A French gambler, the Chevalier de Mere, noticed (after losing large sums of mon

Question

A French gambler, the Chevalier de Mere, noticed (after losing large sums of money) that the total number of spots showing on three six-sided dice thrown simultaneously turns out to be 11 more often than it turns out to be 12, although from his point of view both events should occur equally often. He reasoned that 11 occurs in just six ways: 6:4:1, 6:3:2, 5:5:1, 5:4:2, 5:3:3 and 4:4:3 and that 12 also occurs in just six ways: 6:5:1, 6:4:2, 6:3:3, 5:5:2, 5:4:3, and 4:4:4. And therefore rolling 11 and rolling 12 on three dice thrown simultaneously should have the same probability. a)What is the fallacy in de Mere’s argument? b)How many total outcomes are there for throwing three six-sided dice simultaneously? c)Calculate the probability of rolling 11. d)Calculate the probability of rolling 12.

Explanation / Answer

there are total 28 ways to come 11 as sum when 3 dies are rolled

and 25 ways for 12 as sum when 3 dies are rolled

b) total outcomes =6*6*6 =216

c)  probability of rolling 11=28/216

d)  probability of rolling 12=25/218

1st Die 2nd Die 3rd die 1 6 4 1 4 6 1 5 5 2 3 6 2 6 3 2 3 6 2 4 5 2 5 4 3 2 6 3 6 2 3 3 5 3 5 3 3 4 4 4 1 6 4 6 1 4 2 5 4 5 2 4 3 4 4 4 3 5 1 5 5 5 1 5 2 4 5 4 2 5 3 3 6 1 4 6 4 1 6 2 3 6 3 2

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