Correctly Calculating Pot Odds - Does Your Bet Count?

lasvegaspokerchick

lasvegaspokerchick

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I recently posted a hand analysis thread to which AlienGenius responded. Part of his response included the following brief statement about calcualting pot odds:

"Also, you don't get to "add" your call into the pot, and then make your calculations. The main pot is $102 ($12 + $45 + $45), and the side pot is $55 ($100 shove - $45 that went in the main pot), not $135 and $110 as given in the above analysis."

Good point. I am curious, however, whether or not this is true. I have been playing poker for a long time and I have heard this rule stated both ways. I am not ashamed to admit that I don't know for sure which is correct: that you DO get to add your bet to the pot total in your calculations or that you DON'T.

I looked at it myself and it would seem to me that you DO get to add your bet into the pot when calculating pot odds. Let's take a simple example where we will include our bet in the pot total when making the decision. Let's say that you have a draw that will hit 1 in 4 times (1:4). You need four times your bet in the pot in order to call and break even. Let's say that the pot is $200 and the action to you is a $100 bet. If you call, there will be your $100, the $100 bet, and the $200 in the pot for a total pot of $400. If you do this exact same scenario four times, let's say that you win exactly what you are "supposed" to...so one time out of the four. On the three hands that you lose you will lose $100 each time for a total loss of $300. On the hand that you win, you will win a $400 pot minus the $100 bet you called for a total win of $300. Even.

You will notice that I removed your $100 bet when calculating the profit, but left it in when calculating the odds. Although I hate to disagree with AlienGenius (and maybe I am misunderstanding him) this seems right to me.

I would really love everyone's thoughts.
 
ChuckTs

ChuckTs

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You never include your own bet into pot odds calculations.

Pot odds = [pot size]:[amount to call] where [pot size] includes any and all bets on the current street as well as the amount in the middle.

Ex: pot is $200, and there is a $100 bet in front of you. Your pot odds are:

[pot size]:[amount to call]
= [$200+$100]:[$100]
= $300:$100
= 3:1
 
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maltz

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Right you don't include your call amount, since at the time of your decision the money belongs to your stack, not the pot.

Just a minor correction:
"Let's say that you have a draw that will hit 1 in 4 times (1:4)."
It is actually 1:3 --- 1 win vs. 3 loss.
That might help clarify the concept a bit further.
 
lasvegaspokerchick

lasvegaspokerchick

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Maltz - Correct...my bad in stating the odds as 1:4. A 25% chance of winning would be 1:3. However, if you have a 25% chance of winning, or one in four, than you only need four times the size of your bet (including your own bet) in the pot in order to correctly call. Or another way of saying it is that if you have a 25% chance of winning, 1:3, you need three times your bet (not including your own bet) in order to correctly call.

I learned the odds using the idea that you include your own bet, but the odds that I learned for each hand were probably skewed to include your own bet. So maybe we are saying the same thing a different way?
 
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maltz

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Yeah I see what you are saying.
If we express win as a precentage "W"%, we can then express loss as "100-W"%.

The pot odds are expressed as:
"Call vs. Pot (current)"

Now people will tell you to call when:
W/(100-W) > Call/Pot

In your example, your winning chance is 25%, losing chance 75%. You call $100 to a current pot of $300. The two sides are the same (1/3) so you can either decide to call or not.

Now for your alternative notion:
W/100 > Call/(Pot+Call)

Are the two equations equivalent?
You can actually prove it... :D

W/(100-W) > Call/Pot
(100-W)/W < Pot/Call (reverse)
100/W - 1 < Pot/call
100/W < Pot/Call + 1
100/W < (Pot+Call)/Call
W/100 > Call/(Pot+Call)

So yes you can do it either way.
 
zachvac

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Let's look at the concept of pot odds. If you are 4:1 to win the hand (20%), you win 1 every time you lose 3. When you calculate pot odds, say the pot is 300. Someone else bets 100 into it. The pot is 400, and you have to put 100 in. If you fold, that 100 is yours. If you call, it's not. So if you are paying 100, you need to get at least 100 back including your bet. BUT... in this case you are paying 100 into a 400 pot, meaning you are getting 4:1 on your money if you don't count your bet, 5:1 on your money if you count yours. If the first version is correct, you will break even with the call. If counting yours is correct, you will win making this call. So 1 time you win, 4 times you lose, you win and get 500 (net 400), lose 4 times and win nothing (net -100). It adds up to... 0. The key is that when you win 500, you're not really winning 500, your profit is the pot before you call.

I made a recent post about pot equity. This makes pot odds make a lot more sense as well. Same situation, 400 in the pot before your call. If you call you get 500*.2 = 100, or you get your money back, meaning you could fold or call and the same thing happens. Say there was 500 in the pot already. If you call you get .2*600 = 120 and you just have to pay 100 to get 120 back, it's a smart call. With pot equity you can factor in your money in the call, pot odds you can't, because of the concept of pot odds. It's a ratio of loss:win, it would be foolish to use the ratio money won+money paid:money paid. you compare loss:win to money you can win:money you have to pay. Pot equity examines how much of the pot will be yours if you call. That includes the money you put in. Pot odds are easier to do during a game, but equity just makes a lot more sense and whenever I'm confused about pot odds I go back to pot equity and compare the 2. Hope this helped.
 
dj11

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I wrote a long post before I read the replies.

Now my head hurts.
 
lasvegaspokerchick

lasvegaspokerchick

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Good point Zach. I like the pot equity method as well. It does seem to give you a more balanced approach to the hand. After all, we don't want to be risking money in a real way in order to theoretically break even...much better to risk money in order to win money.

Thanks for the analysis.
 
zachvac

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My head hurts, too, and so many hashed calculations like this one did it.


oops, that was a typo, originally I had it as 3:1, but to make it even changed it to 4:1, guess I forgot to change that, I'll do that now.

EDIT: can't edit the previous post
 
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maltz

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Actually, pot equity and pot odds are the same concept. You can write it as:

Pot equity = (Pot + Call) * (Win% / 100)
And this should be greater than your call amount.

(Pot + Call) x (W / 100) > Cal
W/100 > Call / (Pot + Call)

This is identical to OP's concept of including call into the pot. They are just different angles (and their fancy names) to look at the same issue.
 
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simmo5050

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Let's look at the concept of pot odds. If you are 4:1 to win the hand (20%), you win 1 every time you lose 3.


This is wrong! If you are 20%, you win 1 for every time you lose FOUR.
[FONT=NotoSans, Lato, arial, sans-serif]
[/FONT]
3:1 is 25% !
2:1 is 33%
1.5:1 is 40%
1:1 (rare but happens basically if someone goes all inwith just the blinds as happens in microstakes not infrequently ) is 50%
[FONT=NotoSans, Lato, arial, sans-serif]
[/FONT]
I'm not sure how reliable your info is after this . No, I don't know why my text is highlighted either.
 
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Dark Army

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Maltz - Correct...my bad in stating the odds as 1:4. A 25% chance of winning would be 1:3. However, if you have a 25% chance of winning, or one in four, than you only need four times the size of your bet (including your own bet) in the pot in order to correctly call. Or another way of saying it is that if you have a 25% chance of winning, 1:3, you need three times your bet (not including your own bet) in order to correctly call.

I learned the odds using the idea that you include your own bet, but the odds that I learned for each hand were probably skewed to include your own bet. So maybe we are saying the same thing a different way?

The bold part is where you're not thinking about it correctly. Your looking at the entire pot as profit. It's not.

When you make a bet and win the pot, that bet is NOT a part of your profit. You're getting the bet back when you win, but it's not profit. In fact, until you win it's a loss.

Let's say you have a $2 bankroll to gamble with. You decide to place a bet on your favorite sports team that has 2:1 odds of winning.

If your team wins, it will also pay you 2:1 odds (The same odds of them winning)

Now, if you make a $1 bet, that doesn't change the payout odds. The payout doesn't go up to 3:1. It's still 2:1.

But that's exactly what you think is happening in poker. You think that if you add your bet to the pot it increases the payout. It doesn't. The payout is the same as it was before you made your bet. The bet is not part of your profit.
 
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