What are the odds?

A funny thing happened in a tourny I was in.

I am extremely short stacked, 254 chips, blinds are 50/100, I am UTG.

I get dealt Q9 :d: and I figure this might be my best chance to at least double up. Not likely going to steal the blinds, but maybe. So I push in.

Everyone folds except the SB and BB.

Flop comes rags, SB and BB check to each other.

Turn comes another rag, they check again.

River comes a rag again and I am looking at an early exit from the tourny.

The cards flip over..........

As you know, I have Q9 :d:

SB shows Q9 :s:

And BB shows Q9 :h:

What are the odds of that?

None of us hit a flush and the pot was chopped 275 each and I lived for another few hands. :D

Comments

  • I have asked Brian Alspach this question. He is smarter than I am. I will post the answer if he sorts it out. Anyone else?

    Lets assume that the question is: "What is the chance of three out of four Q-9s occuring in ten hands?"

    In other words, "Three out of four specific hands being dealt out in ten hands out of a possible 1326 hands?"
  • In other words, "Three out of four specific hands being dealt out in ten hands out of a possible 1326 hands?"

    Although it's hard to imagine someone having designed an a priori statistical experiment designed to search for three occurances of the exact hand Q9s, the probability of this happening is probably not too hard to work out.

    My initial intuition is that the probability of this occuring is astronomically small. But probably not as small as someone sitting down at a poker table and deciding that they monitor for an occurance of this event before the event has occured. :)

    Dave, you have already done the hard part. There are 1326 ways of dealing someone a holdem hand, disregarding the order in which the cards are dealt.

    Label the players 1 though 10. The probability that players 1, 2 and 3 are dealt a Q9s is approximately (4*3*2/1326^3). The probability that players 4 through 10 are dealt a non-Q9s hand is approximately (1322/1326)^7, or roughly 1. (It's actually ~0.997.) Note that I'm doing some very crude things here to make the calculations smoother, such as replacing 1326*1325*1324 by 1326^3, etc.

    So, the probability that the hands are dealt as indicated is

    (4*3*2/1326^3) * (1322/1326)^7

    which is about 1.0 * 10^(-8), or 0.00000001

    Now, the probability of any three players (not necesarrily exactly players 1, 2, and 3) being dealt the Q9s is just the probability we've already calculated, multiplied by (10 choose 3) = 120

    That works out to about 1.2 * 10^(-6), or 0.0000012. Rounding a bit, odds against the occurance are about 800,000 to 1. Once again, this is very rough due to the crude approximations above, but I'm sure we're really only interested in the order of magnitude here. For example, does this compare with someone winning 20 hands in a row in blackjack or someone winning 20 Lotto 6/49 tickets on a row.

    Note that it is only possible to compute the probability of being dealt the relevant cards. The probability of seeing three copies of this cheese at a showdown is going to be many orders of magnitude smaller. :cool:

    ScottyZ
  • Number of total non-ordered 10 handed deals = (52c20) * 19!! = 8.24923E+22

    Ways to deal four Q9 suited hands and 6 others = 1 * (44c12) * 11!! = 2.19238E+14

    ways to deal three Q9 suited hands and 7 others = (4c3) * (46c14) * 13!! = 1.29663E+17
    however this over counts the ways to deal four Q9 by a factor of 4 so subtracting
    = 1.29663E+17 - (4 * 2.19238E+14) = 1.28786E+17

    so the odds of three Q9 hands occuring are = 1.28786E+17 / 8.24923E+22 = 1.56119E-06 = ~1 in 640536
    So, the probability that the hands are dealt as indicated is

    (4*3*2/1326^3) * (1322/1326)^7

    which is about 1.0 * 10^(-8), or 0.00000001

    Now, the probability of any three players (not necesarrily exactly players 1, 2, and 3) being dealt the Q9s is just the probability we've already calculated, multiplied by (10 choose 3) = 120

    That works out to about 1.2 * 10^(-6), or 0.0000012. Rounding a bit, odds against the occurance are about 800,000 to 1.

    Scotty.... just a few changes gets you real close

    (4*3*2/1326^3) * (1322/1326)^7 * 120

    should be

    4*3*2/(1326 * 1225 * 1128) * (1034/1035)^7 * 120 = ~1 in 640523
  • I am a philosophy graduate. I am not sure that Q-9s actually exists. The question, therefore, is irrelevant.
  • Dear Dave,

    As a philosophy graduate, you should be sure of Q9s existance or not, and therefore the questions relevance. Since you're unsure, I must deduce that this is what kept you from moving on to graduate studies in this field, although I see you keep up your research into it's existance (or not) through regular poker activities. I wish you continued success in your endeavours.

    Yours Sincerely,
    MickeyHoldem B.A, M.B.A., Ph.D. (all in B.S.)

    for continued laughs check out Piled Higher and Deeper ... Link
  • Scotty.... just a few changes gets you real close

    I was away for the weekend when I realized I botched some major parts of the calculations. :(

    But even if some things were okay, I was making some serious round offs (rounds off?), etc.
    4*3*2/(1326 * 1225 * 1128) * (1034/1035)^7 * 120

    [smacks own forehead] Makes sense. Once you deal a hand, you don't get to start with a fresh deck.

    Thanks for cleaning up my mess. :)

    ScottyZ
  • I am a philosophy graduate. I am not sure that Q-9s actually exists. The question, therefore, is irrelevant.

    I'm not sure that philosophy graduates exist.

    ScottyZ
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