1. ## Basic probability...

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

2. Originally Posted by struck
What is the probability that, out of these 24 throws, at least one is a double 6.
$1 - \left(\frac{35}{36}\right)^{24}$
WHY?

3. Originally Posted by struck
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
Hello. Try this approach. The event of getting at least one double 6 in 24 tosses and the event of getting zero double 6's in 24 tosses are complements of one another. Therefore, the sum of their respective probabilities equals 1. Use this axiom of complements to your advantage.

Now, this is a rather typical binomial probability distribution problem. There are 6^2 possible permutations of die pairings and only one of these yields a double 6. Ergo, the probability of rolling a double 6 is 1/36. If the probability of rolling a double 6 is 1/36, then the probability of not rolling a double six is 35/36. The probability of not rolling a double 6 in 24 tosses is 24 choose 24 times (35/36)^24. Therefore, the probability of rolling a double six at least once is 1-((24 choose 24) *(35/36)^24) = .4914

4. Originally Posted by Bilbo Baggins
There are 6^2 possible permutations of die pairings and two of these yield a double 6.
Ergo, the probability of rolling a double 6 is 1/18.
Please be sure that you understand the correct process before so you don’t reply using faulty information.
You are correct saying that there are $6^2=36$ pairs in the cross product of two sets of six each.
However, there is only one pair of double sixes, $(6,6)$.
Thus, the probability of getting a pair of double sixes on a throw of two dice is $\frac{1}{36}$.

5. Originally Posted by Plato
Please be sure that you understand the correct process before so you don’t reply using faulty information.
You are correct saying that there are $6^2=36$ pairs in the cross product of two sets of six each.
However, there is only one pair of double sixes, $(6,6)$.
Thus, the probability of getting a pair of double sixes on a throw of two dice is $\frac{1}{36}$.
Whoops! Thanks for catching my mistake. I'll amend my post.