# The probability of the ace with the best kicker depending on the number of opponents #### BelFish

##### Legend
Loyaler
I calculated the probabilities for the general case when we have an ace with some kind of kicker, that at least one of the opponents will have an ace with the best kicker.

For example, if we have 8 opponents at the table and our hand is A2, then the probability is 66.492%

Prior to that, poker sources believed that this probability is equal to 60.706%.

I deduced the exact formula and calculated for any hand (from A2 to AK) and any number of opponents (from 1 to 22).

I will not write the formula yet, because this is a rather interesting task, maybe someone wants to try to solve it himself  #### Edu1

##### Legend
since you are the math guy, I think is better you shows and everybody will learn, I have no idea how this formula is made 😎

• Tracid #### BelFish

##### Legend
Loyaler
• alien666dj #### Tracid

##### Squirrel Tiger
Community Guide
Well, your calculations make for interesting comparison (if little else)...

It seems at a glance that your results start from pretty much exactly the same and end up at a roughly ~6% difference.

Here's the alleged incorrect 'standard' for anyone interested in any correlations... ~EDiT:- I'm not sure when you're going to find yourself at a table with 22 opponents though! :icon_scra #### BelFish

##### Legend
Loyaler
Okay, i'll give a hint. It is necessary to count through combinations. This concept is from the section of probability theory - combinatorics.

The easiest way is to start counting this probability as the difference between the probabilities that opponents have any aces at all, and that opponents have aces with the same or worse kicker:

P(N, n) = P(Ax) - P(A2..An)

N is the number of opponents, n is the face value of the card (rank, for two n = 2, for king n = 12).

And then it is already calculated through combinations of C(k, m)

For example, combinations of 2 cards out of 52 total C(2, 52) = 1326 - so many different variants of pocket cards. #### BelFish

##### Legend
Loyaler
I looked at the case with 22 opponents - for fun ))
There can be no more than 22 opponents in a home live game Only 23 players. 23 two cards each + 5 cards on the board = 51 cards. And one card to "burn". #### BelFish

##### Legend
Loyaler
I didn't notice the screenshot with the calculation right away (it didn't load on the page at the beginning).
Yes, it’s this table that i’m mean up to when i say that it’s wrong in poker sources.

My results are absolutely accurate. They were checked by the programmer by numerical methods, after i calculated the table using the formula. #### Tracid

##### Squirrel Tiger
Community Guide
I'm not in a position to confirm nor deny your formula/calculations but the difference is marginal enough for me not to be overly worried. A couple of % in the scheme of things is practically nothing.

Out of interest, where do you think the original calculations went wrong? #### 0546474

##### Legend
Loyaler
I am not good at mathematics, so do not judge strictly !!! 1326 different variants of pocket cards - as I understand this calculation for 52 cards in a deck ??? If this is so, then it is not entirely clear to me why you take into account these two pocket cards that cannot be present in the deck in any way, since they are in your hands and there are 50 cards left in the deck !!! #### BelFish

##### Legend
Loyaler
Out of interest, where do you think the original calculations went wrong?
I can even give the formula by which the author of that table calculated. Because according to this formula, all the numbers from his table are obtained with an accuracy of thousandths. I'll post it later. #### BelFish

##### Legend
Loyaler
I am not good at mathematics, so do not judge strictly !!! 1326 different variants of pocket cards - as I understand this calculation for 52 cards in a deck ??? If this is so, then it is not entirely clear to me why you take into account these two pocket cards that cannot be present in the deck in any way, since they are in your hands and there are 50 cards left in the deck !!!

I do not take them into account, i just gave an example for the number of permutations for the case of dealing 2 cards out of 52. Just to show for those who do not know what the number C(k, m) is.

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I'm not in a position to confirm nor deny your formula/calculations but the difference is marginal enough for me not to be overly worried. A couple of % in the scheme of things is practically nothing.

Yes, an inaccuracy of 6% doesn't hurt much. Just for the sake of academic interest, i wanted to get a true table, in which completely correct results 1

#### 1nsomn1a

##### Visionary
Congratulations on the more correct calculations  #### Tracid

##### Squirrel Tiger
Community Guide
I can even give the formula by which the author of that table calculated. Because according to this formula, all the numbers from his table are obtained with an accuracy of thousandths. I'll post it later.
That doesn't really answer my question... Yes, an inaccuracy of 6% doesn't hurt much. Just for the sake of academic interest, i wanted to get a true table, in which completely correct results Well, to be accurate, it varies from 0% difference up to 6% in the extreme. #### BelFish

##### Legend
Loyaler
I have not seen the formula of the author of the table, but here is how he almost certainly calculated for case A2 against 9 opponents:

P = 1 - {[C(2, 47) + C(1, 3)*C(1, 3)]/C(2, 50)}^9 = 1 - (1090/1225)^9 = 0.65037 or 65.037%

So he calculated for the case of A8 against 5 opponents:

P = 1 - {[C(2, 47) + C(1, 3)*C(1, 27)]/C(2, 50)}^5 = 1 - (1162/1225)^5 = 0.23202 or 23.202%

By analogous calculations, he obtained the rest of the results.

Last edited: #### BelFish

##### Legend
Loyaler
For now, i will not give my formula for the general case, giving a chance to those who want to try to get it on their own. Because for those who love mathematics, it is a great pleasure to solve some relatively difficult problem. I'll write a formula at the end of the week if no one solves this task  #### Tracid

##### Squirrel Tiger
Community Guide
I have not seen the formula of the author of the table, but here is how he almost certainly calculated for case A2 against 9 opponents:

P = 1 - {[C(2, 47) + C(1, 3)*C(1, 3)]/C(2, 50)}^9 = 1 - (1090/1225)^9 = 0.65037 or 65.037%

So he calculated for the case of A8 against 5 opponents:

P = 1 - {[C(2, 47) + C(1, 3)*C(1, 27)]/C(2, 50)}^5 = 1 - (1162/1225)^5 = 0.23202 or 23.202%

By analogous calculations, he obtained the rest of the results.
Don't get me wrong, I'm impressed with your calculus...

Just enquiring as to where is the discrepancy between the previous formula and yours..?

Preferably without having to dig out a protractor & compass! :laugh:

For you to state your formula is correct and arrive at different results must mean the previous equation is incorrect.

That is the nature of my intrigue!  #### BelFish

##### Legend
Loyaler
It's just better to always know the true result. The formula of the author of the table is very approximate, but with fairly good accuracy. But a completely accurate calculation is possible, which i received  #### BelFish

##### Legend
Loyaler
Another thing that i thought then, realizing that the percentages in the table do not coincide with reality: these percentages are indicated with an accuracy not even to tenths, but to thousandths. Therefore, i assumed that the author considered the formula to be absolutely correct.

I'm not saying that he is a bad mathematician. Quite the opposite, good mathematicians very often make mistakes on the simplest things )))

For example, i met in the books of very famous mathematicians examples with "preference" (card game), where the calculations were (3/4)*(3/4), instead of (3/4)*(2/3), this is an example of the same discrepancy as in example with aces: 0.563 vs 0.5

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Since it's almost weekend, i decided that i would give more time (until the end of next week) to try to solve the problem on my own before posting the exact formula.

Try it, this is a really interesting example to solve, and it is not nuclear physics, but combinatorics, so everything is relatively simple  #### darthjewel

##### Rock Star
This is the forbidden science, spreaders of which should be burned on stake.
Everybody knows that poker is nothing but fun,fun,fun and trying to inject some math into our pure, undamaged brains is a sin.
Besides, this is sheer heresy, everybody plays against of 5 or 8 opponents,
yet in their secret formulas they use number 9.
Why, you may ask - as this is the inverted 6 - they symbol of we all know who.
Beware.

• Tracid #### BelFish

##### Legend
Loyaler
This is the forbidden science, spreaders of which should be burned on stake.
Everybody knows that poker is nothing but fun,fun,fun and trying to inject some math into our pure, undamaged brains is a sin.
Besides, this is sheer heresy, everybody plays against of 5 or 8 opponents,
yet in their secret formulas they use number 9.
Why, you may ask - as this is the inverted 6 - they symbol of we all know who.
Beware.

Any opinion is valuable ))

I also wanted to say that there are no online games against 9 opponents now. But in live poker you can even play against 22 opponents  #### darthjewel

##### Rock Star
But in live poker you can even play against 22 opponents For sure the tables are useful in such a case.
You will be like:
-Oh, it's my turn. Sorry, wait a moment please, I need to check one thing on my magic chart.
Ooops, I have forgotten, Gary is in the bathroom, thank makes 21 players,
aha, it is in the other pocket. Hmmm it seems that...
What ? Ted and Bob went to the loo as well ?
That makes 19, or 18 ? It is better to recount the people.
Please don't move and put that sandwich down, I am trying to count here.
Finally, thank you , 19!
Oh, hi Garry you are back, shoot, now there are 20 , where is my cue card ? -

Overall it is worth to remember that the more players are at the table, the weaker the hand is.
So the pocket aces AA are very strong in heads game, but weaker in 6 players table and weaker again with 9 players. #### BelFish

##### Legend
Loyaler
No. I think quite the opposite: you don't need to count anything at the table in the game, you just have to play by intuition And intuition will be good if you often solve math problems. I haven’t used any materials or charts for 17 years, but i just pressed the buttons and won )) #### BelFish

##### Legend
Loyaler
Okay, i'll give a hint. It is necessary to count through combinations. This concept is from the section of probability theory - combinatorics.

The easiest way is to start counting this probability as the difference between the probabilities that opponents have any aces at all, and that opponents have aces with the same or worse kicker:

P(N, n) = P(Ax) - P(A2..An)

N is the number of opponents, n is the face value of the card (rank, for two n = 2, for king n = 12).

And then it is already calculated through combinations of C(k, m)

For example, combinations of 2 cards out of 52 total C(2, 52) = 1326 - so many different variants of pocket cards.

If someone wants to try to solve, i give one more hint to the one i have already given.

The probability that at least one of the opponents will have at least one ace is made up of three parts: that one will have aces, two will have, and three will have:

P(A2..An) = P1(A2..An) + P2(A2..An) + P3(A2..An)

Using these hints, it becomes much easier to solve the problem. #### BelFish

##### Legend
Loyaler
I decided that i would extend the period allotted for the possibility of solving the problem until the end of the summer. I will definitely publish the answer on August 31, if no one tries to solve this task before. #### BelFish

##### Legend
Loyaler
P(N, n) = 1 - C(2*N, 47)/C(2*N, 50) - c(1, N)*{[C(1, 3)*C(1, 4*(n-1)-1)]/C(2, 50)}*[C(2*N-2, 47-1)/C(2*N-2, 50-2)] - c(2, N)*{[C(1, 3)*C(1, 4*(n-1)-1)]/C(2, 50)}*{[C(1, 2)*C(1, 4*(n-1)-2)]/C(2, 50-2)}*[C(2*N-2-2, 47-1-1)/C(2*N-2-2, 50-2-2)] - c(3, N)*{[C(1, 3)*C(1, 4*(n-1)-1)]/C(2, 50)}*{[C(1, 2)*C(1, 4*(n-1)-2)]/C(2, 50-2)}*{[C(1, 1)*C(1, 4*(n-1)-3)]/C(2, 50-2-2)}*[C(2*N-2-2-2, 47-1-1-1)/C(2*N-2-2-2, 50-2-2-2)]