Chevalier de Mere's Paradox

Factual Questions
1

In the 17th century there were 2 popular games, one to obtain at least one 6 in four throws of a fair die, and the other to bet on at least one double 6 in 24 throws of 2 fair die. Chevalier de Mere a math dude, said that the probabilities of a 6 in one throw of a fair die and a double 6 in one throw of 2 fair die are 1/6 and 1/36. Therefore the probability of at least one in four throws of a fair die and at least one double 6 in 24 throws of two fair dice are 4 X 1/6 = 2/3 and 24 X 1/36 = 2/3. However experiance had proven that a gambler would win on the first game was higher then winning the second game a contradiction. Obviously Chevalier de Mere the math guy was wrong so what are the odds of the two games.

I came across this in a history book and began to look at it further, however being a history guy i am not to keen on probability and therefore I request the infinite + 1 knowledge of you the reader, please offer thoughts thanks

The mighty quinn
“The pigeons are going to run to him” ← yes even the pigeons :wink:

2

Yeah, the probabilities are wrong.

For the first one, it’s easier to find the probability that none of the rolls is 6, and then subtract that from one. The probability that none of the rolls is 6 is (5/6)^4, so the probability of winning the first game is 1 - (5/6)^4, which is about .5177.

The second one can be done similarly. The probability that none of the rolls is double 6 is (35/36)^4, so the probability we’re looking for is 1 - (35/36)^4, which is about .1066.

3

Oops, didn’t notice that the second game had 24 rolls instead of 4. So the probability for the second game is 1 - (35/36)^24, which is about .4914.

4

Good luck finding fair dice in the Middle Ages, my friend.

5

Cabbage is right.

de Mere’s assumption that the chance of a 6 showing up in 4 throws is 1/6 (the odds of a 6 showing up) * 4 (the number of throws) is faulty- by that math, if I were to roll a die 6 times, a 6 is guaranteed to show up (1/6 * 6 = 1).