different numbers form the partition, then there are three different ways of permuting these numbers.

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