Straight probabibilities

    • luitzen
      luitzen
      Bronze
      Joined: 03.04.2009 Posts: 664
      - What is the probability of hitting a straight when you have 3 connected cards on the flop?
      - And what is the chance of hitting the other side of a straight when you have a gutshot draw with three connected cards on the flop.

      I couldn't find an answer to these question anywhere.
  • 9 replies
    • martizzo
      martizzo
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      Joined: 10.06.2007 Posts: 957
      If you have OESD on 3connected board then you have 8 outs.
      So straight probability is 8: (52-5) -> 8:47 -> 1:47/8 -> 1:5.875 -> 1:~6

      If you have gutshot 1:11,75
    • luitzen
      luitzen
      Bronze
      Joined: 03.04.2009 Posts: 664
      I think you misunderstand me. In the first example you have to hit twice and in the second one you can hit either once or twice.

      Scenario 1 (example): you have 7, 8, 9, T:6 R:5 or T:6 R:T or T:T R:J (in all examples T and R can be swapped, but that doesn't matter, I suppose)
      Scenario 2 (example): you have 5, 7, 8, 9, T:6 R:x or T:T R:J (T and R can be swapped again, of course)
    • TheBu11d0g
      TheBu11d0g
      Bronze
      Joined: 25.07.2008 Posts: 2,019
      Hello luitzen,

      If you hit an OESD on the flop then you have odds of 2:1 with 2 cards to come and 5:1 with one card to come.

      Likewise for a Gutshot Draw you have odds of 5:1 with 2 cards to come and 11:1 with 1 card to come.

      The following articles may be of some help to you:

      Mathmatics of poker: Odds and Outs

      Mathmatics of Poker: Implied Pot Odds

      The Odds Chart

      Kind Regards,
      -Steve
    • nafar84
      nafar84
      Bronze
      Joined: 20.09.2008 Posts: 546
      Originally posted by luitzen
      Scenario 1 (example): you have 7, 8, 9, T:6 R:5 or T:6 R:T or T:T R:J (in all examples T and R can be swapped, but that doesn't matter, I suppose)
      Scenario 2 (example): you have 5, 7, 8, 9, T:6 R:x or T:T R:J (T and R can be swapped again, of course)
      I think you're being impossibly difficult :D

      What are your hole cards? Or are you talking about the board making a straight??
    • Schnitzelfisch
      Schnitzelfisch
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      Joined: 08.11.2008 Posts: 4,952
      lol you've all got it wrong. the probability of hitting a backdoor straight after you have 3 card OESD on flop is (8/47) * (8/46) = 2,9%. basic maths!

      oh and for the 2nd question, even though i didn't exactly understand it, but i guess you want to make a straight out of like 78T, right? well then it's (4/47) * (8/46) = 1,5% (first you hit the gutshot to make OESD and then complete a straight).

      Regards,

      Primzi
    • luitzen
      luitzen
      Bronze
      Joined: 03.04.2009 Posts: 664
      Why wouldn't it be (8/47)*(8/46) since there's one card left?
    • luitzen
      luitzen
      Bronze
      Joined: 03.04.2009 Posts: 664
      Regarding scenario 2: I haven't thought of the scenario you just created, but I was talking about a situation where you have 3 connected cards and a 4th card so that there would be one gap between the 3 connected cards. You can create the straight my filling the gutshot as well as by adding two cards at the other end.
    • Schnitzelfisch
      Schnitzelfisch
      Bronze
      Joined: 08.11.2008 Posts: 4,952
      oh ok... then it's just

      (4/47)+(4/46) - gutshot (17,2%)

      +

      (4/47)*(4/46) - two cards to complete straight. you have same chances with AKQ to hit JT. (0,7%)

      =

      17,9%

      your chances don't improve much :( .
    • luitzen
      luitzen
      Bronze
      Joined: 03.04.2009 Posts: 664
      These situations are interesting if you hold, for example J 8 and the flop shows 9 3 2 and you suspect your opponent has an over pair or AK and he raises.