Chevalier de Mere's Paradox

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:

The probabilities you expressed are correct; odds are that if the dice are not loaded, you will win 2 out of 3 times in either. Of course, being that it will take six times as long to throw double sixes, it might seem that you don’t win as much with the second game simply because it is slower.