Pot odds enough to justify the call

[CroMagnon]

In the thread below nsiderstrate says that having 5:1 pot odds justifies calling with P QQ's because you will flop a set 1 in 8 times.

I am sure I am missing a great deal, but wouldn't the pot odds need to be 8:1 to justify this call soley on the pot odds issue. In the thread there are many other considerations as to why you would call. Perhaps it is those other consideratiosn that compensate for the lack of pot odds or am I misunderstanding the way the odds are being stated?

Thanks,
Cro

http://www.internettexasholdem.com/phpbb2/stars-180-already-itm-vt46479.html

[rayrns]

I hope I am correct here: the reason you could take 5:1 pot odds with a pair and not necessarily just Q's is the "implied odds" when you hit your set. The odds of hitting a set are 7.5:1 against but if you get 5:1 "immediate" pot odds you can usually count on making more than 2.5 BB's extra when you hit your set.

You will normally make enough extra BB's to make taking 5:1 "immediate" pot odds profitable even on the times you do not hit your set.

Ray

[Cript]

[torch]

[CroMagnon]

Ah, implied odds. While I have read about implied odds I don't have a complete grasp of them, but that makes sense that you would add your potential gain if you hit.

How is it that you have a 1 in 8 chance of hitting the set when you have a pocket pair? At a 9 person table there will be 34 cards left and you have two outs to the set . How does that translate to 1 in 8?

Thanks,
Cro

[nsidestrate]

You have to figure your odds with 50 unknown cards, because you don't have any idea what cards the other players hold. 2 in 50 three times is about 1 in 8.

[Fumseck]

Cro,

[SPAM]
I can recommend a great book on poker odds:

Texas Hold'em Odds and Probabilities - by Matthew Hilger

I have read it and I think it covers everything you ever wanted to know about poker odds and probabilities.
[/SPAM ]

[CroMagnon]

I'm a bit of a math dummy, could you show a bit of the calculation that turns "2 in 50 three times" into a 1 in 8 chance? Or point me to a source. (I am asking the wife to get me Matt's new book for christmas )

Thanks,
Cro

[Fumseck]

The odds of you flopping your set is 11.8% or roughly 1 in 8.5 (7.5:1 as a ratio).

To calculate this, the odds of each flop card not giving you a set is as follows: (48/50) * (47/49) * (46/48 ) = 88.2% which is the chance of all three cards not giving you a set. Subtract that from 100% and you get 11.8% as the chance of you flopping a set. 11.8% is roughly 1 / 8.5.


To understand the (48/50) * (47/49) * (46/48 )

48/50: When the first card of the flop is dealt, there are 2 cards in 50 (there is 52 cards in a deck and you already have 2 of them) that will give you a set so 48 in 50 do not give you a set.

47/49: When the second card of the flop is dealt, there are 2 cards in 49 (there are only 49 cards left now) that will give you a set so 47 in 49 do not give you a set.

46/48: When the third card of the flop is dealt, there are 2 cards in 48 (there are only 48 cards left now) that will give you a set so 46 in 48 do not give you a set.

When you multiply all 3 probabilities you get the probability (in %) that you will not flop a set. So to get the probability that you will flop a set you just have to subtract this number from 100%.


To get the 1 in 8.5 you do this 1 / 11.8% which equals to 8.47 that we round up to 8.5 and that Nside rounds down to 8 .

[CroMagnon]

Fumseck, do you teach math? You're very good.

Grasshopper now understand how number is arrived at.

Thanks everyone.

Cro

[Fumseck]

Nope,

but I read Matts book . This is explained in the first chapter if I remember correctly.

Glad I could help a little.

[ciaran]

[seanof30306]