I got some extremely challenging "Combinations" problems, I am freaking out! I simply don't have the clue of how to approach such problems. It's really too hard for me, but I will try to learn it! Please help me, you "Combinations" experts Please Register or Log in to view the hidden image! 1) How many different 5-card poker hands contain a three-of-a-kind in Aces and two cards of other ranks? [My first questions is, can the last two cards have the same rank (e.x. 7) as each other as soon as they are not Ace? But no matter the answer to above is yes or no, I still don't know how to do it! Please Register or Log in to view the hidden image! ] 2) How many different poker hands are there that contain a full house? (Full house-3 cards of the same rank, 2 other cards of the same rank as each other) 3) Find the number of poker hands that contain only a straight, not a royal or straight flush. (Straight-5 consecutive cards, not all the same suit Royal Flush- AKQJ10 of one suit Straight flush-5 consecutive cards of one suit, Aces can't be the high card, but can be A2345
I'm learning a surprising amount of physics and maths answering your posts Please Register or Log in to view the hidden image! For each 2-card combination of cards that AREN'T aces, how many 3-card combinations are there out of the four aces? How many selections of three same-rank cards are there out of the pack? For each one, how many selections of 2 rank cards are there out of the remaining 12 ranks? How many different sequences are there (card ranks only)? For each sequence, how many variations are there because of the four suits each card can have? Then subtract the royal and straight flushes.
"I'm learning a surprising amount of physics and maths answering your posts " Good times, przyk! Please Register or Log in to view the hidden image! 1) I think I need to understand the question first, which unfortunely, I don't! Can the last two cards have the same rank (e.x. 7) as each other as soon as they are not Ace. Say, for example, are AAA77, AAA44, AAAQQ possible hands?
The question is slightly ambiguous, since as you say, we're not sure whether to dicount full houses (e.g. AAA 77). However the question does say that the hand contains a 3-of-a-kind, which a full house does, so we might as well allow full houses. The solution isn't too difficult. Firstly we want to know how many ways there are of getting 3-of-a-kinds with aces. That is simply 4 choose 3 = 4 ways. We have 2 cards remaining, and we have 52-4=48 possible cards left (we can't choose any of the aces). So there are 48 choose 2=48.47/2=1128 ways of choosing two cards from this set. So the total number of hands is 4.1128=4512. This is easy if you modify the above argument. Firstly, we have A2345, 23456,....,9TJQK, which gives 9 straights. So consider straight A2345. Pick one of the cards. This card can be any of the 4 suits. Pick the second card, again, it can be any suit. The third card can be any suit, the fourth card can be any suit, and finally the last card can be any of the 4 suits. This gives 4^5 possible ways. However, somewhere in there we have a straight flush. In fact a straight flush can be chosen in 4 ways (the four suits) so we have 4^5-4 non-flush straights. Now simply multiply by the 9, to get (4^5-4).9 possible straights, excluding flushes.
Don't know - its not clearly stated in the question. If the two cards can't have the same rank, you'll have to find the total number of possibilities first, then subtract the full houses. In both cases you have to find the total number of hands allowing the other two cards to have the same rank, either as the final answer or as an intermediate solution.
