let's just say we're dealt JJ UTG at 9player table
According to cardfight.com, JJ has:

winning equity vs 95.8% of villain's range
coinflipping vs 2.9% range (AQo+, KQo, AQs+, KQs+, JJ)
and dominated by 1.4% range (AA, KK, QQ)

how do i calculate (roughly)
A.) odds that at least one remaining player holds a hand that dominates JJ (1.4% range)
B.) odds that at least one remaining player holds a hand that is coinflipping vs JJ (2.9% range)


I just need the basic formula so I can test out the value of certain hands based on position and players left to act. Let's not account for the two jacks i'm holding in the calcs b/c that's going to confuse the crap out of me

is this close enough for a rough estimate?

2.9% range x 8 remaining players = 23.2% chance of flip
and 1.4% range x 8 players = 11.2% chance of domination


sorry, i'm like math ******ed

(50*49)/2 = 1225 total combos
18 total combos of QQ+

So the probability that a given player will be dealt a hand in the range {QQ+}, given that you've been dealt JJ is 18/1225 = ~0.0147 = 1.47%

The probability that no players are dealt a hand in said range is

p(plyr1 not dealt QQ+) * p(plyr2 not dealt QQ+) * p(plyr 3 not dealt QQ) * ... etc.

The probability that at least one player is dealt QQ+, is the compliment of this probability, or 1 minus the above expression.

The probability that plyr1 is not dealt QQ+ is 98.53%

It gets pretty complicated if you take blockers into account, because if plyr1 is dealt a hand in the weak range, it slightly affects the probability of the next player being dealt a hand in that range, due to combinations. So I won't worry about that. This calculation will therefore be slightly inaccurate.

So... the probability that at least one player is dealt QQ+ at a 9 player table, given that you've been dealt JJ, is...

1 - (0.9853)^8 = 0.1117 = 11.17%

I wouldn't bother with this, as the result will not be a useful metric. Are you having trouble dealing with players 3 betting you? I ask because JJ is a standard open raise at any table from any position even for beginners.

Yeah, your math is wrong.

It's a coincidence that 1 - (0.9853)^8 ≈ 1.4 * 8.

The probability that 2 events with probability 0.5 occur is not 0.5 * 2, rather, it is (0.5)*(0.5), or (0.5)^2.

On the remote chance that someone is interested in how close this approximation is, a simulation of 10 million hands gave a result of 11.27%. Of course a simulation is also an approximation, but I doubt very much if we're both off by any meaningful amount.`

jj was just an arbitrary example, i was just looking for the basic formula so i can calc odds of being ahead/coinflipping/dominated during late stages of tourney with certain hands based on players left to act.

thanks for the clarification, i think i understand the basic process now for rough calculations.

For the exact answer, it's not a simple formula. For an estimate, what you and timidcynic did are both good.

The reason OP's estimate is good is, its error is the overlap, and the overlaps have low probabilities (therefore the error is low). Timid showed that you can't say .5*2 when flipping a coin twice, but the reason that's a bad estimate for the coin example is that the overlap has a 25% chance.

That makes a lot of sense heehaww. Thanks.

Just for fun, here is a graph of the two functions f(x) = n*x and f(x) = 1 - (1 - x)^n

where n represents the number of players not including hero (in OP's case, n = 8)

and x represents the probability of one player being dealt a hand in a particular range (in OP's case, the range is QQ+ and x =1.47% = 0.0147)



As you can see, n*x (blue line) is a reasonable approximation for 1 - (1 - x)^n (red line) for x < 0.05

and the lower n is, the better the approximation is

this formula is going to help immensely with open shove decisions late in tourneys once i do the calculations.

now i'll be able to see exactly where i'm at for each hand relative to position. sweet.

Weird question on slightly on topic but seems like thread is over so no point in starting a new topic.
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Its been about ten years since I read anything serious on holdem pot odds etc. Got into a discussion with a fish who asked me this the other day.

Can someone explain whether we should ever deduct the volume of hero's hole cards + villains hole cards as well as board cards when doing out to pot odds probability calcs? And some rationale as to which method would be correct and why?

Obv we only know our own cards at any point in time, i get that.

What?

What do you mean by VOLUME of hole cards?

Did you mean OUTS to pot odds calculatIon?

Are the two methods-- include volume and do not include volume?.

Maybe with some more explanation, you can also include a specific example. More likely to get a response that way.

Sorry I dont think very much. Plus I speak my own language very often. Which I sometimes confused myself with.

Volume ie number of villain hole cards from deck. to reduce the start probability milestone.

Basically should we start from 49 cards to calc pot odds via outs? (just flop)
- OR is its correct to start from 47? (flop+hero)
- OR is it moar accurate to begin from 45? (flop+hero+villain)
- etc

There are varying misleading charts out there, theoretically speaking - which method is correct from a math perspective.

Ok. You are using outs to estimate your equity or winning chances and you will compare that with pot odds to determine how to proceed. Assume you do this after the flop. Then, you know at least 5 of the 52 cards, so you use 47 cards as unknowns. If you make an assumption as to what villain specifically has, you would subtract 2 more and end up with 45.

Example: You hold Qs Js and flop is 8s 5s Ad. Villain goes all in. You have a four flush giving you 9 outs plus a backdoor straight draw. You believe you need to flush to win as you think villain has either AK or a set. Since you have no idea as to villain’s suits, you estimate your winning chances with 9 outs (and maybe 1/2 more for the straight) using 47 unknown cards. If you were to use the 4x rule and estimate equity of 36%, in actuality that is equivalent to using 50 cards; in this case, it is a very good approximation

Thanks for the reply. I understand the outs >> probability >> odds process very well.

What I don't know and I am trying to get a respected view on; is if it is good practice to factor in villains cards to the calculation for a small edge to increase the odds by ~20% to create a "tolerance" if you will?

Or if using 45 unknown cards instead of 47 is a flawed view?

I understand the process you mention, ie.
= 12 outs
= (12*3)+6 (or multiple by 4 when <6 outs for improved accuracy based on 49 unknown cards)
= (100-42):42 OR 58:42 OR 1.4:1

Thank you.

I'm not certain what you are asking but as far as known cards are concerned, you always remove those when doing calculations. Those would be your hole cards and any revealed board cards or accidentally revealed cards in live games.

As far as Villains range goes, you need to assign the Villain a range based on his actions, history (frequencies), assumptions of player level, etc. then make your best guess. Then you would remove the known cards from his range to see how likely he is holding certain combos.

If villain hasn't acted yet, you can't make any guesses so his range would be all combos that didn't include the known cards.

Is this what you were asking?


BTW, I have worked out a method where you can make estimates based on multiple ranges (even folds) from multiple other players who have acted as well.

http://forumserver.twoplustwo.com/15...olved-1418266/

Unfortunately, I've been side tracked from getting closer to releasing a final product so it could be some time before getting something out for the public.