Ask Spoony About Poker Math

[spoonitnow]

It's been a while since I had a decent BC thread. Gogogo.

[Ragnar4]

I want the math behind the 3-bet and why it's the size it is.

What are our boundries to a 3-bet, and cover multiple situations.

what about the 4-bet?

[Keith]

I've worked out that flush draws are fractionally more likely to come in 6 max than full ring and I'm assuming that when you get the flush that it is more likely to hold up. Some confirmation of my maths would be very handy.
assume you are dealt two hearts. At a 6 max table 12 hole cards are dealt and 1 in 4 will be a heart on average, therefore 3 hearts on average will be in the hole cards. If two hearts are dealt on the flop this then leaves 8 hearts in the deck of 37 cards giving the chance of hitting the flush before the turn at (8/37)+(8/36)=43.84% or at the river is 8/36= 22.2% with the chance that people dealt a single card on the flop hitting their flush is (8/37)*(7/36)=4.2% before the turn and 19.4% (7/36) before the river if they hit the heart on the turn.

at a full ring 10 player table using the same calculations , 20 hole cards , 5 of which will be hearts, 2 hearts on the flop , leaves 6 hearts in a deck of 29 cards giving (6/29)+(6/28)= 42.12% before the turn,and 6/28=21.4% before the river. The single cards will hit (6/29)*(5/28)=3.69% before the river and 17.85% before the river if they hit a heart on the flop.
using the same principles for a flop with 3 cards of a suit it comes up with the following figures.

                                                    chance of hitting flush with 
                       hearts in             2 hearts                 1 heart
            hole cards    flop    deck      turn     river         turn       river
6max        3               2        8        43.84% 22.2%                    4.2%
            3               3        7          100%  100%           38.36%   19.4%
10 max      5               2        6        42.12% 21.4%                    3.69%
            5               3        5          100%  100%           27.40%  13.88%

These figures assume average numbers , whereas in reality number of individual suit cards in the hole cards will i expect have bell curve distribution but as a rough approximation can we rely on these figures . Are my numbers correct and if so when we hit the flush on a 6 max table can we assume that in general that the flush will hold up?.

[jackvance]

I'm pretty sure your math is wrong Keith, because you have to keep all unknown cards as one big pool of unknown, whether they're in someone else's hand or in the deck. Since you are assigning a discrete number of hearts to people you're basically invoking a rounding error.

I want the math behind the 3-bet and why it's the size it is.

What are our boundries to a 3-bet, and cover multiple situations.

what about the 4-bet?

I don't think there's any solid math here because it's a combination of a lot of factors. You don't want to give people odds to draw out on you (which is why minraising sucks), but you don't want to raise too much (like 10x their raise, that'll sure cut their odds) because then you're like never getting a call from any worse hands, or if you don't have a premium (AA/KK) and you have to lay it down if you get 4bet it cost you more, etc. If you squeeze over a raiser and some callers, you will want to raise more because they're all getting better odds to draw out on you, especially if one of them calls, the next ones are getting better and better odds since the pot grows larger compared to what they have to call to see the flop.

Now what these odds are that you want to cut, is not so easy to determine: it has to do with stack sizes, how often you'll stack off (with say AA if they draw out on you and hit a set), etc..

For a very simple example, 100bb stacks, villain has 44 and raises 4bb. You have AA and reraise to 12bb. Let's assume you stack off 100% of the time if he hits his set, and he folds the flop every time he doesn't hit a set. (we already made an incorrect assumption, but somewhat plausible - and we're ignoring the redraw to your A to overset him) He has like 12% to hit his set I guess, so his EV if he calls is (0.12*104bb)-(0.88*8bb)=+5.44bb so he has odds to call under these assumptions. If stacks were 50bb, (0.12*54bb)-(0.88*8bb)=-0.56bb, he can't call profitably. But what if you had AK instead of AA, you're not stacking off unless you hit your A or K (maybe).. or in general if you add whole ranges of hands to people, and then try to make assumptions about how much you win/lose in this or that case, it becomes rather vague. Well maybe spoonitnow can add more here

edit: I wanted to recalculate with the odds of you oversetting villain and stacking him because it's rather significant. Odds he hits his set are (2/48+2/47+2/46)=0.1277 and the odds you spike your ace when he spikes his set are (2/47+2/46+2/45+2/44)=0.176. So our equation becomes: (0.105*104bb)-(0.022*96bb)-(0.88*8bb)=+1.76bb so it's a lot more marginal if he calls with 100bb stacks. For 50bb it's (0.105*54bb)-(0.022*46bb)-(0.88*8bb)=-2.38bb.

[spoonitnow]

Keith_MM, your ideas are off for a few reasons, and I'm going to try to explain why as clearly as I can. I'll just look at your 6-max example only since the ideas work similarly at tables with more or less people.

assume you are dealt two hearts. At a 6 max table 12 hole cards are dealt and 1 in 4 will be a heart on average, therefore 3 hearts on average will be in the hole cards.

The first problem is in this premise. If we assume that we are dealt two hearts, then there are only 10 hole cards left because we know what two of them are. That leaves 11 hearts out of the 50 cards left in the deck that could be distributed amongst the remaining five players. If we also know that we flop a flush draw, then we know that there are 9 hearts left out of 47 cards that we haven't seen yet. That would be your starting point if you were to do this type of analysis.

I'll let you go from there.

[spoonitnow]

I want the math behind the 3-bet and why it's the size it is.

What are our boundries to a 3-bet, and cover multiple situations.

what about the 4-bet?

We size our 3-bets typically to balance risk/reward with not giving our opponent good implied odds (but still tempting them to make the incorrect call). It's much like any other bet or raise.

[Keith]

if the 12 hole cards are dealt blind , on average there will be 3 of each suit randomly distributed between each of the players.I'm struggling to see how picking your cards up and seeing that they are suited can then affect the chances of other peoples cards being of individual suits since they have already been dealt.

[spoonitnow]

if the 12 hole cards are dealt blind , on average there will be 3 of each suit randomly distributed between each of the players.I'm struggling to see how picking your cards up and seeing that they are suited can then affect the chances of other peoples cards being of individual suits since they have already been dealt.

Because you're not concerned with the average case before you see your cards, you're concerned with what the other players will have on average after you're already dealt 2 hearts.

A slightly different example which is a lot easier to understand is something like this: suppose you're playing heads up Omaha, and you are dealt AAAA. On average, how often does your opponent have an Ace? Now ask the same question when you're dealt AAAx, AAxx, Axxx, and xxxx.

[jyms]

Whats the chances that if you have 2 hearts and your opponents hold 3 avg) that you will see 2 more hearts on the flop? What's the chances that if they hold none, that you see 2 on the flop. Maybe having 2 and getting two makes your chances of having your opponents hold none greater.

[JR9477]

What does X/X mean over (whatever) number of hands mean?

Is it VPIP%/PFR%? I see it all the time and feel silly for not knowing for sure.

What does X/X mean over (whatever) number of hands mean?

Is it VPIP%/PFR%? I see it all the time and feel silly for not knowing for sure.

yes, also X/Y/Z might mean vpip/pfr/af
although I use afq myself

[TheScientist23]

AF? AFQ?

Not exactly on topic, but I don't know it.

For the flush draw scenario:

KNOWN: 2 suited hole cards
KNOWN: 2 suited flop cards

2+2 = 4 suited cards already known
13-4 = 9 suited cards left in deck and opponents hole cards (they are grouped together all as unknown)

52(total) - 2(hole) - 3(flop) = 47 (left in deck and opp. hole cards)

Chance of hitting flush on TURN or RIVER:
(9/47) + (9/46) = 38.7%

Change of hitting flush on RIVER if TURN was a miss:
(9/46) = 19.6%

I'm pretty sure that is all correct. Am I right?

What chances of getting dealt 2 suited cards?

(13/52)X(12/51) = 5.8%

not sure if that's the right way to do it.

What about figuring out what your chances of being dealt 2 suited cards, one of which is an Ace? this is what I'm not sure about at all.

are either of these correct?

13/52 chance of being dealt a specific suit.
1/52 chance of being dealt the Ace of that suit.
(13/52)X(1/51) = .5%

OR

4/52 chance of being dealt an Ace
12/51 chance of being dealt a card of the same suit
(4/52)X(12/51) = 1.8%

They can't both be right, obviously, so which one is it, if either?

Also, what is the math to figure out if you would be dealt 2 connect cards? What about suited connectors?

[spoonitnow]

Also, what is the math to figure out if you would be dealt 2 connect cards? What about suited connectors?

There are (52*51)/2 = 1326 possible starting hand combinations in hold'em. So count out the combinations of hands you want to know, and you can figure out the % by dividing that by 1326.

For example, if we want to know what percent of starting hands {AK, KQ, QJ, JT, T9, 98, 87, 76, 65, 54, 43, 32} is, then we know there are 12 hands and each hand has 16 possible combinations, so that's 12 x 16 = 192. Now we do 192/1326 and get 0.1448 = 14.48% of starting hands.

If you don't know how to count hand combinations, read this http://www.flopturnriver.com/phpBB2/...tc-t75711.html

[Parasurama]

The probability of being dealt suited cards is 23.5%, this is calculated by multiplying the probability of being dealt a card by the probability of the second card being of the same suit, that is: (52/52)*(12/51)=23.5%.

The probability of being dealt a suited ace is more complicated since the order you get dealt the ace matters. You might think that it's just (52/52)*(1/51), which is the probability of being dealt a card times the probability of being dealt the ace suited to that card. But this discounts the probability of being dealt the ace first. You might also think that it's (4/52)*(12/51), which is when you get dealt the ace first, but this of course discounts the possibility of getting dealt an ace second. To solve a complicated conditional probability it helps to calculate the probability of the desired event not happening, which is often much easier, and subtract the result from 1.

So: probability of being dealt two non-aces: (48/52)*(47/52)=.851
probability of being dealt one ace and an unsuited second card: (4/52)*(39/51)=.059
probability of being dealt one non-ace and a unsuited ace second: (48/52)*(3/51)=.054
Probability of NOT being dealt a suited ace=.851+.059+.054=.964, therefore the probability of being dealt a suited ace=1-.964=.036=3.6%

A much simpler way to do both of these problems is to simply calculate the number of ways to arrange these hands and divide into the total number of hands possible. There are (52*51)=2652 possible starting hands in hold 'em.

So, the probability of being dealt suited cards is the number of ways to arrange suited cards/2652. The number of ways to arrange a suited hand is (13*12*4)=624. Thus, the probability of being dealt a suited hand is 624/2652=.235 or 23.5%, the same result as above.

The probability of being dealt a suited ace is the number of ways to arrange a suited ace/2652. For each ace in each suit there are 12 cards to suit with. For each non-ace (12) in each suit there is 1 ace to suit with. (4*(1*12+12*1))=96. The probability of being dealt a suited ace is 96/2652=.036=3.6%, the same as the result above.

Hope this helps. I'm not completely sure that I did the second one correctly but since two different approaches yielded the same exact answers (the unrounded solutions are exactly the same), I felt confident enough to post it. I'm absolutely certain that the probability of being dealt suited cards is 23.5% though.

[jackvance]

You might also think that it's (4/52)*(12/51), which is when you get dealt the ace first, but this of course discounts the possibility of getting dealt an ace second. To solve a complicated conditional probability it helps to calculate the probability of the desired event not happening, which is often much easier, and subtract the result from 1.

Just multiply it by two, you can flip the cards it won't change the probability. So (4/52)*(12/51)*2=3.6%.

Or another simple way is that for the second case you deal a non-ace and then the ace of the same suit, so (4/52)*(12/51)+(48/52)*(1/51)=3.6%

[Parasurama]

true haha I feel dumb now

[Thunder]

1) How do you work out your post flop equity against an opponent's range before you run out of time?


2) Should you always call pre flop ATC if the odds offered are better than 2:1? Eg: 1 min raise, 3 callers, and you're sat in the BB with Q7os. Mathematically you're priced in but.......

[spoonitnow]

1) How do you work out your post flop equity against an opponent's range before you run out of time?


2) Should you always call pre flop ATC if the odds offered are better than 2:1? Eg: 1 min raise, 3 callers, and you're sat in the BB with Q7os. Mathematically you're priced in but.......

1) It depends on the situation, but here's an example way to think about it if you're making a call. Put your opponent's range into groups that have similar equity against your hand, keeping in mind how weighted each group is. From that point it's just a weighted average.

2) The simple answer is no, not always. It depends on the players, your position, and the playability of your hand. In the spot above, I would call with 72s before I would call with Q7o.

I'm losing money in NL with final hand: two pair
so I don't know about this whole Q7o business

[Keith]

I'm losing money in NL with final hand: two pair
so I don't know about this whole Q7o business

funny you should mention this , but from personal experience , if i flop two pair , I seem to lose the hand an awful lot more often than if I make two pair on later streets. No stats to back this up and it could just be that I'm now noticing my losses on the flopped two pair and the winners go through unnoticed.

I'm losing money in NL with final hand: two pair
so I don't know about this whole Q7o business

funny you should mention this , but from personal experience , if i flop two pair , I seem to lose the hand an awful lot more often than if I make two pair on later streets. No stats to back this up and it could just be that I'm now noticing my losses on the flopped two pair and the winners go through unnoticed.

actually when I flop one pair or better I win money
but FINAL hand two pair is a loser

[jackvance]

actually when I flop one pair or better I win money
but FINAL hand two pair is a loser

A bunch of them are when the board pairs so you really only have one pair, and also two pair is often a pay-off hand for stronger ranges. In my database 2p final hand is a winner could be variance or some subtle influence from your playstyle..

actually when I flop one pair or better I win money
but FINAL hand two pair is a loser

A bunch of them are when the board pairs so you really only have one pair, and also two pair is often a pay-off hand for stronger ranges. In my database 2p final hand is a winner could be variance or some subtle influence from your playstyle..

well that's because I set it as a final hand
a lot of the time when you have two pair, you don't go to the river

[Robb]

There are (52*51)=2652 possible starting hands in hold 'em.

You're entire post suffers from a "first this happens, then that happens" approach to combinations. There are "52 choose 2" combinations of starting hands, which is math-speak for the following formula:

"52 choose 2" = 52! / (2! 50!)

If the factorial notation is not familiar, let me know. The key is that the total number we're choosing from (in this case, a deck of cards) is the factorial component in the numerator, and the two components in the denominator sum to it.

Anyway, a good spreadsheet or scientific calculator has combinations, so you don't need to remember the formula to do the calculations. You just need to know what they mean.

Let me step through your post above using "choose notation" with some short explanations.

The probability of being dealt suited cards is 23.5%, this is calculated by multiplying the probability of being dealt a card by the probability of the second card being of the same suit, that is: (52/52)*(12/51)=23.5%.

The probability of being dealt 2 hearts, for example, is:

"13 choose 2" / "52 choose 2"

But there are four suits, so we multiply this by 4 to cover all the possibilities:

(4 * (13 * 12 / 2 )) / (52 * 51 / 2) = 12 / 51 = .235294118

Some folks use "4 choose 1" to select the 1 suit used of the four, the "13 choose 2" to select the two cards.

The probability of being dealt a suited ace is more complicated since the order you get dealt the ace matters. You might think that it's just (52/52)*(1/51), which is the probability of being dealt a card times the probability of being dealt the ace suited to that card. But this discounts the probability of being dealt the ace first. You might also think that it's (4/52)*(12/51), which is when you get dealt the ace first, but this of course discounts the possibility of getting dealt an ace second. To solve a complicated conditional probability it helps to calculate the probability of the desired event not happening, which is often much easier, and subtract the result from 1.

This is easier to do with counting. There are 12 card values that can be placed with an ace, so there are 48 total Axs hands (12 in each suit). So take 48 and divide by 1326 = 0.036199095.

So: probability of being dealt two non-aces: (48/52)*(47/52)=.851
probability of being dealt one ace and an unsuited second card: (4/52)*(39/51)=.059
probability of being dealt one non-ace and a unsuited ace second: (48/52)*(3/51)=.054
Probability of NOT being dealt a suited ace=.851+.059+.054=.964, therefore the probability of being dealt a suited ace=1-.964=.036=3.6%

Correct answer. Unsure of work.

A much simpler way to do both of these problems is to simply calculate the number of ways to arrange these hands and divide into the total number of hands possible. There are (52*51)=2652 possible starting hands in hold 'em.

No, there are "52 choose 2" possible starting hands, or 1326.

So, the probability of being dealt suited cards is the number of ways to arrange suited cards/2652. The number of ways to arrange a suited hand is (13*12*4)=624. Thus, the probability of being dealt a suited hand is 624/2652=.235 or 23.5%, the same result as above.

Both of your calculations are 2x the actual answer. So division canceled out two proportional errors into the correct answer.

4 * "13 choose 2" is 4 (13 * 12 / 2) = 312. See above for total starting hands.

The probability of being dealt a suited ace is the number of ways to arrange a suited ace/2652. For each ace in each suit there are 12 cards to suit with. For each non-ace (12) in each suit there is 1 ace to suit with. (4*(1*12+12*1))=96. The probability of being dealt a suited ace is 96/2652=.036=3.6%, the same as the result above.

There are 48 suited ace hands (4 suits, 12 non-ace cards per suit). Again, your double-counting error happened both here and in your calculation of overall starting hands, so they cancel to the correct answer.

Hope this helps. I'm not completely sure that I did the second one correctly but since two different approaches yielded the same exact answers (the unrounded solutions are exactly the same), I felt confident enough to post it. I'm absolutely certain that the probability of being dealt suited cards is 23.5% though.

No offense, but you're making the same error when calculating both your numerators and denominators, so the answers are correct while the calculations are faulty.

he forgot in his calculation of starting hands that AK is the same hand as KA

[bigred]

I always laughed in my discrete mathematics class when we were talking about "choose" combinatorics because I'd want to yell pikachu I choose you. Always thought of triptanes too...

I always laughed in my discrete mathematics class when we were talking about "choose" combinatorics because I'd want to yell pikachu I choose you. Always thought of triptanes too...

let me guess, you laughed out loud when the number 69 came up

[jackvance]

No offense, but you're making the same error when calculating both your numerators and denominators, so the answers are correct while the calculations are faulty.

He's not necessarily wrong, it's human intuition that for example AcKc=KcAc but from a purely mathematical point of view they could be different, which is why his calculations yield correct results.

[Robb]

No offense, but you're making the same error when calculating both your numerators and denominators, so the answers are correct while the calculations are faulty.

He's not necessarily wrong, it's human intuition that for example AcKc=KcAc but from a purely mathematical point of view they could be different, which is why his calculations yield correct results.

This is a good point. I'm probably more concerned than I should be with mathematical elegance.