This is a discussion on $1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five? within the online poker forums, in the Cash Game Hand Analysis section; First off, I'm new here and I'm not really sure if this belongs in this area so forgive me if this should be somewhere else. 


#1




$1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five?
First off, I'm new here and I'm not really sure if this belongs in this area so forgive me if this should be somewhere else.
Let me dive right in. I have a situation I'm trying to get around that's giving me problems and has to do with mathematics. I'm trying to find an equation to calculate the odds of a card being dealt in a certain number of cards. For example: The flop comes out 5s, 5h, 2d. I have maybe a high pair... say kings or aces. Obviously if another player has a 5 he has me beat. What are the odds of him (or the other 8 players) having the 5? (This is not taking into account the fact that I could outdraw a set on the turn or river. I would just like to know how to calculate this for future reference.) I'd love it if anyone could provide me with a good answer. Thanks in advance! Frodo 
#6




2 of 47 cards are 5s which is 0.0425, multiply by opponents, in this case 8 which equals 0.34. So 2 to 1 for a 5 to be held.
You also have to factor in the fact that i just made that up so there's a strong chance it makes no sense at all but it really doesn't matter as any action is going to make any equation pointless. So no point wasting time really. 
#7




re: Poker & $1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five?
Yeah, wvhs right.
If you're lookking to figure out the odds that someone HAD the 5 (preflop), simple math. But realistically, useless because most hands with a 5 are folded preflop, especially if you had aces or kk and raised preflop. That's where you really have to give one specific situation including player types, stack sizes, how much was raised, etc to get a realistic answer. 
#9




Are you saying that they have a random hand? It's fairly simple maths but it gets complicated iof you want to know with 8 other people to the flop (since you have to calculate every scenario, the maths is simple but it would be really complicated. Ask on a maths forum because 'cos I'm sure there is a simpler way to calculte for 8 others to the flop but I've only done stats to Alevel standard.
For 1 person it would be combos of possible hands with 5/total combos of hands. If you have AA there is 1 combo of 55, 88 combos of 525k and 4 combos of A5. So total 93. There are 47 unknown cards and he as 2 in his hand so 47c2 gives us 1081 combos he can have. 93/1081 gives 8.6% chance. 
#10




Yeah, as in real life you have no idea what your opponents have. But you know what you have and what's on the table in front of you... This exact situation came up the other day and I've been wondering for the past couple of weeks if there was a way to calculate the odds of a player getting a certain card. I know I'm way over complicating the scenario but I guess that's part of the process of growing. This is question could probably be better answered by a mathematician but I wanted an answer that I could understand :P I'll take this to a math forum and see what they come up with.

#11




Quote:
This is where table dynamics, and betting pattern reads come into play.  
#12




Quote:
Yea if you do post on a maths forum and get a decent answer put a link up in this thread please 'cos I'm really struggling to see a method since calculating for like 8 players would be an insane manual calculation. 
#14




re: Poker & $1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five?
Quote:
Oh yeah and raise preflop! 
#15




Quote:

#16




Quote:

#18




Quote:
I know where you're going. If they had 1 card then your way would be true, they have two cards though so it isn't. The chances of them not having a 5 is 45/47 *44/46, which equals 91.6% . So the chances of them having a 5 is 10091.6 = 8.4%. Probably an easier way to approach it than my original combos method. Edit: Didn't see your last post before I posted this. I'm not understanding why you get 8.6%, I got 8.6% with the combos calcuulation but this metyhod gives me 8.4% 
#21




re: Poker & $1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five?
Using what I believe is the correct method described above, I think the answer when facing 8 opponents is 82.14% that someone holds a 5.

#23




Quote:

#25




Quote:

#26




I know the way to work this out. If there are 8 other people at the table therefore 16 cards the chances of everybody not having a 5 are (45/47)*(44/46)*...(30/32). That calculation is (45!/30!)/(47!/32!) = 0.4588
so the chances of somebody having a 5 are 10.4588 = 0.5412 = 54% To test this works you can workout that if there were 23 people sitting at the table therefore 46 cards somebody must have a 5. So it would be (45/47)*(44/46)*...(0/2) which clearly = 0 since 0/2 = 0 If there were 22 other people at the table the chances of everybody not having a 5 would be (45/47)*(44/46)*...(2/4) = (45!/2!)/(47!/4!) = 0.0056 so the chances of someone have a 5 would be 10.0056 = 0.9944 = 99.44% 
#27




Quote:

#28




re: Poker & $1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five?
Quote:

#29




js520 is extremely close, tough he made a very slight mistake (when introducing the factorials) that alters the result in ~2%. He also considered only 8 villains when the OP asks for 9 villains. His calculations do handle the case where the two fives are out there in the villain's hands. I'll reexplain, with the calculations fixed:
First, out of the 52 cards of the deck, we know 5 of them, and 2 of them are fives. This means that there's 47 unseen cards, and 2 of them are fives. We're going to deal a total of 18 cards (9 villains, 2 cards each), and we want to know the chance that one or more of them are fives. Logically, what we want to calculate is: (Note how if two of the cards are fives, the logical condition still holds true). Code:
P(Card1 is a five OR Card2 is a five OR ... OR Card18 is a five) Code:
1P(Card1 is not a five AND Card2 is not a five AND ... AND Card18 is not a five). Code:
1(45/47)*(44/46)*...*(28/30) BTW, this calculation is useless anyway because if you have a high pair, you should raise, and any nondonk opponent will almost always fold hands with a 5 except pocket pairs, connectors and gappers, and which ones of those he does fold depends on his style, skill and the size of your raise, so it is a completely different problem. PS: You can't use tricks like the rule of 2 or approximating ORs by doing sums in this calculation, because the calculation is long and too many error accumulates. 
#33




Number of cards remaining = N = 47 here
Number of cards we are interested in (the 5s) = m = 2 here Number of "spots" (villains times 2) = s = 9 villains x 2 = 18 here Probability that the cards we are interested are NOT in the s spots = (N  m)!/(N  m  s)! _________________ (N)!/(N  s)! EZ game. So probability is just 1  that. For 9 villains (a 10handed table) this is 62.44%, and for 8 villains, this is 56.98%. Edited: Had a slight error. 
#34




Quote:

#35




re: Poker & $1000 NLHE Full Ring: flop comes 5,5,2. How do I calculate the odds of another player having a five?
Yea my calculation was slightly wrong, should have been 1 (45!/29!)/(47!/31!) which does indeed = 56.98% for 8 villains
