Suppose myself and 3 friend are playing Texas Hold 'Em poker. Each player gets 2 hidden card and then 5 cards are dealt to the centre which any player can use to build the best possible hand.
Each player is dealt two cards, face down. I look at mine and see 2 hearts. No cards have been dealt to the centre yet. I want to figure out what my odds are of getting a flush. At first I thought it was easy, but the more I think about it the more confused I get.
[ul]
[li]Since you have 2 hearts, there are 11 of them remaining in a partial deck of 50 cards. Note also that this deck has 39 “non-heart” cards.[/li]
[li]Since the deck is well shuffled and no one’s cheating (right?), the order in which the cards are dealt is irrelevant to the mathematics. We can ignore the other players’ hands (since we have no information about them anyway) and imagine that we deal the center hand right after dealing yours.[/li]
[li]Therefore, there are 50-choose-5 (2,118,760) possible hands that can show up in the center. Of all these hands, you’d like one with at least 3 hearts in order to make your flush.[/li]
[li]There are 11-choose-5 (462) ways of getting 5 hearts in the center hand.[/li]
[li]There are 11-choose-4 (330) ways of getting 4 hearts, with 39-choose-1 (39) ways of getting a single non-heart — and therefore 330 x 39 = 12870 ways total for the center hand to have exactly 4 hearts.[/li]
[li]There are 11-choose-3 (165) ways of getting 3 hearts, with 39-choose-2 (741) ways of getting 2 non-hearts — or 122265 ways total for the center hand to have exactly 3 hearts.[/li]
[li]Summing the last three results together, there are 135597 ways for the center hand to have 3 or more hearts.[/li]
[li]Since there are 135597 desirable combinations, and 2118760 possible combinations, the probability you’ll get your flush is the ratio of these numbers — about 6.4%.[/li][/ul]
Hope that helps. I hope, even more, that I didn’t slip up.
After you get your two cards, there are 50 left in the deck. Five cards are dealt to the center; there are (50 choose 5) = 50! / (45! 5!) = 2118760 such five card combinations.
Eleven hearts remain, and you need at least three of them represented in the five cards; there are (11 choose 3)*(39 choose 2) = [11! / (8! 3!)] * [39! / (37! 2!)] = 165 * 741 = 122265 combos with exactly three hearts; (11 choose 4) * (39 choose 1) = [11! / (7! 4!)] * 39 = 12870 with exactly four hearts, and (11 choose 5) = 11! / (6! 5!) = 462 with all five of them hearts.
So your chances of getting a flush (or straight flush) are:
(122265 + 12870 + 462) / 2116760 = 135597 / 2116760 = 6.4058750165347039815567187588579 %
Three and a half hours without an answer, then a simulpost! 
At least we agree on the answer, aside from my typo in the last line, which should read:
135597 / 2118760 = 6.3998282014008193471653231135192 %
I think there is about a 11% chance of hitting say three hearts out of five cards with two hearts already out. But my probability is very weak, and I probably calculated it wrong.
Just so a real mathematician can come along and tell me where I am wrong
with two cards out there are 50 left. there are 50^5 (or 312500000) combinations of cards (assuming each card position is unique)
there are 10 ways of getting three hearts eg. hhxhx, xxhhh etc. In each of these ways there are 11 hearts left and 50 all cards (hearts included) so there are 11x11x11x50x50 combinations (permutations?) in each way or 3327500 x 10 ways = 33275000 total three heart combinations. the ratio is then 0.106 or about 11%.
There is a very good chance that I messed up there somewhere.
I would beleive the first two people as they seem to know what they doing. Will write a little simulation program to check though. Report back later
guys - one thing - did you include the probability of getting 4 or 5 hearts (not just exactly three?) as both the former also satisfy the OP
I appreciate both replies. I understand the process you used, and it sounds correct but only for when I’m playing with myself (get your mind out of the gutter, it is a sick mental image anyway 
). I.e. you assume 50 cards are left. However, there are only 44 cards left since my three friends also have 2 cards. If I use the same method I get a better chance of having my flush, i.e. 44 choose 5 is a lower number then 50 choose 5 resulting in a higher percent. So, that means that as a I add players the chance of getting the flush will be improved. I guess what I’m saying is, how can we ignore the other players? Shouldn’t some of those combinations result in them having hearts in their hands?
There are 50 cards left to deal out. We’re only concerned about what combination of them could end up in the center. Think of it like this: the deal goes two to you, which you then look at, then five to the center, then two to all the other players. Since the shuffle is presumed random, the dealing order doesn’t really matter, you know? Without additional information, you are just playing yourself.
scm1001, yeah, both Bytegeist and I included the possibility of 3, 4, and 5 hearts.
Glitch, while in reality 8 cards have been dealt out, you only have knowledge of the identity of two of those cards. Given your limited knowledge, the upcoming five cards could be any of the 50 cards that are not in your hand.
erislover’s comment about the order of the deal being irrelevant is spot on. It may also be easier to see if you think of this simpler example. Say you and 50 other people are playing high card. Each person gets one card, let’s say you get the queen of spades. After the deal, one card is left; what’s the probability it’s the ace of spades? You have no knowledge of its location, it could be the card left in the deck, or one of your 50 opponents may hold out. The chance of it being left in the deck is 1/51.
On the other hand, say you’re playing with yourself again (;)). Again, you get the queen, now what’s the probability of the top card on the deck being the ace of spades? The ace is one of 51 cards left, so the probability once again is 1/51.
In short, if you have no knowledge of what other people are holding, it’s the same as if they didn’t exist.
please ignore all my posts - i have recalculated and completely agree with cabbage and bytegiest. Its about 20 years since I last did probablilty, but its actually coming back to me now
Cabbage, it’s good to have a confirmation. Thanks for slogging it out with me.
Because, you have no information about the states of those other cards — aside from the noises of glee or despair that your fellow players make, but that’s a psychological calculation, not mathematical.
We assume of course that the game is played fair, which would mean, among other things, that the deck is well shuffled, and that no one has knowledge about the ordering of the deck. This means that the order in which the cards are dealt is unimportant. The dealer could just as easily deal the center hand first before the players’ hands. Or he could deal your hand, then the center hand, then all the other players’ hands. The various ways in which the cards can fall are the same, regardless.
In a way, we are taking into account the other players, by figuring the probability that the hearts are in the center hand versus the rest of the deck. It’s just that the rest of the deck includes all the other players’ cards, as well as the cards not dealt to anyone. But that total number of cards is the same, regardless of the number of players.