The Chevalier du Mere's Problem. A seventeenth-century French gamlber, the Chevalier du Mere, had run out of takers for his bet that, when a fair cubical dice wass thrown four times, at least one 6 would be scored. He therefore changed the game to throwing a pair of fair dice 24 times. What is the probability that, out of these 24 throws, at least one is a double 6.
It's basic probability here - since I am at the basic stages yet, so I am not sure if there's an easier / shorter way to solve it.
I think I could so something like:
Let the event that both fair dice in a throw are 6 = A
Let the event that both fair dice in a throw are not 6 = B
P(A) = 1/6 x 1/6 = 1/36
P(B) = 5/6 * 5 / 6 = 25/36
Then I would have to use something like this: (doesn't sound right to me?)
P( (A1 and B2 and B3 and B4 and B5 ... and B24) or (B1 and A2 and B3 and B4 and B5 ... and B24) ... etc.) = 23 x (1/36) x 25/36
That is not right - I know