Combinatorics/Statistics/Basic poker knowledge (Long post)

Poker background knowledge:

There are 52 cards in a deck

9 players at a table

Question:
If a player is dealt KQ what is the probability of any of the other 8 players being dealt AK or AQ?

I have two methods of solving this both give different answers, a combinatorics approach and a statistics approach, just wondering what everyones thoughts are, and why which one is correct and which one is wrong.

Combinatorics method:

1- (\frac{\displaystyle \binom{50}{2}-4*3*2}{\displaystyle \binom{50}{2}})^x.

x = number of players left to act

50C2 = total # of preflop combos, given we were already dealt a hand

432 = combos of AK/AQ, given we were dealt KQ

So say for example there's 8 players left to act its to the power 8

OR the next person folds, 7 players left to act it's to the power 7...etc

Statistics approach

probability of AK or AQ is the same

(4/50)x(3/49)+(3/50)*(4/49) The probability or getting dealt an ace, then a king(or queen) + the probability of getting dealt a king(or queen), then an ace.

I multiply this by two because the probability or AK or AQ is the same.

Multiply by X where X is the amount of players remaining, i.e

(\frac{4}{50}*\frac{3}{39}+\frac{3}{50}*\frac{4}{49})*2x

Reply 1

DFranklin

18

Without spending much time on this, neither of your answers look right. In both cases you seem to be assuming independence where it definitely doesn't hold.

In method 1, knowing that players 2-to-X didn't have AK or KQ is going to affect the probability that player X+1 has AK or KQ.

In method 2, you can't simply multiply by the number of players, because the events aren't disjoint. You will be counting the case where player 2 and player 3 both get AK twice, when you should only count it once. (c.f. "toss 2 coins, what's the probability of getting a head? Answer is NOT (1/2) * 2").

In practice I think this is very hard and probably only reasonably solvable computationally.

Reply 2

J_W-x

OP

13

Original post by DFranklin

Without spending much time on this, neither of your answers look right. In both cases you seem to be assuming independence where it definitely doesn't hold.

In method 1, knowing that players 2-to-X didn't have AK or KQ is going to affect the probability that player X+1 has AK or KQ.

In method 2, you can't simply multiply by the number of players, because the events aren't disjoint. You will be counting the case where player 2 and player 3 both get AK twice, when you should only count it once. (c.f. "toss 2 coins, what's the probability of getting a head? Answer is NOT (1/2) * 2" :wink:

.

In practice I think this is very hard and probably only reasonably solvable computationally.

I appreciate the fast response!

That's an interesting perspective on method 1 that is definitely correct, but that calculation is used through-out the poker community.

Will report back to the gambling degenerates with this new news :wink:

thank you!

Reply 3

J_W-x

OP

13

Original post by DFranklin

Without spending much time on this, neither of your answers look right. In both cases you seem to be assuming independence where it definitely doesn't hold.

In method 1, knowing that players 2-to-X didn't have AK or KQ is going to affect the probability that player X+1 has AK or KQ.

In method 2, you can't simply multiply by the number of players, because the events aren't disjoint. You will be counting the case where player 2 and player 3 both get AK twice, when you should only count it once. (c.f. "toss 2 coins, what's the probability of getting a head? Answer is NOT (1/2) * 2" :wink:

.

In practice I think this is very hard and probably only reasonably solvable computationally.

I didn't feel like writing out latex again.
http://imgur.com/a/IMATe
how do you feel about this, where n=1 is the first player after me to act, n=2 is the second etc

edit:
http://imgur.com/a/HtojZ

(edited 7 years ago)

Reply 4

DFranklin

18

Original post by J_W-x

I didn't feel like writing out latex again.
http://imgur.com/a/IMATe
how do you feel about this, where n=1 is the first player after me to act, n=2 is the second etc

edit:
http://imgur.com/a/HtojZ

They both look unlikely to be correct, but without some kind of explanation of why they are supposed to be correct it's purely gut feeling.

Couple of general comments:

What makes this question hard is that you don't just need to worry about "did player X get AK/KQ/etc. or not?", but also "did player X get a K / Q / etc. which is going to reduce the chances of a downstream player being able to get AK/KQ/etc?". I'm not seeing an easy way of getting round this, which is why *any* formula that doesn't look faintly horrific feels unlikely to be correct without a convincing argument to go with it.

I honestly don't think this is an appropriate topic for the forum; at least not unless you're able to sensibly discuss the calculations. I'm pretty certain the only appropriate approach here is computational; either simple Monte-Carlo type simulation, or by running a markov chain-like machine which keeps track of the key probabilities (e.g. you'd want to keep track of the probability distribution for number of K still in the pack, number of A, etc.). The latter approach is considerably more work but should be able to give exact figures, even if not an exact formula.

I'm capable of following such discussion, but to be honest, I'm not that interested (*). Unless there are other people with both capability and interest, I think you're kind of wasting your time here.

(*) An old friend of mine who is pretty successful mathematically (has a PhD, for example), seems to spend a lot of time calculating various poker odds like this. To be honest, my feeling is "what a total waste of time and talent". YMMV

Combinatorics/Statistics/Basic poker knowledge (Long post)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you