Pot Odds Question

[Stocks]

Hey

I got a question about calulating the odds to make a call.

Heres the formula I was looking at.

(opponents bet + pot) x (outs/number of cards) = number you need to call

Example

Blinds are $100/$200

You have KQ of harts

Someone raises $500 preflop your the only caller so theres $1300 in the pot.

Flop comes A,4,7 with 2 harts

Opponent goes all for $4000

Lets assume oppenent has an A and you have a flush draw so you have 9 outs.

So using our forumla

($1300+$4000) x (9/47)

($5300) x (.19) = $1007

Now am I correct to assume that the $1007 to call for the right odds is only for one street so for our example theres 2 cards to come so we can double it to $2014. So in this example calling the $4000 would be a horrible call.

Is all this correct?

[pokeraussie]

Of course its a horrible call don't over complicate things (assuming you are playing as deep as your opponent). He is always going to have an Ax hand in this spot at the very least, so you will need to hit your flush to make the winning hand. You know you have about 2:1 odds to make the flush by the river (35% if you want to be exact), but the pot is giving you only about 1.3:1 odds. Since the pot odds < hand odds you should fold.

[Stocks]

Ya I know its a bad call and a flush is 35% it was just an example for using the forumla. what I ment is the actual forumla correct.

(opponents bet + pot) x (outs/number of cards) = number you need to call

With the number of outs devided by number of cards part it don't give the right answer.

9 devided by 47 is .191 or 19.1% to hit the flush on the turn
9 devided by 46 is .196 or 19.6% to hit flush on the river

So if you add them both together its 38.7% to hit your flush not 35%

[jason@bp]

...

I don't really know what's going on in a lot of your post just due to cloudy terminology, so forgive me if I answer a question you haven't asked. It sounds like you want to figure out the expected value of your call, and you want a way to figure out what betsize would make a call +EV.

If that's the case you need to use this formula:

EV = ($pot)(Probability of Win) - ($to call)(Probability of Loss)

So in your above scenario, we get:

EV = ($5300)(1 - ((38/47)*(37/46))) - ($4000)((38/47)*(37/46))
EV = ($5300)(0.35) - ($4000)(0.65)
EV = $1855 - $2600
EV = -$745

So you expect to lose $745 every time you make that call.

Now, if you want to figure the breaking point for EV in this situation, you just do a bit of substitute-and-solve for X, where X is the size of the call:

0 = ($1300 + X)(0.35) - (X)(0.65)
0 = $455 + 0.35X - 0.65X
0.3X = 455
X = $1516.7

So $1516.7 is the breaking point where a call is exactly neutral EV. Any amount below $1516.7 is +EV, and any amount above $1516.7 is -EV.

To confirm this, let's plug in a call amount less than the breaking point -- $1000 -- and see our call EV:

EV = ($1300 + $1000)(0.35) - ($1000)(0.65)
EV = $805 - $650
EV = $155

And now plug in a call amount greater than the breaking point -- $2000 -- and see our call EV:

EV = $1300 + $2000)(0.35) - ($2000)(0.65)
EV = $1155 - $1300
EV = -$145

So in sum, whenever you want to figure out the breaking point of a call across two streets, you use the following equation:

A = Pot amount up to the call
X = Call Amount
Y = Probability of a Loss

0 = (A + X)(1 - Y) - (X)(Y)

And solve for X.

Keep in mind, though, that equity calcs and pot odds aren't the holy grail in tournament applications. I mention this because it seems like you're asking about a tournament situation. In the case of a tourney, you've got to take your prize equity into account -- your share of the prize pool based on your skill and your stack size -- whenever you make a decision. That's where ICM comes into play.

Hope this helps.

-J

[nik rossi]

If you want to make it exactly it looks something like this:

Let´s assume we have a 5 outer for 2 pair+:

We hit on the turn but not on the river:
(5/47)x(42/46) = 0,097 ~ 9,7%

We don´t hit on the turn but on the river:
(42/47)x(5/46) = 0,097 ~ 9,7%

We hit both the turn and the river (full house):
(5/47)x(4/46) = 0,009 ~ 0,9%

9,7 + 9,7 + 0,9 = 20,3%

20,3% probability to hit our 5 outer either on the turn or on the river or both.


We could simplify this and calculate -> (100% - we don´t hit at all)

we don´t hit at all: (42/47)x(41/46) = 0,797 ~ 79,7%

100% - 79,7% = 20,3%


So in the end our calculation looks like this:

(opponents bet + pot + our call) x {1-[(number of cards turn - outs)/(number of cards turn)]x[(number of cards river - outs)/(number of cards river)]} = max amount to justify a call


That´s a bit messy so just use the standard shortcut:

probability to hit on the next street -> outs * 2 + 2
probability to hit either on the next or the street after (turn or river) -> outs * 4

(opponents bet + pot + our call) * (outs * 4 / 100) = max amount to justify a call



I hope this is not too confusing.

[nik rossi]

It seems there was an answer in the meantime.

[Stocks]

Thanks fellas

Although one thing I can't seem to wrap my head around is why this is'nt right.

If you have 9 outs and 47 cards left its 9/47 = 19%

Is that right or am I missing something?

If it is correct a flush would be 38% not 35%

[nik rossi]

Thanks fellas

Although one thing I can't seem to wrap my head around is why this is'nt right.

If you have 9 outs and 47 cards left its 9/47 = 19%

Is that right or am I missing something?

If it is correct a flush would be 38% not 35%

The thing is you can´t just calculate 19+19. Let´s assume your probability is not 19% but 50%. If you calculate 50%+50% you would hit by the river every time which is not the case. You have to calculate [1-(0,5*0,5)]=0,75 = 75%. Let´s take the 19% -> [1-(0,81*0,81)]=0,35 = 35% (81% is the propability we don´t hit).

Google for converse probabilities. Hope this helps.

[jason@bp]

19% is correct, but it only takes one street into account -- the turn. You can't just double the probability to figure out P(HitEitherStreet) because on the river you've got 46 cards unseen and not 47.

So the way I calculated it was to figure out the probability of you missing on both the turn and the river:

( (cards unseen on turn - outs) / cards unseen on turn) * ( (cards unseen on river - outs) / cards unseen on river) = Pmiss

((47-9)/47) * ((46-9)/46) = 0.65032374

Thus to find the probability of you not missing turn and river -- i.e. hitting either turn or river -- you just find the complement of Plose. Do this by subtracting the P of all possible outcomes (100%, or 1) by Plose:

1 - 0.65032374 = 0.349676226

And you have your probability of hitting either turn or river: ~35%.

[GregW]

Thanks fellas

Although one thing I can't seem to wrap my head around is why this is'nt right.

If you have 9 outs and 47 cards left its 9/47 = 19%

Is that right or am I missing something?

That is correct. There's a 9/47 chance of hitting your flush on the turn (if you are on the flop), and a 9/46 chance of hitting your flush on the river (if you are on the turn).

If it is correct a flush would be 38% not 35%

Close, but you can't just double the percentage probability because there are 2 cards to come, although that does seem quite logical at first.

You'd think that because you want to hit a heart on the turn OR the river, that you just ADD 9/47 and 9/46, which gives you somewhere in the region of 38%. The small problem is you don't just want to work out the probability of hitting a heart on the turn OR river.

Instead of just adding those two probabilities, you have to find the probability of not getting a heart on the turn AND river. The opposite of this probability gives you the chances of hitting your flush across the turn and river.

No heart on turn AND river = 38/47 x 37/46 = 65%
Opposite = 1 - 0.65 = 35%

But yeah, just doubling the percentage is an okay way of doing it, although it's quite uncommon to want to work out the probability of hitting your flush across 2 streets instead of 1.


EDIT: Looks like there were two replies in the time it took me to post haha.