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  #1  
Old 10-20-2007, 08:25 PM
rakewell rakewell is offline
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Default Doyle Brunson is wrong about straights and free cards

In Super/System, Doyle Brunson makes this observation:

"[W]hen an ace comes on the flop...unless the board is paired, the next card can always make a straight."

This struck me as a peculiar point on which to make a blanket rule like he does. I thought about it for a while, played around with a deck, and did a little math to explore whether it really made sense.

The short version is this: With an ace on the flop, it is 100% certain that an opponent holding just the right hole cards could make a straight with just the right turn card. But without an ace on the flop, it is 98.2% certain! In other words, I can't see any reason that Doyle makes a distinction here. It is virtually always the case that at least one possible turn card could make an opponent's straight, whether or not there is an ace on the flop.

For my full exploration of Doyle's advice on this point, see:

http://pokergrump.blogspot.com/2007/...-is-wrong.html

Comments are enabled on the blog, and also welcome here.
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  #2  
Old 10-20-2007, 08:45 PM
jay_shark jay_shark is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

It's more probable that if the board contains an ace , then your opponent can hit a straight by the turn . This should be obvious for the simple reason that we could have an ace-high straight or an ace-low straight . No other card has this flexibility that the ace has .

So if the flop comes k-7-2 , then there is absolutely no way for your opponent to hit a straight by the turn .

If the flop comes A-7-2 , then it's possible for your opponent to hit a straight if he were holding 3-4 , 3-5 or 4-5 .
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  #3  
Old 10-20-2007, 09:01 PM
jay_shark jay_shark is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

Mathematically , it works out like this .

Suppose the flop contains an ace . We know that the other 2 cards cannot be {2-5} or {10-k) since it's always possible for your opponent to make a straight by the turn . This means that the other 2 cards must belong to {6-9} . But this means that your opponent may still be able to hit a straight by the turn .

So with 100% certainty , there is a greater than 0% chance that your opponent may hit a straight if the flop contains an ace .
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  #4  
Old 10-20-2007, 09:30 PM
Flip-Flop Flip-Flop is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

Don`t put words like Doyle Brunson and "wrong" in same sentence.
It makes you look pretty foolish.
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  #5  
Old 10-20-2007, 09:33 PM
rakewell rakewell is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

If I'm wrong, please feel free to show me exactly how, why, and where. But if you're deciding that I'm wrong just because of who I'm disagreeing with, well, don't you think that's a tad shallow on analysis?
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  #6  
Old 10-21-2007, 07:27 AM
Vetgirig Vetgirig is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

[ QUOTE ]
If I'm wrong, please feel free to show me exactly how, why, and where. But if you're deciding that I'm wrong just because of who I'm disagreeing with, well, don't you think that's a tad shallow on analysis?

[/ QUOTE ]

Doyle uses the fact that its a 100% chance that when an A flops that the opponent can get a straight on the turn to emphasise that its bad to give a free card. Hes entirely corrent in that assessment as you yourself point out. The reason for that is that he wants to make sure that the reader thinks about the flop to see if any opponent can make a straight on the turn if the perfect card drops so the reader dont give too many free cards.

So hes not wrong.
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  #7  
Old 10-21-2007, 08:40 AM
btmagnetw btmagnetw is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

[ QUOTE ]
If I'm wrong, please feel free to show me exactly how, why, and where. But if you're deciding that I'm wrong just because of who I'm disagreeing with, well, don't you think that's a tad shallow on analysis?

[/ QUOTE ]
title says brunson is WRONG
then your post asks why he makes a distinction

durrr?
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  #8  
Old 10-21-2007, 09:23 AM
jogsxyz jogsxyz is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

Brunson wrote that book like 30 years ago. At the time all previous poker books was total junk. Now even Doyle admits some of his advice is dated and doesn't work in today's game.
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  #9  
Old 10-21-2007, 11:06 AM
AaronBrown AaronBrown is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

I agree that the Ace observation is true, but not very useful. It's misleading in the sense you mention, a better statement is a straight possibility is 100% certain with an unpaired flop unless the flop is 2, 7, Q; 2, 7, K; 2, 8, K; or 3, 8, K.

Sure an A guarantees a straight possibility, but so does 4, 5, 6, 9, T or J. And even if flop consists only of 2, 3, 7, 8, Q and K, there's a 75% chance of a straight possiblity.

It's not a bad pedagogical point for getting people to think about how many straight possibilities there are, and how dangerous free cards can be. But it's a bad point in that it can misleading about the situation without an Ace.

As a practical matter, you don't worry too much about straights that require someone to hold 63 and hit an inside card on the turn. You worry a lot more if the flop contains JT because KQ is a hand that will likely see the flop, and an open-ended high straight is a likely draw.
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  #10  
Old 10-22-2007, 05:40 AM
br.bm br.bm is offline
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Default Re: Doyle Brunson is wrong about straights and free cards

[ QUOTE ]
Don`t put words like Doyle Brunson and "wrong" in same sentence.
It makes you look pretty foolish.

[/ QUOTE ]
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