Suppose there are 10 players who each receive 5 cards. What is the average best poker hand among these ten players?

I believe that the question can be illuminated by thinking about the following questions:

1. What is the average 5-card hand?

2. What is the average best 5-card hand over 10 hands dealt with replacement when hands are numbered from 1 to N where N is the total number of possible 5-card hands? (That is, we introduce 10 players and take the maximum.)

3. What is the average best 5-card hand over 10 hands dealt without replacement when hands are numbered from 1 to N where N is the total number of possible 5-card hands? (That is, we introduce the concept of without replacement.)

4. What is the average best 5-card hand over 10 hands dealt without replacement when hands are dealt from an actual 52-card dealt? (That is, we introduce the concept of card removal effects.)

I have some ideas on some of these questions but welcome any thoughts on any of the above.

Do you want to know what hand is even money to beat the other nine or do you want to know the median hand of the winner?

I am interested in this question in all its glory, so I welcome any thoughts on either aspect of David's post.

- The break-even ("average") hand is interesting to think about.

- The median hand is also of interest.

Based upon my preliminary explorations, I believe that deriving the break-even (average) hand is much more tractable.

Here is my top-of-mind thinking:

There are C(52,5) = 2,598,960 possible 5-card hands. Number them from 1 (7c 5c 4c 3c 2d) to 2,598,960 (As Ks Qs Js Ts). Then the midpoint hand would be number 1,299,480. Where does that fall?

By my count there are:

4 Royal flushes
36 Straight flushes
624 Four of a kind
3,744 Full house
5,108 Flushes
10,200 Straights
54,912 Three of a kind
123,552 Two pair
1,098,240 One pair
1,302,540 No pair

So I think hand number 1,299,480 is around Ac Kc Qc Jc 7d.

Suppose you select K integers out of {1, ..., N} with replacement where both K and N are large and K<N.

Let M be the Maximum of these K integers.

Then it is straightforward to derive:

E(M) = (KN+1)/(K+1)

We have K=10 and N=2,598,960 so that:

E(M) = 2,362,691.

I will not try to find what hand corresponds to that number at this point.

Suppose you select K integers out of {1, ..., N} without replacement where K<N.

Let M be the Maximum of these K integers.

Then it is straightforward to derive:

E(M) = (KN+K)/(K+1)

We have K=10 and N=2,598,960 so that:

E(M) = 2,362,691.8181818181

which I believe corresponds to a pair of aces (I think AAQ84).

Alas, I do not have any thoughts on how to go about deriving this answer. Other than doing a simulation over millions of deals.

My intuition is that the "card removal" effect is very tiny and is unlikely to budge the average found in the previous post.

Moving into the realm of the median:

Median of the maximum of K iid uniform variables in {1, ..., N} is given by:

Median(M) = N * (1/2)^(1/K)

Here we have K=10 and N=2,598,960 so that:

Median(M) = 2,424,915

which I believe corresponds to two pair 88663.

I coded up a simulation of dealing 10 5-card hands from a 52-card deck to take into account card removal effects.

The median of the best of the ten hands over 1,000,001 trials was 7755T.

I am not sure how "stable" this result is so I will post further results later.

I should have said above that it is almost certainly true that the median of the maximum of 10 samples with replacement is virtually equal to the median of the maximum of 10 samples without replacement when the samples are drawn over 2,598,960 possibilities. This is probably obvious and is demonstrated by the fact that the respective expected values are less than one apart.

Continuing from above, here are the results of the simulations from dealing 10 5-card hands from a 52-card deck, with the number of simulation trials given first:

1,000,001 7755T (posted previously)
1,000,001 7755T
1,000,001 7755Q
1,000,001 77559

So it seems like card removal effects are not insignificant as there is a noticeable difference between this median hand and the median hand found above based upon the pure math of sampling from 1 to N (that is, not real cards) of 88663 two pair.

I am a little surprised by this.

I'm currently running a simulation of the same. It's taking about 8 minutes per Million on my cheap laptop. I asked it to do 10M. Currently at 6.5M, Median is 7755Q

I got:

10,000,001 7755Q

10 hands dealt, odds you have the best hand if you're holding:

7H,7S,5D,5C,QS: 51.4%
7H,7S,5D,5H,QS: 51.3%
7H,7S,5H,5S,QS: 51.2%

(based on 50,000 simulations each)

No....

I believe 2,424,915 corresponds to 88225

88224 and below = 2,424,828
88225 144 combinations

---------------------------------------------

tried to confirm with simulations, but a fresh shuffle for each hand is running much slower for me. for 1M runs I get:

1,000,001 88227

Very nice.

Yes, I agree that hand 2,424,915 actually corresponds to 88225.

Thanks for doing that. And all of your simulations to capture the median too.

No problem. Teaching myself Python, so good exercise.

(left the simulation running, hit 88225 at 2.05M)

out of curiosity, I decided to check for Median best hand out of 7 - 7card stud hands (normal deal, meaning no replacement), I get:
500,001 76543

We can easily deduce the approximate median hand, if we are talking about an ordered list of all hands from lowest poker value to highest. Since no-pair hands are very slightly over 50% of all hands, we know for sure the median will be one of the highest no-pair hands possible, which is certainly going to be AK high plus some other 3 cards.

For the average I'm not sure the thread has agreed on a definition. You can't average abstract poker hands, they need a point value assigned or something similar. I'd suggest using one of the common point value systems to find the "type" of hand that is sure to be the average. Then we can probably narrow that down further using the distribution.

The thread topic isn't Median single 5 card poker hand.

Obviously my post above was referring to the median hand dealt, not the median winning hand among 10. That's a harder problem.

Edit: and I was writing this at the same time as your post above. Good catch.

10 hands dealt from a standard deck, odds you have the best hand if you're holding:

6H,6S,5H,5S,TD: 49.97% best hand equity

based on 10,000,001 simulations
4,996,694 wins
72 ties
5,003,236 losses

good timing

This is neat. So the median winning hand is 7755Q but the hand with 50% equity is 6655T.

Very interesting.

Handy straight blockers it would seem