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westside_eh
04-28-2005, 03:50 AM
First post. Found out about the forum from a friend, some pretty good information here.

Anyway, I know the trick to calculating odds where after the flop you multiply your outs by 4, and on the turn you multiply them by 2.

Are there any tricks to calulating the odds after the flop when 2 cards are needed to complete the flush/straight?

olavfo
04-28-2005, 08:24 AM
Welcome to the forums.

Before the turn there 47 unknow cards and 10 spades. If a spade comes on the turn, there are now 46 unknown cards and 9 spades.

P = (10/47)*(9/46) = 90/2162 = 0.0416 = 4.16%

In odds form:

(1/P -1) : 1 = (1/0.0416 - 1) : 1 = 23 : 1

So the math is simple. Calculate the chance of hitting on the turn and multiply it with the chance of hitting again on the river. (Remember to adjust the number of outs and unknown cards on the river. You have one less out, since the turn card hit, and you also have one less unknown card.)

olavfo

binions
04-28-2005, 12:01 PM
[ QUOTE ]

Before the turn there 47 unknow cards and 10 spades. If a spade comes on the turn, there are now 46 unknown cards and 9 spades.

P = (10/47)*(9/46) = 90/2162 = 0.0416 = 4.16%

In odds form:

(1/P -1) : 1 = (1/0.0416 - 1) : 1 = 23 : 1

[/ QUOTE ]

Correct - backdoor flush draws are always 23:1. Backdoor straight draws are 21.5:1 when you start with 3 in a row open-ended.

Bottom line, each of these should count as 1 to 1.5 outs on the flop. After all, if on the flop, you needed 1 specific card to win (Ace of Spades let's say), you are 23:1 to hit the Ace of Spades by the river. So, backdoor draws are like 1 to 1.5 outs on the flop.

If you miss the turn, the backdoor out disappears. If you catch a suit you need on the turn, now you have 9 outs to the flush. If you catch a straight card you need, you either have 4 or 8 outs to the straight.