We were playing "trash". If you're not familiar with "trash", each player is dealt 7 cards. Then each player choose 3 cards to pass on to the next player (assume clockwise). We were trying to figure out if the probability of flush is higher than the probability of a straight and got this far:

For flush (5 cards of same suit, non-consecutive):

- - the probability you are holding 4 cards of the same suit out of 7 dealt is: 52/52 (can be any one card) * 12/51 (12 remaining in the suit) * 11/50 * 10/49, I guess. then you add up the same total for the number of combinations of 4/7 there are (didn't bother with this step for now).

- for flush, assuming you are now holding 4 cards of the same suit after passing on your "trash". If someone were to pass you one card, the random probability of it being the same suit would be 9/52 (there are 9 more cards of that same suit). Since we don't have any information on whether that player or another is trying to go for a flush of the same suit, we won't try to account for that possibility for now. The probability that one of the 3 cards is the same suit is 9/52+9/52+9/52 = 27/52.

For straights (5 cards in sequence, any suit):

- - the probability you are holding 4 consecutive cards out of the 7 dealt is: 52/52 (any card is fine) * 4/51 (very specific next 4) * 4/50 * 4/49 * 4/48, maybe. Then add that total however many times for the 4/7 combinations (did not bother with this step).

- assuming you are holding 4 consecutive cards (different suits is fine), the probability of getting the next card in the sequence is 4/52. Same with the card before (unless the current cards begin or end with an Ace). So the probability of one of the 3 trash cards passed to you being the needed card is 4/52+4/52+4/52 = 12/52.

- If you almost have a straight but need a card in the middle of the sequence, at best the probability of getting the correct card is 4/52.