Odds Question

[22pajo]

Anyone know the odds of flopping a straight if you hold 2 connected cards preflop in Hold'em?

[arahant]

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Anyone know the odds of flopping a straight if you hold 2 connected cards preflop in Hold'em?

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1.306% (+10.449% of flopping open ender)
Edited for niceness [img]/images/graemlins/smile.gif[/img]

[22pajo]

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[ QUOTE ]
Anyone know the odds of flopping a straight if you hold 2 connected cards preflop in Hold'em?

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1-(odds you post this in the right forum)
50%
49
try GOOGLE, lazy ass...
etc...

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ok wrong forum, i didn't see "probability"
and you think its lazy to post a question on these boards?
using google would have been less work. i post here because i get most of my questions answered by people who show more respect than you.

mods feel free to move this to probability

[samsonite2100]

100% if you're playing me and I have aces.

[jay_shark]

You have a 0 gap connector like 4-5 .

The straights are a-5 , 2-6, 3-7 , 4-8 .

(4*4*4)*4/50c3 =256/19600 =0.01306 or 1.306% .

[arahant]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Anyone know the odds of flopping a straight if you hold 2 connected cards preflop in Hold'em?

[/ QUOTE ]

1-(odds you post this in the right forum)
50%
49
try GOOGLE, lazy ass...
etc...

[/ QUOTE ]

ok wrong forum, i didn't see "probability"
and you think its lazy to post a question on these boards?
using google would have been less work. i post here because i get most of my questions answered by people who show more respect than you.

mods feel free to move this to probability

[/ QUOTE ]

You're right man...sorry. I was feeling snippish.
mea culpa.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Anyone know the odds of flopping a straight if you hold 2 connected cards preflop in Hold'em?

[/ QUOTE ]

1.306% (+10.449% of flopping open ender)
Edited for niceness [img]/images/graemlins/smile.gif[/img]

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10.449% is not the probability of an open ender. You got this from 3*4*4*48/C(50,3) minus the 1.306% for the straights. The 3*4*4*48 double counts half of the straights, and it double counts all of the draws with paired flops. For example, if you hold JT, the flop 9QK will be counted once with Q9 as 4*4 and the K as one of the 48, and then again with KQ as one of the 4*4 and the 9 as one of the 48. For offsuit connectors this should be

3*(4*4*34 + 2*6*4)/C(50,3) =~ 9.061%.

The 34 is the number of 3rd cards excluding 8 cards that complete the straight and 6 cards that pair the board (48-8-6 = 34). The second term in parentheses counts the paired flops, where either one of the 2 cards can pair, times 6 ways to make the pair, times 4 ways to choose the unpaired card. Note that if we add to this number half of the straights (0.5*1.306%) plus the paired flops (3*2*6*4/C(50,3)), we get your number 10.449%.

For suited connectors we must also exclude the flushes, and if we then also include the 8-out double belly-buster draws, this comes to 9.6%. For a detailed explanation of this calculation, see this post.

[22pajo]

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You're right man...sorry. I was feeling snippish.
mea culpa.

[/ QUOTE ]
Apology accepted. Thanks to bruce and the others for their explanations.