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Puzzle 1: Occurance rate of sets

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Fnord

07-06-2004, 01:07 PM Post subject: Puzzle 1: Occurance rate of sets

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Fnord is an unknown quantity at this point

If no one ever folds a pocket pair pre-flop, what is the expected occurance rate of a set on the flop for 2-10 player tables? What is your expected rate of flopping a set if you never fold a pocket pair pre-flop?

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http://www.youtube.com/watch?v=w-Xz8aLuD7Y

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koolmoe

07-06-2004, 04:37 PM

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Here's my best shot at it:

An individual player can expect a pocket pair every 1 in 17 hands and upon playing that PP will flop a set or better 11.75% of the time. If you play every pocket pair, you'll flop a set 0.69% (11.75%/17) of the time (once every 145 hands or so).

For two players, the probability of flopping a set is the sum of their individual probabilities less the probability of both flopping a set on the same hand:

P[A U B] = P[A]+ P[B] - P[AB]

Unfortunately, the two events aren’t independent, so there are a few hoops to go through to calculate AB. First, assume that player A has a pocket pair (1/17). There are 72 ways the second opponent can be dealt a pocket pair that can yield a set (combin[4,2]*12) out of 1225 hands (combin[50,2]). There are 184 (2*2*46) ways to flop each player a set or better out of 17296 (combin[48,3]) possible flops. This gives P[AB] = P[AB|A]P[A] = (72/1225)*(184/17296)*(1/17) = 3.678x10^(-5). Thus,

P[A U B] = 0.006915 + 0.006915 – 0.00003678 = 0.013793

This gets a little more difficult as the number of players increases. For example, for three players:

P[A U B U C] = P[A] + P[B] + P[C] – P[AB] – P[AC] – P[BC] + P[ABC]

We already computed P[AB] (= P[AC] = P[BC]), but to get P[ABC], we need to assume that all three players have PP’s of different rank (probability is (1/17)*(72/1225)*(66/1128)) and that one of the 6 (2*2*2) three set flops occurs out of 17296 possible flops. Thus P[ABC] = (1/17)*(72/1225)*(66/1128)*(6/17296) = 7.02x10^(-8). So,

P[A U B U C] = 3*(0.006915) – 3*(0.00003678) + 0.0000000702 = 0.020634.

For more than three players, you will encounter terms for the probability of four (or more) players flopping a set, which is not possible. As a result, the formula reduces to:

P[at least one of N players hitting a set on the flop] = combin(N,1)*(0.006915) – combin(N,2)*(0.00003678) + combin(N,3)*(0.0000000702).

Here is a table summarizing the results:

Players...................Probability of a set
2..............................0.013793
3..............................0.020634
4..............................0.027373
5..............................0.034207
6..............................0.040938
7..............................0.047633
8..............................0.054292
9..............................0.060915
10............................0.067501

Fnord

07-06-2004, 06:00 PM

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...well spank my ass and call me Sally.

The only nitpick I see here is that probability of each player receiving a pocket pair is not an independent event (but pretty close.) However, the difference is probably not significant for the numbers we're looking for.

http://www.youtube.com/watch?v=EBvge-c2_Xg
http://www.youtube.com/watch?v=w-Xz8aLuD7Y

koolmoe

07-06-2004, 06:31 PM

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Quote:

Originally Posted by Fnord

That's accounted for in the computation of P[AB] where the possible ways for the second player to be dealt a PP is 72/1225. That's 6 possible ways to deal each of the 12 other PP's. There is also one way to deal the same PP, but in that case neither player can make a set. If I had assumed independence, I would have simply set P[AB] = P[A]P[B] (and save quite a bit of work). In that case, the results are slightly lower.

Incidentally, the P[A U B] = P[A] + P[B] - P[AB] is the inclusion/exclusion principle (http://mathworld.wolfram.com/Inclusi...Principle.html).

Warpe

07-22-2008, 01:28 AM

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No poker in the commune, pls.

a500lbgorilla

07-22-2008, 02:33 AM

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JESUS TAKE THE KEYBOARD

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yer so old [thread, warpe]

Smithers, use the amnesia ray.
You mean the revolver, sir?
Precisely.

Renton

07-22-2008, 02:47 AM

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set value is so 2004

Lukie

07-22-2008, 04:23 AM

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Most people hit sets with a frequency of approximately 1 in 8 or 1 in 9 times. That seems well enough, however, when I tweaked my game a bit and started hitting them about 1 in 4 or 1 in 5 times, my winrate sky-rocketed.

bigspenda73

07-22-2008, 01:55 PM

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bigspenda73 will become famous soon enough

I remember when koolmoe used to post cool shit like this

DrivingDog

07-22-2008, 04:41 PM

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Koolmoe's analysis is close enough for practical purposes, but a more precise analysis for numbers of players > 1 needs to take into account the the possibility of more than one player having a PP as well as the (remote) possibility of two players sharing the same PP.

E.g., with two people, there are three possibilities to consider preflop:

a) player 1 and player 2 both get the same PP (e.g., both get AA)
b) player 1 and player 2 get different PP (e.g., AA v. KK)
c) player 1 gets unmatched cards (UMC), player 2 gets a PP, OR player 1 gets a PP, player 2 gets UMC

a) will occur 1/17 * (2/50 * 1/49) = 0.000048
b) will occur 1/17 * (3/50 * 11+1/49) = 0.003348
c) will occur 1/17 + (3/50 * 11+1/49) - a - b =0.109098

Sets or better (SOB*) can occur in conditions b) and c) above, but not a).

For b) a SOB will occur 4/48 + 4/47 * 44/48 +4/46 * 44/48 * 43/47 * b = 0.0007732

For c) a SOB will occur 2/48 * 46/47 *45/46 * c = 0.013057

Adding these two values gives the probability of SOB for 2 plyrs: 0.0007732 + 0.013057 = 0.01383.

But really not much different from his value of 0.01379

For 3 players, there's even more complications, as now up to three players can be dealt a PP, and two of those three can be dealt a PP of the same rank.

For 4 players and beyond, the possibility exists of two different pairs of equal PPs preflop being dealt (e.g., AA, AA, KK, KK), though this will be uncommon with a probability of 0.000000000030, or one time in about every 338 million four handed deals.

For 6 players and beyond, the possibility exists of three different pairs of equal PPs preflop, though this would be an exception, occurring with a probability of 0.00000000000023996, or one time for every 4 billion or so six handed deals.

For 8 players and beyond, the possibility exists of four different pairs of equal PPs preflop, although this would not be likely, about once every 38 quadrillion eight-handed deals.

For 10 players, it's possible that five different pairs of equal PPs could be dealt preflop, but do not expect to see this anytime soon. The probability of this happening is 0.00000000000000000000401 or about one every 250 quintillion ten-handed deals. However, the chance of ten different PPs being dealt out at the same table (Fnord's dream hand, in which case the probability of at least one player flopping a set or better is .9565) is much greater, about once every 4.5 billion hands.

* This assumes the flop includes at least one card matching his PP, and ignores the possibility of 3 matching cards coming on the flop and giving the player a FH - e.g., a player holds 88 and the flop comes 555.

"You can fool some of the people all of the time, and those are the ones you want to concentrate on." (George Bush).

tetleyboy

07-24-2008, 12:46 PM

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the simple answer is 'not enough'

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Puzzle 1: Occurance rate of sets

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