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Pot Odds - Which Ratio is Right?

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Jason

02-14-2006, 03:26 PM Post subject: Pot Odds - Which Ratio is Right?

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I just realized I might have been calculating pot odds incorrectly. If there is $75 in the pot and a $25 bet is made, I have been calculating the odds as $25/$100=25% and then generally proceeding if I feel I have a 25% chance or better of winning. Similarly, if I was faced with a $40 bet into a $60 pot, $40/$100=40%. Lastly, a bet of $30 into a $70 pot equated to $30/$100=30%.

However, from watching television and reading a chapter from Phil Gordan’s latest book, I realize this might not be the right way to do it. For the three previous examples, if the odds are 25%, which is 1 divided by 4, that means in a trial of 4 hands, I’d win 1 and my opponent would win 3, so the ratio is actually 1:3 or 33%. Continuing that logic to the other two examples:

40/100=2/5, so out of five trials, I’d win 2 and my opponent would win 3 or 2:3=66.7%

30/100=3/10, so out of 10 trials, I’d win 3 and my opponent would win 7 or 3:7=42.86%

So, which method is the right way to do it? For these three examples, the differences are significant 25% vs. 33%, 40% vs. 67%, and 30% vs. 43% respectively.

- Jason

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Staresy

02-14-2006, 03:32 PM

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no, the way you said at the start of your post is correct.

3:1 ratio still means you have a 1 in 4 chance. I think you are just mistaking the ratio factor. Instead of thinking that 3:1 means you have a 1 in 3 shot, you need to add the two together, as this is the total number of trials and each of your respective chances of winning. i.e. your opponent 3 and you 1, from a total of 4 trials.

think of it like this; If you and I flipped a coin, our chances would be 50% each. or 1:1, which means even money. It does not mean 1/1=100%, otherwise one of us would win all the time and the other would never win.

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Jason

02-14-2006, 03:36 PM

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Ah, yes. Good call. I should have taken it that far to see why the second method doesn't make much sense. That's a relief since it keeps the math much easier.

- Jason

arkana

02-14-2006, 03:40 PM

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Sounds like you are confusing odds and probabilities.

Odds are defined as (good outcomes) : (bad outcomes)
Probability is defined as (good outcomes) / (total outcomes)

Which means 2:3 is the same as 2/5.

Jason

02-14-2006, 04:00 PM

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Ok, I just confused myself again.

If we were flipping a coin and you made a bet of $10, calculating the odds the way I would for poker, I'd get:

$10/$10 = 100% which means I'd need a 100% odds to call, but I know that's not right, it should be:

$10/($10 + $10) = 50%, but that takes into account money I put in in the denominator whereas my examples before never did that.

So, if someone makes a $25 bet into a $75 pot, should it be:

$25/($100+$25)=20%

- Jason

Staresy

02-14-2006, 04:10 PM

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no. It should be $25/($75+25)

remember, the pot only has $100 in it after their $25 bet. It then costs you $25 to call it.

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arkana

02-14-2006, 04:13 PM

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If i understand you correctly the pot is $75, the opponent bets another $25 which makes the pot $100 and you have to call $25.

Doing it with odds:
Pot is offering you 100:25 or 4:1 so you need 1:4 or better odds to win the hand.

Doing it with probabilities:
Your bet divided by the sum of the pot and your bet is 25/125 or 1/5 so you need at least a 20% chance of winning the hand.

20% is the same as 1:4

arkana

02-14-2006, 04:15 PM

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Quote:

Originally Posted by Staresy

no. It should be $25/($75+25)

remember, the pot only has $100 in it after their $25 bet. It then costs you $25 to call it.

You have to call $25 to win $100, not $75. The pot size includes your opponent's bet.

Jason

02-14-2006, 04:32 PM

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But, apply that logic to a bet made for a coin flip as I mentioned above. If a bet of $10 is made to a coin flip, the way we apply pot odds doesn't seem to work:

$10/$10=100%, which means you need 100% odds to call, but that obviously isn't right. Where is the flaw for the poker case of $25 bet into a $75 pot and the coin flip case where $10 is bet into a dry pot.

- Jason

Pelion

02-14-2006, 05:07 PM

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Quote:

Originally Posted by Jason

there is no division to be done.

If op makes a $10 bet on Heads there is $10 in the pot.
Your pot odds are $10 to $10 i.e. 1:1

You look at the chance of winning and decide that you will win 1/2 of the tosses.

i.e. for every 1 time you win, your opponent wins 1 also.

that means you have 1:1 odds of winning.

(odds of winning are expressed as number of losses:number of wins so a flush draw is said to be about 4:1 against winning with one card to come. i.e. you will win 1 time for every 4 that you lose)

You compare these odds to the pot odds and realise that 50% of the time you will win $10 (the pot) for an EV of +$5
but 50% of the time you will lose $10 (your bet) for an EV of -$5

You notice that the total EV of the bet is exactly $5-$5 = $0 but you are board so you take the bet anyway....and lose.

gabe: Ive dropped almost 100k in the past 35 days.

bigspenda73: But how much did you win?

chardrian

02-14-2006, 06:28 PM

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Actually you both are wrong.

Pot odds are figured out by your bet divided by the total pot to be won.

Let's say the pot is $25 and your only opp. goes all-in for $25. The way you are figuring percetnages would give you $25/$50 (the pot with opps bet) = 50%. But actually it should be $25/$75 (the total pot plus how ever much your call will make it) = 33%.

As an example to show this is true. Consider:

You have :Kh:

on a board of :Ah:

You know you have 9 outs - or about a 1 in 3 shot to win the hand by the river.

Your opp goes all-in with $25 in a pot that already has $25. Should you call?

According to your theory no. You would need a 50% chance to win to call and we don't got it.

But actually calling is basically even money because you really only need a 33% chance to win here to call.

1 time you will call and lose $25 = -$25
1 time you will call and lose $25= -$25
1 time you will call and win $50 = +50
----------------------------------------------------------
Total Ev = even stevens.

Make sense?

http://chardrian.blogspot.com
come check out my training videos at pokerpwnage.com

Pelion

02-14-2006, 08:09 PM

#12 (permalink)

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Quote:

Originally Posted by chardrian

Actually you both are wrong.

How am I wrong?

$50:$25 is correct. You bet $25 for a chance to win $50. If you win the 50 you get your 25 back aswell. Thats what it means at a bookies and thats what i means here too.

You are confusing the chance to make your hand with the odds against completing your hand.
You are also confusing the fraction of your bet to the pot with the pot odds.

I will say it once again.

The odds of something happening are expressed as the number of times it will not happen for every 1 time it does.

e.g. 9:1 against means it will not happen 9 times for every 1 time it does.

The total number of trials is therefore 9+1 = 10.
So there is a 1/10 chance of it happening (or 1 in 10).
and a 9 in 10 chance of it not happening.
Now to your example.

If the pot is $25 and op goes alin for $25 then there is now $50 in the pot.
You have to call $25 to win this $50.
The pot odds are therefore 2:1 meaning the pot is twice as big as your bet has to be to call it.

You are about a 1 in 3 shot of winning the hand by the river.
So you will win 1 time in every 3. The other 2 out of every 3 times you will lose.
This means the odds against you winning are 2:1.

Comparing the 2:1 odds of winning with the 2:1 pot odds shows that you are even money (as you said).

gabe: Ive dropped almost 100k in the past 35 days.

bigspenda73: But how much did you win?

Jason

02-14-2006, 08:41 PM

#13 (permalink)

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So, it sounds like I was right when I used the bet with the coin - you do have to consider the pot after you put money in.

So for all my examples, the numbers change:

Code:

Pot Bet	     Math     Right%  Wrong%
75  25   25/(25+25+75)  20.00%  25.00%
60  40   40/(40+40+60)  23.08%  40.00%
70  30   30/(30+30+70)  28.57%  30.00%

(Note: Pot is the pot size before the opponent bets, so the total pot by the time it is your turn to call, fold, or raise is the Pot + Bet)

By Right%, I mean, if you are faced with that bet and pot, that's the % of the time you need to win or better to make it worth the call - otherwise, you'll lose money in the long run.

By Wrong%, I mean, the other way I and many people use of simply dividing the bet in front of us by the pot size, which doesn't take into account the money you will win with the bet itself - see my example with the coinflip where it comes out to 100% when we clearly know it's 50%.

Agree? Disagree? Discuss ...

- Jason

Demiparadigm

02-14-2006, 09:29 PM

#14 (permalink)

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lol....

Pelion is right.

You are all silly.

1:1 = 1/2 = 50%
2:1 = 1/3 = 33%
3:1 = 1/4 = 25%
4:1 = 1/5 = 20%

Quote:

If there is $75 in the pot and a $25 bet is made, I have been calculating the odds as $25/$100=25% and then generally proceeding if I feel I have a 25% chance or better of winning.

$25 : ($75+$25) = 25:100 = 4:1 = 1/5 = 20%

100:25 = 25/125

Quote:

Similarly, if I was faced with a $40 bet into a $60 pot, $40/$100=40%.

$40: ($60+$40) = 40:100 = 5:2 = 2.5:1 = 2/7 = 28.5%

100:40 = 40/140

Quote:

Lastly, a bet of $30 into a $70 pot equated to $30/$100=30%.

$30

$70+$30) = 30:100 = 10:3 = 3.33:1 = 3/13 = 23%

100:30 = 30/130

Notice I switched the odds around, because odds are typically stated as losses:wins or in the case of a bet, profit: investment.

To win in poker you only need to be one step ahead of your opponents. Two steps may be detrimental.

chardrian

02-15-2006, 01:33 AM

#15 (permalink)

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Sorry pel - I shoulda clarified. Both Jason and staresy are wrong.

Jason u gots it right in your last post.

http://chardrian.blogspot.com
come check out my training videos at pokerpwnage.com

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