Chances of having strong hand

    • ragney
      ragney
      Bronze
      Joined: 02.08.2010 Posts: 2,417
      Does anyone have the math or calculation for this?

      In HU games, the chances to get a very strong hand like JJ+ / AK is much lower compared to SH games, where 12 out of the 52 cards are being dealt preflop. And even lower in FR games because 18 out of the 52 cards are being dealt preflop.

      So all in, how do we calculate the chances for someone in HU/SH/FR to have a hand like AA (6 combos)?
  • 8 replies
    • Hackett77
      Hackett77
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      Joined: 02.02.2009 Posts: 372
      chances of being dealt AA are 1 in 220 hands, HU FR or 6max
    • Schnitzelfisch
      Schnitzelfisch
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      Joined: 08.11.2008 Posts: 4,952
      Hey there,

      I think you're talking about the probability that your opponent has a strong hand, not you yourself (as it's always the same probability ;) ).

      I guess you can easily calculate the probability of someone else having some hand (let's say AA, and let's say that you don't have any aces in your hand) by multiplying the probability of the hand (in our case AA) with the number of opponents.

      P(AA) = number of possible AA combinations / number of all combinations

      P(AA) = (4 choose 2) / (52 choose 2) = 0.0045

      P(someone else has AA) = P(AA) * number of opponents

      P(opponent has AA in HU) = 0.0045 * 1 = 0.45%
      P(opponent has AA in SH) = 0.0045 * 5 = 2.25%
      P(opponent has AA in FR) = 0.0045 * 8 = 3.6%

      I hope that I didn't make some stupid mistake, but that should be it :) .

      Does that make sense?

      -SF
    • Schnitzelfisch
      Schnitzelfisch
      Bronze
      Joined: 08.11.2008 Posts: 4,952
      Originally posted by Hackett77
      chances of being dealt AA are 1 in 221 hands, HU FR or 6max
      :)
    • Hackett77
      Hackett77
      Bronze
      Joined: 02.02.2009 Posts: 372
      Originally posted by Schnitzelfisch
      Originally posted by Hackett77
      chances of being dealt AA are 1 in 221 hands, HU FR or 6max
      :)
      pretty close tho on the end of a 9 hour drinking sesh:)
    • ragney
      ragney
      Bronze
      Joined: 02.08.2010 Posts: 2,417
      Originally posted by Schnitzelfisch
      Hey there,

      I think you're talking about the probability that your opponent has a strong hand, not you yourself (as it's always the same probability ;) ).

      I guess you can easily calculate the probability of someone else having some hand (let's say AA, and let's say that you don't have any aces in your hand) by multiplying the probability of the hand (in our case AA) with the number of opponents.

      P(AA) = number of possible AA combinations / number of all combinations

      P(AA) = 4 choose 2 / 52 choose 2 = 0.0045

      P(someone else has AA) = P(AA) * number of opponents

      P(opponent has AA in HU) = 0.0045 * 1 = 0.45%
      P(opponent has AA in SH) = 0.0045 * 5 = 2.25%
      P(opponent has AA in FR) = 0.0045 * 8 = 3.6%

      I hope that I didn't make some stupid mistake, but that should be it :) .

      Does that make sense?

      -SF
      Yeah this is what I wanted, because I'm not used to play FR at all and I noticed people having good preflop hand waaaaaaaaaay more often than HU/SH (logic but want a calculation).

      So the chance that an opponent has AA, KK and QQ is 3.6% * 3 = 10,8%?
      In SH 6,75%?

      How about adding AK, which has way more combos than AA/KK/QQ?
    • Schnitzelfisch
      Schnitzelfisch
      Bronze
      Joined: 08.11.2008 Posts: 4,952
      Yeah, that seems about right.

      Well, there are 4*4 = 16 combinations of AK (4 aces, each with 4 different kings), and to calculate the probability of AK, we simply use the following formula:

      P(AK) = 16/(52 choose 2) = 0.012

      P(someone else has AK) = P(AK) * number of opponents

      so for FR, this is 0.096 = 9.6%

      to get the probability that someone has a range of {AK,QQ+} while you don't have any A, K or Q, you simply calculate:

      P(AA) + P(KK) + P(QQ) + P(AK) =
      3.6% + 3.6% + 3.6% + 9.6% = 20.4%

      Of course it gets a bit more complicated if you for example hold an A, which blocks some aces from AA and AK... In that case there are 3 combinations of AA left (3 choose 2) and 12 combinations of AK (3 aces times 4 kings).

      In that case, the probabilities for AK and AA are the following:

      P(AA, 1A blocked) = 3/(52 choose 2) = 0.0022 (half less than if you didn't have an A)
      P(AK, 1A blocked) = 12/(52 choose 2) = 0.0090

      so the probability of any of the opponents having something from the range of {AK, QQ+} while you hold one A would be:

      P = 1.8% + 3.6% + 3.6% + 7.2% = 16.2%

      You should be able to calculate similar stuff by yourself now, although it might get a bit more complicated if you start wondering about what the probabilities are that two people have something from that range at a time, etc.. :)

      -SF
    • ragney
      ragney
      Bronze
      Joined: 02.08.2010 Posts: 2,417
      Thanks, that explains those damn FR tournies always 3betting me (I play loose) and they always have the nuts lol :D
    • IngridN
      IngridN
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      Joined: 02.03.2011 Posts: 12,162
      Hi guys,

      I've moved this to the relevant board.

      Ingrid