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  1. skatter skatter is offline

    Posts: 56
    Joined: Mar 05

    Mar 7th, 2005 (3/7/2005 7:22pm)

    Hello, I have been playing online for a little over a year now, and I feel that usually I am pretty good with odds when taking one additional card into account, but taking two cards throws me off. For example:

    You hold JsTs, and the flop is 4d-8s-9h, the odds of hitting a straight on the turn is easy, you have 8 outs and there are 47 unseen cards (5.875:1 dog), but what are the odds of hitting one of those 8 outs by the river?

    47:8 (5.875:1)
    46:8 (5.75:1)

    How do you combine the two?

    Also, as a side question, what are the odds of flopping a straight flush, not turning/rivering, flopping.

    You hold As2s, flop must be 3s-4s-5s.. to start out I'd do something like this..

    50:3 (16.7:1)
    49:2 (24.5:1)
    48:1 (48:1)

    But I'm not quite sure how to combine themall, do you multiply them all together (16.7*24.5*48=19639.2)?

    Thanks,
  2. Cal Cal is offline Co-Founder of PocketFives

    Posts: 3,658
    Joined: Dec 04

    Mar 7th, 2005 (3/7/2005 8:11pm) in reply to skatter

    Hey skatter, welcome to the site.

    In your first example, you simply add the two fractions. You are looking for one of those eight outs in two different events, either the turn or the river. Since the denominators are different, think of them in terms of percentages. You have a 17% chance on the turn and a 17.3% chance on the river. Add those together and your chances of hitting the straight are 34.3%

    The second example is more complicated because three events have to happen in succession for you to hit your straight flush. You are right in this case - you multiply the odds of each event happening to find the true odds of flopping the straight flush. You will flop the straight flush one in every 19369 times you are dealt A2 suited.
     
  3. BettinBenny BettinBenny is offline

    Hinton, IA
    United States

    Posts: 169
    Joined: Jan 05

    Mar 7th, 2005 (3/7/2005 8:31pm) in reply to Cal

    ...and you will flop a straight flush one in every 19369 times you are dealt AK suited.

    I thought it might be worth going through the probability of being dealt a straight flush with middle-suited connectors, a hand like JT.

    The math here will allow us to use both principles of statistics that you have come accross.

    1) Add any percentages where you use the statement "or" (i.e., either x must happen OR y must happen).

    2) Multiply any percentages where you use the statement "and" (i.e., both x AND y must happen).

    In order to flop a straight flush when you have JT, the board must come either ((7 AND 8 AND 9) OR (8 AND 9 AND Q) OR (9 AND Q AND K) OR (Q AND K AND A)).

    Plugging in numbers gives us our desired answer:
    (3/50 * 2/49 * 1/48) * 4 = (6/117600) * 4 = 24/117600 = 1:4900.

    So one in every 4,900 times you are dealt a specific suited connector, where there are 4 possible ways to flop a straight flush, you can expect to do so.

    -Bettin Benny

    *Note that for our AND clauses we use 3/50, 2/49, 1/48 and NOT 1/50, 1/49, 1/48. This is becuase the three cards we are talking about can be flopped in ANY order.
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