1- Four cards are dealt off the top of a well shuffled deck. You have a choice: (a) To win $1 if the first card is a club, and the second a diamond, and the third is a heart, and the fourth is a spade, in this order. (b) To win $1 if the four cards are of four different suits, in any order. Which option is better? Or are they equivalent? Justify by showing what is the probability in each case.

in option A you have 1/52 ways to fill the first card, 1/51, 1/50, 1/49 for the rest, multiply it all out you have a chance in 6.5 million. Option B, you have 4 ways of 52 to fill the first one, 3 of 51 to fill the second, 2/50, 1/49... multiply it all out you have a chance in 270,000 or so. Maybe I'm wrong but that's what I think. Why am I doing this math?

I'm pretty sure your math is wrong here. You don't have a 1/52 chance in the first example, you have a 13/52 chance.

ok my guess A- 13/52 x 13/51 x 13/50 x 13/49 = .0043% B-52/52 x 39/51 x 26/50 x 13/49 = .105% of course B is better

This is how I see it: Option (a) and option (b) are very similar. They both require the player to draw 4 cards in different suits. The only different is that for option (a), you need to get four different suits in a defined order. So, option (b) would be less difficult given one less criteria to fulfill. PA