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View Full Version : Super System/ Hold'em Poker F.A.P. odds discrepency.

Poker21
12-19-2003, 10:37 PM
I was looking over the Hold'em Probability charts in Doyle Brunson's SUPER SYSTEM and in Sklansky &amp; Malmuth's HOLD'EM POKER FOR ADVANCED PLAYERS and I have a question.

On page 575 of SUPER SYSTEM on the chart labeled ''Holdem- basic data'' it says: if you have four parts of an Open Ended Straight-Flush after the flop you will make at least a straight 54.12% of the time and the odds against you are 0.85 to 1.

In HOLD'EM POKER F.A.P.on page 309, 54.1% is also the percentage listed to complete the above stated 15 out straight-flush draw.

Now to my question. How do you get 54.1 percent? the formula for converting a percentage into ''odds to 1'' is to subtract the percentage 54.1% from 100 equaling in this case 45.9. Then dividing the result 45.9 by the percentage subtracted (54.1)which equals .85 to 1.

100 - 54.1 = 45.9

.8482 = 45.9/54.1 or .85 to 1

Now with the straight-flush draw you have 47 unseen cards and 15 outs which is a 32% shot on fourth street and a 30% shot(46 unseen cards and 14 outs) on the river to complete your hands. Would that be correct?

If it is then you would have 2.1 to 1 odds on fourth street and 2.3 to 1 odds on the river. How do you get 54.1% or .85 to 1 out of this? What am I doing wrong?

bigpooch
12-19-2003, 11:14 PM
It's best to consider the combinations of turn and river
cards that miss. For the turn to miss, the chances are
32/47 and after that, to miss on the river is 31/46 so that
the chances of missing is the product 32/47 x 31/46 =
496/1081.

Thus, the chances of not missing is simply 1 - 496/1081 =
585/1081 = 0.54117.

Forget about using the odds to one if that proves confusing.
Just consider the raw chances or probability.

mosch
12-20-2003, 07:37 PM