Math riddle

Consider the problem in reverse. Deal out a two card hand first then deal out the three ten card hands. What is the probability that the two card hand has two aces? This should be statistically equal.

The total number of two card hands is 32!/30!/2! = 465
The number of hands with two aces is 4!/2!/2! = 6 so there are 459 hands without two aces.

For the remaining 30 cards, the number of hands should be 30!/10!/10!/10!. The product of that and 459 should be the total number of hands without a dual ace widow.

I can't promise that answer is worth more than you paid.

Unfortunately it is not the good answer.

What do you think the answer is?

There are 4C2 ways of choosing two aces out out the 32C2 combinations

of the (32 x 31) / (1 x 2) = 496 combinations

paired aces form (4 x 3) / (1 x 2) = 6 of those combinations

hence there are 490 ways to deal two cards out of 32 cards which are not a pair of aces

yoohoo greek spider, i have a friend who's still waiting for you to confirm

the answer to your 30 july question

or don't you know the answer?

Thanks! Silly math error undoes everything else.