Odd's for Outs

[bosulli]

I'm trying to recalculate probabilities for outs, then I was going to convert them to odds for pot odd calculations. In the book "Ken Warren Teaches Texas Hold'em" he calculates that the probability for drawing with 20 outs, 2 cards to come is 67.5%. I cannot figure out how it is calculated. Help please. Thanks in advance.

[Nottom]

For future reference: you will probably get a quicker response to this in the Probability forum, but since you are here ...

You have 20 outs and 2 chances to catch one.

The easiest way to calculate this is to fine the chances of missing twice and then subtract that from 1 (the alternative is so add the odds of hitting on each street and then subtract the odds of hitting twice).

For 20 outs this works out like this.

Chance to miss turn: 27/47
Chance to miss river: 26/46
Chance to miss both: 27/47*26/46 = 0.3247 = 32.47%

Chance to make your draw: 1 - 32.47 = 67.53%

[bosulli]

Since I am a begginer and this was fundamental prob question, I put in the begginer forum.

Anyway, thanks a bunch.

I calculated the prob of hitting a four flush with 2 to go as 35%. Now, on to odds. Therefore the odds of not hitting my flush is 2:1, right. So, to make a longterm profitable bet I need the pot to be twice as big as my bet?

[Henke]

Therefore the odds of not hitting my flush is 2:1, right.

I think it should be closer to 1.86:1.

So, to make a longterm profitable bet I need the pot to be twice as big as my bet?

Well, not really... If your opponent bets, and you have to go all in when you call and the pot is offering you 1.86:1, you make no money. EV=0. If you're first to act against your only opponent, and you know he will not fold, then the best situation you can hope for is to see the turn for free, because then about 35% of the pot is yours. If you are heads up, 50% of the money going in is yours and 50% belongs to your opponent. But, only 35% of the new money is yours and the rest (about 65%) will now belong to your opponent. Because of this, you need the pot (including his bet) to be about 1.86 times what you have to call to make the call even money. Also, for this equation to be correct, you will need to go all-in on your call. Because if we don't consider the money going in for you and your opponent on the turn (which is unknown at this time) then we cannot use the odds for two cards to come. For instance, assume two persons are playing NL-holdem. The pot is 0.86 units and your opponent bets 1 unit. You have a nut flushdraw and call. On the turn you get no help. Your opponent now bets 100 units into a pot that contains 2.86 units. The actual effective pot odds you had on the flop here where 101.86:101 and not the apparent 1.86:1.

But, if you're up against more than 1.86 opponents with a nut flush, and you know you will see the river, then if you make them call one bet each you gain! Because now you put in 33% of the new money and your opponents put in 67%. And as earlier you're "entitled" to 35% of the pot, which is more than you put in.

[bosulli]

Thanks alot.

That is the kind of analysis I have been looking for. Now to get a handle on EV (expected value?). I am going to learn the concept you depicted, and I am sure I will have a follow up question, but first I got to do the day job.

I appreciate your helping me out on this. I believe it is an important concept, and I have not been able to make practical application from this theory put forth in several books I have read.