# Any Two Cards

## Any Two Cards

Is there a magic number that we can look for that will indicate when i can open up to ATC in different situations? kinda like you min raise/folding chart.

If I have it right, to RFI(with antes) ATC UTG we need our opponents to be folding 90% of hands.

I would like to do this for Iso's, 3b's, etc., but not sure if im wasting my time doing so?

lets say its a 9 handed game, blinds are 100/200 25a, 20bb eff

UTG opens for 400, making the pot 925. For me to 3bet ATC AI on the BTN I would need the 3 plrs left in hand to fold 90% of the time.

If I have it right, to RFI(with antes) ATC UTG we need our opponents to be folding 90% of hands.

I would like to do this for Iso's, 3b's, etc., but not sure if im wasting my time doing so?

lets say its a 9 handed game, blinds are 100/200 25a, 20bb eff

UTG opens for 400, making the pot 925. For me to 3bet ATC AI on the BTN I would need the 3 plrs left in hand to fold 90% of the time.

**RWPKRPLR1**- Posts : 882

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## Re: Any Two Cards

well if it were true you still aren't considering the player that opened.

Whats in the pot already vs what you risk will tell you what the remaining players need to fold which includes the opener.

Whats in the pot already vs what you risk will tell you what the remaining players need to fold which includes the opener.

**JodaB.**- Posts : 1327

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## Re: Any Two Cards

ya i did SB+BB+UTG=3

ok so it always 90% of deck or is mor like it ..

lets use my example

utg/sb/bb all need to fold 90% of the deck or utg needs to fold 90% of his range & sb/bb90%of the deck

ok so it always 90% of deck or is mor like it ..

lets use my example

utg/sb/bb all need to fold 90% of the deck or utg needs to fold 90% of his range & sb/bb90%of the deck

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## Re: Any Two Cards

nope but talk this out you'll get it...

risk vs reward gives you total needs folds

so you compare it to the summation (added) of all 3 players responses.

So read that over until you next question is "how do we add probabilities"

risk vs reward gives you total needs folds

so you compare it to the summation (added) of all 3 players responses.

So read that over until you next question is "how do we add probabilities"

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## Re: Any Two Cards

lmfao

ok how do we add probabilities

ok how do we add probabilities

**RWPKRPLR1**- Posts : 882

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## Re: Any Two Cards

yes go find out quick (google) cause your gonna realize something i think.RWPKRPLR1 wrote:lmfao

ok how do we add probabilities

**JodaB.**- Posts : 1327

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## Re: Any Two Cards

to be clear you want how to add probabilities, not how to add 'these' probabilities.

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## Re: Any Two Cards

lol

so like whats the probability of rolling a 2 or 5 on a die

2=1/6 + 5=1/6, both 2/6 =1/3

so like whats the probability of rolling a 2 or 5 on a die

2=1/6 + 5=1/6, both 2/6 =1/3

**RWPKRPLR1**- Posts : 882

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## Re: Any Two Cards

bah on tables so i can only think so hard.RWPKRPLR1 wrote:lol

so like whats the probability of rolling a 2 or 5 on a die

2=1/6 + 5=1/6, both 2/6 =1/3

you didn't address the question in relation to op.

if there are 3 players that fold x y or z

we want the probability of x y or z

maybe we want the or answer.....

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## Re: Any Two Cards

yikes.. if youre on tables just let me know i can wait

**RWPKRPLR1**- Posts : 882

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## Re: Any Two Cards

our risk vs reward in a vacuum determines the %

it either determines how often we need folds or how often our opponents can call on average (remember average folds + average calls = 100%, so if we have one we can get the other by subtracting it from 100)

The amount we

If we just do simple addition of the %'s we are double counting times that 2 or 3 players fold.

for example: If player A calls and B and C fold, that

so if each player folds 30% of the time, they don't only call 100 - (30 + 30 + 30) which is 10%. we still intuitively know we get through more than 10%, its more like 60% or something we get through.

I'll let you sit on the for a moment and then we'll find the formula.

**we need**to get through (so our bet size vs the pot determines how often we need to get through)it either determines how often we need folds or how often our opponents can call on average (remember average folds + average calls = 100%, so if we have one we can get the other by subtracting it from 100)

The amount we

**actually**get through is all the times on average**every**player folds.If we just do simple addition of the %'s we are double counting times that 2 or 3 players fold.

for example: If player A calls and B and C fold, that

**shouldn't**count as a success right?so if each player folds 30% of the time, they don't only call 100 - (30 + 30 + 30) which is 10%. we still intuitively know we get through more than 10%, its more like 60% or something we get through.

I'll let you sit on the for a moment and then we'll find the formula.

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## Re: Any Two Cards

so we have 3 seats behind and we want to know how often they all fold (or 100-call).

its like we have 3 coins and we want to know the probability that all three are heads/call (or 100-tails/call)

So what i meant we need is how to do the probability that all three coins are Heads given a toss of them all.

its like we have 3 coins and we want to know the probability that all three are heads/call (or 100-tails/call)

So what i meant we need is how to do the probability that all three coins are Heads given a toss of them all.

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## Re: Any Two Cards

here this will explain, read it, get scared confused and overwhelmed then ill straighten it out and show you the revaluation if there is one to be had.

http://www.mathgoodies.com/lessons/vol6/independent_events.html

lemme know once you skim it.

http://www.mathgoodies.com/lessons/vol6/independent_events.html

lemme know once you skim it.

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## Re: Any Two Cards

went through it

**RWPKRPLR1**- Posts : 882

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## Re: Any Two Cards

alright so with coins if heads is folds and we want the probability of all folds we just multiplyRWPKRPLR1 wrote:went through it

.5 x .5 x .5 = 12.5 %

or

1/2 x 1/2 x 1/2 = 1/6 = 12.5%

the issue here is all our 'coins' (each players % they fold) do not flip at equal percent.....so multiplying the percents is tough to do on the fly.

With me so far?

**JodaB.**- Posts : 1327

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## Re: Any Two Cards

JodaB. wrote:alright so with coins if heads is folds and we want the probability of all folds we just multiplyRWPKRPLR1 wrote:went through it

.5 x .5 x .5 = 12.5 %

or

1/2 x 1/2 x 1/2 = 1/6 = 12.5%

the issue here is all our 'coins' (each players % they fold) do not flip at equal percent.....so multiplying the percents is tough to do on the fly.

With me so far?

yup

**RWPKRPLR1**- Posts : 882

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## Re: Any Two Cards

Meh I'm having an issue because i thought the problem was each % needs to be the same, but I think the issues is if they are 35% x 42% x 25%, the issue is its too hard to do on the fly.

So I solved that by assuming everyone calls the same % and using the root function (that divides them into equal parts the multiply together to get the whole).

So if we need 40% folds and we have 3 left to act we do root 3 and get 73.6

so that .736 x .736 x. 736= equals 40% of the time 3 players fold.

But now I'm wondering if i can avoid this by learning how to multiple two digit number by two digit numbers on the fly.

I'll have to think about that...

anyways...

Yes i think we just multiply the %s and compare it to our risk/reward (pot odds) and determine if its +ev ignoring post. If it is then ATC is plus ev, provided we don't spew post.

The issue is utg isn't on atc so they don't fold very often compared to the players left to act (they still have atc).

So if its not +ev with atc then we have to only use cards that have enough value when called either preflop allin or that we gain postflop often enough to make the entire hand +ev (not often prob).

I think that explains it, but its pretty mathematical to calculate irl.

We can also look at the min raise/3bet nash equilibrium that often gives us a clue to what hands to 3bet shove.

The hands we 3bet in short stack poker are quite related to 3bet spots in deeperstack play although im not sure players on the whole have realized that yet.

So I solved that by assuming everyone calls the same % and using the root function (that divides them into equal parts the multiply together to get the whole).

So if we need 40% folds and we have 3 left to act we do root 3 and get 73.6

so that .736 x .736 x. 736= equals 40% of the time 3 players fold.

But now I'm wondering if i can avoid this by learning how to multiple two digit number by two digit numbers on the fly.

I'll have to think about that...

anyways...

Yes i think we just multiply the %s and compare it to our risk/reward (pot odds) and determine if its +ev ignoring post. If it is then ATC is plus ev, provided we don't spew post.

The issue is utg isn't on atc so they don't fold very often compared to the players left to act (they still have atc).

So if its not +ev with atc then we have to only use cards that have enough value when called either preflop allin or that we gain postflop often enough to make the entire hand +ev (not often prob).

I think that explains it, but its pretty mathematical to calculate irl.

We can also look at the min raise/3bet nash equilibrium that often gives us a clue to what hands to 3bet shove.

The hands we 3bet in short stack poker are quite related to 3bet spots in deeperstack play although im not sure players on the whole have realized that yet.

**JodaB.**- Posts : 1327

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## Re: Any Two Cards

RWPKRPLR1 wrote:

lets say its a 9 handed game, blinds are 100/200 25a, 20bb eff

UTG opens for 400, making the pot 925. For me to 3bet ATC AI on the BTN I would need the 3 plrs left in hand to fold93% (fix)of the time.

OK so going back to the OP...

1. 100+200+225+400 = 925

2. 925/4000 = .23

3. .23+1= 1.23

4. 1/1.23 = .81

5. .81^3= .93

I added the min/raising chart to show u that my calc does what your chart describes( 2x/fld i know the 1 opponent is off)

so now following you in this thread you say we need to

6. .93/3 = .31

7. .31^3= .68

8. .68*.68*.68 =.31 or (.68*.67*.69) or (.70*.66*.68)

**(an avg of .68)**

So if i see a scenario like this and all my opponents and all my remaining opponents fold 68%+ to 3bets, I can think i can exploit this situation with ATC

for some reason i think we may of did a step twice...(thinking we should of skipped 6&7)& and did 8. with .93

I hope i got this all right! if not, thats ok, i will try try again.

Thank you for your time

Last edited by RWPKRPLR1 on Wed Jan 01, 2014 8:46 pm; edited 1 time in total

**RWPKRPLR1**- Posts : 882

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