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set variance?

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  1. CrabClaws CrabClaws is offline

    Guilford, CT
    United States

    Posts: 421
    Joined: Apr 06

    Dec 13th, 2006 (12/13/2006 5:29am)

    What is the normal variance for flopping sets? Standard deviation? Understanding that there are 13 pairs out of 221 hands and that each pair will flop its 3rd 1 in 8 times... Idk, someone run with the math.
  2. Acoplander1 Acoplander1 is offline

    Cambridge, MA
    United States

    Posts: 47
    Joined: Sep 05

    Dec 13th, 2006 (12/13/2006 5:37am) in reply to CrabClaws

    I'll do a relevant statistics problem for you...

    --

    You expect to flop a set 1 in 8.5 times with a pocket pair.

    Let's normalize this to 2 in 17 times.

    --

    Suppose you receive 17 pocket pairs. This is a simple binomial distribution problem.

    Expected Number Of Sets = 2

    Variance = (17)*(.118)*(.882) = 1.77

    Standard Deviation = 1.33

    --

    So... Every 17 pocket pairs you receive, you can be

    ~68% sure of flopping between 0.67 to 3.33 sets

    ~95% sure of flopping between 0 to 4.66 sets

    ~99.7% sure of flopping between 0 and 6 sets.

    --

    -Acoplander1
  3. CrabClaws CrabClaws is offline

    Guilford, CT
    United States

    Posts: 421
    Joined: Apr 06

    Dec 13th, 2006 (12/13/2006 6:04am) in reply to Acoplander1

    Thats a good post, but it only partially answers my question. I've probably gone 40+ pairs tonight without a set. Given that I should excpet an average of one ever 8, I'm curious as to how likely it is that I get ZERO sets.
    Thread Starter
  4. Staple Gun Staple Gun is offline

    Lake Orion, MI
    United States

    Posts: 3,166
    Joined: Aug 05

    Dec 13th, 2006 (12/13/2006 6:23am) in reply to Acoplander1

    I flopped 0.67 sets last night. Lost to 1.33 two-pairs though.
  5. Acoplander1 Acoplander1 is offline

    Cambridge, MA
    United States

    Posts: 47
    Joined: Sep 05

    Dec 13th, 2006 (12/13/2006 6:36am) in reply to CrabClaws

    OK, just plug the relevant numbers in the binomial formula.

    Let n = total number of pocket pairs

    k = number of sets you flopped (in your case, 0)

    p = probability of flopping a set = .118

    q = 1-.118 = .882

    --

    P = (n C k) * [(.118)^k] * [(.882)^(n-k)], where (n C k) is a combination.

    Your case is easy. 40 pairs without flopping a set is (.882)^40 = ~0.6% (1 out of 154 times this will happen)

    X pairs without flopping a set is (.882)^X.

    --

    Another (nonzero) example...

    You get 40 pairs and flop only 1 set. Just use the binomial formula.

    (40 C 1) = 40! / [1! * (39!)]

    40 * .118^1 * .882^39 = 3.5% chance you flop only 1 set out of 40 pocket pairs.

    --

    -Acoplander
  6. Acoplander1 Acoplander1 is offline

    Cambridge, MA
    United States

    Posts: 47
    Joined: Sep 05

    Dec 13th, 2006 (12/13/2006 6:43am) in reply to Staple Gun

    Yeah. I was at Greektown last night...

    I had 99 in the cutoff and raise it up to $15 after a couple limpers.

    Dealer fvcks up the flop twice. Floor is called.

    1st and 2nd flop had a set of 9's, but he used the 3rd flop, which was garbage... so I guess that qualifies as flopping 0.67 sets? I almost had it. :)

    -Acoplander
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