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Silly Odds Question

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Silly Odds Question - Fri Jan 17, 2014, 04:13 PM
(#1)
fp_boss77's Avatar
Since: Jan 2012
Posts: 490
Hi, I'm a little confused when doing some odds calculation.

Let's suppose we're on the River, and villain bets $2 into a pot of $3

So we have to call $2 to win %5 (the total pot).
Given the pot odds, which frequency villain should be bluffing here to make this call break even ?

I'm not sure if we use the $2 from our call to calculate the odds.

Option 1 -> we call $2 to win $3 + $2 + $2 (pot + his bet + our call). Giving 3.5:1 odds

or

Option 2 -> we call $2 to win $3 + $2 (pot + his bet). Giving 2.5:1 odds

So, which calculation is correct ?
And how often should the villain be bluffing here to break even this call ?


Cheers!
 
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Fri Jan 17, 2014, 04:18 PM
(#2)
JWK24's Avatar
Since: Jun 2010
Posts: 24,802
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Hi fp_boss77!

option 1 is correct. For them to be breakeven it will depend on our range and his range. If the hand equities are 50-50 (which they will not), then he needs to bluff 2/7 times to breakeven.

John (JWK24)


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Fri Jan 17, 2014, 04:26 PM
(#3)
ouchbadbeat's Avatar
Since: Mar 2013
Posts: 347
Im pretty tired and wasnt going to post a response but...unless I'm missing something obvious here then Option 2 is the correct way


Getting 5:2 means (except for splits) you either lose $2 - the amount you have to put in - or you win $5 - the amount in the pot - you're not putting in $2 to win that money straight back, that's not the profit on the hand

say you have $10 left on river, pot is $3 he bets $2 for $5 total. if you call $2 and win, you don't gain your own 2 dollars, you gain 5 dollars

if you call and lose, you lose $2

where does the phantom added bet come from
 
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Fri Jan 17, 2014, 04:37 PM
(#4)
ouchbadbeat's Avatar
Since: Mar 2013
Posts: 347
and to work out the % of the time you need to be good to breakeven here, you do this:

betsize / (potsize + betsize) * 100

so in the example you gave:

2 / (2+5) * 100 = 28.5% winrate needed to breakeven


edit: got the maths wrong, tired

2/7 is correct for the equity, but it is not the pot odds

Last edited by ouchbadbeat; Fri Jan 17, 2014 at 05:16 PM.. Reason: sorted
 
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Fri Jan 17, 2014, 05:11 PM
(#5)
adohole's Avatar
Since: Jul 2011
Posts: 1,083
BronzeStar
you have the same answer but answer it difrent both right :P 5:2 is the same as 2/7


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Fri Jan 17, 2014, 05:14 PM
(#6)
ArtySmokesPS's Avatar
Since: Oct 2011
Posts: 7,324
I think there's some confusion because of the terms used in the OP. A bluffer's correct "bluffing frequency" is not the same as the odds required for a caller to call the bluff.

I tend to avoid odds ratios for things like this and use fractions and percentages instead.

If villain has air, he needs you to fold 2 out of every 5 times (2/5 = 40%) for him to make a profit, because...

Fold equity required with a bluff = bet size / (original pot size + bet)

If he suspects you will fold to his bet 40% of the time, then his bet of two thirds of pot is break-even for him.


If you're faced with a bet and have a value hand, you need it to be good 2 out of 7 times (2/7 = 28.6%) to break even, because...

Hand equity required by the caller = cost of calling / (original pot + villain's bet + your call)

If you're the one calling, then the pot odds you're getting are 5:2, which means the same as needing to be right 2/7 = 28.6% of the time.

Note that the bluffer is setting 3:2 on his bluff (he needs you to fold 40% of the time), but the caller is getting 5:2 on a call (needing to win only 28% of the time to break even). It's quite counter-intuitive that each player has different numbers for "how successful" their play needs to be, and this is because there's already $3 of dead money in the pot that they are both fighting for, and one player has to bet to try and win it, while the other only has to call and will also win the "bluff amount" too if he's right.


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Last edited by ArtySmokesPS; Fri Jan 17, 2014 at 05:20 PM..
 
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Fri Jan 17, 2014, 06:15 PM
(#7)
fp_boss77's Avatar
Since: Jan 2012
Posts: 490
What I meant by this 'bluff frequency' is like this.

When facing a bet on the river, I heard one of the instructors say something like this:

' He's betting $2 into $3, so we only need to be right X% of the time to make this call. '


What is the X% ?

Thanks!
 
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Sat Jan 18, 2014, 04:49 AM
(#8)
adohole's Avatar
Since: Jul 2011
Posts: 1,083
BronzeStar
thats 2/7th of the time so 28.5% you need to be right


Triple Bracelet Winner

 
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Sat Jan 18, 2014, 08:58 AM
(#9)
ArtySmokesPS's Avatar
Since: Oct 2011
Posts: 7,324
Quote:
Originally Posted by fp_boss77 View Post
What I meant by this 'bluff frequency' is like this.
When facing a bet on the river, I heard one of the instructors say something like this:
' He's betting $2 into $3, so we only need to be right X% of the time to make this call. '
What is the X% ?
Right. I don't like using the term "bluff frequency" there, because villain might actually be value-betting (albeit with a worse hand).

When all you're doing is calling a bet on the river, then it's a straightforward calculation. You look at what's in the pot, and compare it to how much you need to call. On the table, you can see the size of the pot is $5 and the CALL button says "call $2", so the pot odds are simply 5:2.

The pot odds ratio you're faced with is pot after villain's bet:cost of call, which is 5:2 in your example. You'll break even calling if you're correct 2 times for each 5 bad calls, which means you need to be right 2 times out of 7.

To emphasise, a pot odds ratio of 5:2 is the same as saying "you need to be right two out of seven times".

5:2 = 2/7 = 28.6%


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Sat Jan 18, 2014, 09:18 AM
(#10)
aspok123's Avatar
Since: Aug 2011
Posts: 883
I prefer percentage odds calculations while I know how much in average 1 out card gives me chance of winning and there are many charts with those probabilities eg. flush draw & open ended stright draw on flop and 2 cards to come, so using decision trees it is easy find such equation in the case we have decision to call or fold while faced to some bet (B) and effective pot size (P) , action to us and we know our chance to win this -its probability is (p) :
Bmax < P * p/(1-p)
It is easy to learn a few cases of most common probabilities: p/(1-p) and we know how much we can call in the terms of pot size percentage, however we should adjust this percentage by some other factor to make profitable calls at rate lets say 25% or different one, but this is beyond basic odds calc

Last edited by aspok123; Sat Jan 18, 2014 at 09:25 AM..
 

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