"So, I am a 9 to 5 dog"

[jamartin]

"I am a 9 to 5 dog"

From Doyle Brunson's Super System book, he is an extract and something I just don't "comprehend" at this stage in my poker career.

"Let's say I raised before the flop with a type of hand that's one of my favourites: small connecting cards that are suited. I'm in the pot with one player who called behind me. At this point, I put him on on a couple of big cards or a medium pair. That's all right. It's what I want him to have. Now here's what will happen if the following turn comes up:

Doyle: 6 [img]/images/graemlins/heart.gif[/img] 7 [img]/images/graemlins/heart.gif[/img] - Opponent: A [img]/images/graemlins/club.gif[/img] K [img]/images/graemlins/club.gif[/img]

Flop: A [img]/images/graemlins/diamond.gif[/img] 5 [img]/images/graemlins/heart.gif[/img] 4 [img]/images/graemlins/diamond.gif[/img]

"With that Flop, I am going to lead right off and bet. If plays-back at me, I'm now going to be quite sure he's got two aces (or better). So, I'm about a 9 to 5 dog."

<font color="blue">Herein lies the problem - my naive poker mind doesn't comprehend how he works out that he is a "9 to 5 dog".

Can someone please explain this to me and if they have the time, post other working examples please.

Many thanks in advance.

Kind regards,

jamartin</font>

[ncray]

Plugging those cards into pokerstove gives 6h7h 36.869% equity. 5/14 = .3571

[jamartin]

So what, Doyle is wrong?

I am looking for a manual way (I don't have a pokerstove ;-)) to come up with this type of figure when I am playing live or online.

Many thanks.

[WhiteWolf]

[ QUOTE ]
So what, Doyle is wrong?

I am looking for a manual way (I don't have a pokerstove ;-)) to come up with this type of figure when I am playing live or online.

Many thanks.

[/ QUOTE ]
Doyle is correct. '9 to 5 dog' = Doyle wins 5 hands for every 9 he loses = Doyle wins 5/14 (or about 36%) of the time.

Doyle has 8 outs for the straight here, and a back door draw for the flush which is worth about another out with two cards to come. A simple way to estimate your chances of winning with two cards to come is to multiply your number of outs by 4% (use 2% if there is only one card to come). So here, 9 outs = 36% chance of winning. Doyle is an old-school gambler, so he expresses that value in odds instead of percentages (and undoubtedly has these odds committed to memory). You should use whichever method you are most comfortable with.

[rufus]

[ QUOTE ]

I am looking for a manual way (I don't have a pokerstove ;-)) to come up with this type of figure when I am playing live or online.


[/ QUOTE ]

An approximation technique (This may seem a bit complicated but it's really quite easy.):
If we call a simple out any single card that wins, and backdoor outs cards you need any two of to win you have:
(Simple outs * 400 + backdoor outs * backdoor outs * 8)/10000

8 straight cards (simple) and 9 flush cards (backdoor) works out to:
3200+648/10000=.38
Which is a bit high, but as close a Brunson's estimate.

Alterntively, you can memorize a value for each backdoor out count, and add that to 4% per simple out.

With one card to go, each out is worth roughly 1 in 50.

[ebarnet]

9 to 5 dog means that out of 14 of these situations the villain will win 9 and doyle wins 5.

[ebarnet]

also, get pokerstove. it's easy to use, easy to download(just search google for pokerstove(i think it's just pokerstove.com but i'm not sure)

[BetMe]

This is a not very subtle dig on those of us who have to work for a living isn't it?

[jamartin]

Thank you for the replies, this has been cleared right up by those who contributed. I thank you, this "embarassing" question has finally been answered! [img]/images/graemlins/laugh.gif[/img]

p.s. If the last post before this is directed at myself, the answer is no.