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

$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

Depends. What was the action preflop? How may players saw the flop? What types of players saw the flop? What was the flop action before it was your turn?

In other words there is no way to answer your question as written.

This is purely a mathematical question. I just want to be able to calculate the odds of one of any number of other players having a certain card.

It depends.

If anyone has a legitimate mathematical formula to calculate the answer to this question it would be very much appreciated.

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.

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.

Guys, this was just meant to be an exercise. Yes, most likely any hand with a 5 would be folded pre-flop.. but for the sole task of doing the math, think of this more as a question on a test instead of how you would play this in real life.

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 A-level 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 52-5k 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.

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 I'll take this to a math forum and see what they come up with.

WVHillbilly said:

Depends. What was the action preflop? How may players saw the flop? What types of players saw the flop? What was the flop action before it was your turn?

In other words there is no way to answer your question as written.

Click to expand...

^^ This.

This is where table dynamics, and betting pattern reads come into play.

-

frodobaggins said:

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 I'll take this to a math forum and see what they come up with.

Click to expand...

How many were to the flop? Like I say it's a simple calc for 1 person but 8 people is hard since you have so many different scenarios. e.g. player 1 does not have a 5x so there is high chance of player 2 having 5x, then you'd have to work that out by working amount of hands with 16 combos, amount with 6 combos, amount with 4 combos and amount with 1 combos and weighting it accordinglyso it is reflected in the total combos, I think (not even sure you'd get a whole number, which is confusing. I might try and do this tomorrow and I'll ask my mate who did further maths what he thinks.

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.

JusSumguy said:

^^ This.

This is where table dynamics, and betting pattern reads come into play.

-

Click to expand...

Opponents have random hands, purely a maths question, read thread.

BlueNowhere said:

How many were to the flop? Like I say it's a simple calc for 1 person but 8 people is hard since you have so many different scenarios. e.g. player 1 does not have a 5x so there is high chance of player 2 having 5x, then you'd have to work that out by working amount of hands with 16 combos, amount with 6 combos, amount with 4 combos and amount with 1 combos and weighting it accordinglyso it is reflected in the total combos, I think (not even sure you'd get a whole number, which is confusing. I might try and do this tomorrow and I'll ask my mate who did further maths what he thinks.

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.

Click to expand...

2 of the 47 unknown cards are 5s. So ~4.26% chance for any card to be a 5. 16 cards dealt to 8 people. 16 * .0426 = 68.2% chance.

Oh yeah and raise preflop!

WVHillbilly said:

2 of the 47 unknown cards are 5s. So ~4.26% chance for any card to be a 5. 16 cards dealt to 8 people. 16 * .0426 = 68.2% chance.

Oh yeah and raise preflop!

Click to expand...

It's 8.6% for one person. Your method is effectively saying that the chances of getting aces preflop is 2/52 and not 6/1326. Also your method for 8 people isn't right, it isn't as simplistic as that.

BlueNowhere said:

It's 8.6% for one person. Your method is effectively saying that the chances of getting aces preflop is 2/52 and not 6/1326. Also your method for 8 people isn't right, it isn't as simplistic as that.

Click to expand...

Getting AA preflop can be done the same way EXCEPT that the first card matters. (4/52) * (3/51) = .45% same as 6/1326.

I guess it's actually 2/47 + 2/46 for a single opponent.

WVHillbilly said:

No it's really is. We don't care about anything other than the chances that at least one 5 has been dealt in 16 cards.

Also getting AA preflop can be done the same way EXCEPT that the first card matters. (4/52) * (3/51) = .45% same as 6/1326.

Click to expand...

It isn't. I showed the working earlier with combos.

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 100-91.6 = 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%

frodo i think you are going about this all backwards, it should be a conditional probability that you work with based on each actions street, ie given his flat preflop, what hands contain a 5, or given he c/r me what hand combinations do i lose to and beat.

WVHillbilly said:

I guess it's actually 2/47 + 2/46 for a single opponent.

Click to expand...

This gives me 8.6%. My 1st method was wrong because it didn't account for the greater chances of the cards being dealt after we'd dealt the 1st card.

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.

civilization is crumbling

WVHillbilly said:

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.

Click to expand...

This number seems way to high, how are you arriving at it?

Same way I arrived at 8.6% for a single opponent. 2/47 + 2/46 +.....2/32

WVHillbilly said:

Same way I arrived at 8.6% for a single opponent. 2/47 + 2/46 +.....2/32

Click to expand...

This can't be right. Lets say 11 people were dealt in. Using this method you would calculate that the chance of someone having a 5 in this scenario is 109% despite there still being 27 unknown cards.