In the break at a home game I play we play a game called 7 card turnover. 7 players pay £10 and get 7 cards and the best 5 card hand scoops the pot.

My question is what would be the average winning hand?

I would guess it is in the 2 pair/trips area. I've seen lots of boat over flush over straights etc but also seen it won by a pair of deuces. I presume it is a solvable question but I wouldn't know where to start running any kind of calculation.

A bit pedantic, but I think you'd be looking for the median winning hand, not the average.

I don't think there's any practical way to solve this mathematically; you'd have to find the answer via simulations on a computer.

http://www.propokertools.com/simulations?g=st

Stud Hi Simulation ?
600,000 trials (Randomized)
Hand Equity Wins Ties
6s6h6d7d8cKhAs 49.84% 299,014 0
* 10.06% 60,295 182
* 10.10% 60,479 194
* 10.06% 60,284 190
* 9.98% 59,798 195
* 9.96% 59,642 215

"*" is a random hand.

I don't know if 666 is average, but it does win about 50% of the time. BTW, this is 6-way rather than 7-way. Pro-poker doesn't seem to do 7.

An average is any measure of central tendency and includes median. It isn't wrong to say "the average hand". If you find yourself starting another post with a disclaimer about it's pedantic nature then please just click the x button on your browser and save us all the bore.

The best way I can come up with to answer the OP is using propokertools PQL. Please let me know if there is an easier syntax than:

select count(hiRating(player_1, seventh) >= rateHiHand("3h4c5s6d7h") ) as MEDIAN_IS_50
from game='studhi',
PLAYER_1='*',
PLAYER_2='*',
PLAYER_3='*',
PLAYER_4='*',
PLAYER_5='*',
PLAYER_6='*',
PLAYER_7='*'
where (winsHi(PLAYER_1) OR tiesHi(PLAYER_1))

Results:
Trials MEDIAN_IS_50
75704 38267 (50.55%)

The is not the average hand but instead the hand which achieves 50% win rate if you are dealt it. If you think about it then you will see that 50% of the time that 666 is dealt a higher hand than 666 wins but conversely we can not say that a lower hand than 666 wins 50% of the time. By dealing 666 to one of the hands you are also restricting the high hand capacity of other hands (lots of straights need a 6, lots of full houses need 666, quad 6666 cannot be dealt etc etc).

I hear you run bad!

I think it's a 10 high straight. This is an approximate solution, but I think it's more accurate than the methods suggested to date.

Here are the probabilities of getting various hands with seven cards:

Straight Flush 0.03%
Quads 0.17%
Full House 2.60%
Flush 3.03%
Straight 4.62%
Trips 4.83%
Two pair 23.50%
Pair 43.82%
Nothing 17.41%

It is almost true that knowing what type of hand someone has does not affect the probability of the type of hand another player has. That is, if you have a full house, it doesn't change my probability of getting a flush much. That's not true with specific hands, for example, if you have Aces over 7s full house then you've blocked three of my four Royal Flushes.

Therefore, we can approximate the probability of each of the hand types above being the winning hand. It's simply the probability of at least one player getting that hand type, and all the other players getting the same or lower hand types:

Straight Flush 0.22%
Quads 1.17%
Full House 16.61%
Flush 16.28%
Straight 19.50%
Trips 14.86%
Two pair 28.12%
Pair 3.23%
Nothing 0.00%

Those numbers are only exactly correct if each player is dealing from a different deck. But I think they're very close to what you get all dealing from the same deck.

Adding those up, 34% of the time someone gets a hand better than a straight, and 46% of the time no one gets a straight or better. The other 20% of the time, the hand is won with a straight.

Given than someone wins with a straight, here are the probabilities of the number of people with straights (5 or more have probabilities that round to 0.00%):

1 16.69%
2 2.58%
3 0.22%
4 0.01%

If one player has a straight, the average high card is 9.5 (a random pick from 5 to 14). If two players have straights the average winning hand has a high card of Jack. If all seven players draw straights, it will take a King high on average to win. If you take the weighted average of those possibilities, a Ten high straight is closest to the winning hand.

Note that there are 10,259,268 hands that beat a Ten high straight, and 122,907,290 that lose to it. So the chance that it beats a random hand (technically, the chance that it beats a random hand dealt from another deck, that isn't a Ten high straight; but it's pretty close to the chance that it beats a random hand from the same deck) is 92.30%. If I raise that to the 6th power, I get 61.82%, which is my estimate of the chance of winning the hand if I get a Ten high straight.

So the average winning hand has much better than a 50% chance of winning if you get it. That sounds strange, but suppose I write the numbers 1, 2, 3, 4 and 5 on slips of paper and put them into a hat. Two people draw numbers, higher number wins. The average winning number is 4 (there are 10 possible draws, 1 won by 2, 2 won by 3, 3 won by 4 and 4 won by 5; 1*2 + 2*3 + 3*4 + 4*5 = 40, 40/10 = 4). But if you draw a 4, you win 75% of the time.

The ten high straight is the average hand if you exclude ties.

Let me raise one issue with Aaron's approach (which I think is great). It seems as if Aaron is finding the "average" winning straight since he found that the overall median winning hand is a straight.

But, I think, Aaron is not using the information that 34% of the time the winning hand is better than a straight and 20% (permit me to round for the purposes of elucidation) of the time the winning hand is a straight. The median, of course, is the hand for which the winning hand is better 50% of the time (excluding "chunks").

So shouldn't we be looking for the hand in the straight bucket that is 16/20ths of the way from the top of the straights (34+16=50)?

For simplicity, if we assume only one person can get a straight, and that all straights are equally likely, then, I think, the 16/20th straight would be a 7-high straight (not a 10-high straight).

Hope this makes sense, but, of course, there is always the possibility that I totally missed something.

No, I don't think that's right. If someone has a hand better than a straight, then the rank of your straight doesn't matter.

If only one player gets a straight, and wins the hand, or loses it for that matter, half the time he will have better than the median straight, half the time he will hold worse. It doesn't matter if straights win a lot or a little, or if more hands are won by hands better than straights or worse than straights.

If two players get straights, then the median value for the higher straight is the straight that beats 2/3's of other straights. If three players get straights, then the median value for the higher straight is the straight that beats 3/4's of other straights. This ignores the small deviations due to card subtraction effects. If I have a Jack high straight, that reduces the ways to get anything other than a 5 or 6 high straight.

I am not sure I was clear. Are you understanding what I am saying?

Forget your hand ranking buckets and just write down every possible winning hand:

Royal Flush
King high straight flush
Queen high straight flush
...
Ace high straight
King high straight
Queen high straight
...
5 high straight
AAAKQ (non flush)
AAAKJ (non flush)
...

I thought the exercise was to identify the median winning hand.

Surely you are not saying that the median winning hand (if it is indeed in the straight bucket) is automatically or magically a 10-high straight.

Do you agree with that?

Thinking a little more about this, straights are the main exception to my initial thought that the type of hand I have doesn't affect the type of hand anyone else has.

With five card poker hands, all the other hands use all the cards in the deck equally, only straights and straight flushes deviate from deck frequency. That is, there are 13 x 48 = 624 different quad hands. Each of the 52 cards in the deck is in exactly 60 of them (12 hands where it is the kicker and 48 where it is one of the quads and each of the 48 cards of other ranks are the kicker). The same thing is true of full houses, flushes, trips, two pair and one pair. Hands with nothing will also have slightly skewed frequencies to offset the straights.

There are 10*4^5 = 10,240 straights, including straight flushes, but the cards are not equally distributed among them. As, Ks and 2s are in only 512 straights each (2 ranks, 256 ways to pick the suits of the other four cards, given the rank), Qs and 3s are in 768, Js and 4s are in 1,024 and Ts, 9s, 8s, 7s, 6s and 5s are in 1,280.

Therefore, knowing someone has a straight makes it more likely that there are more As, Ks and 2s remaining in the deck. That makes it very slightly easier to get pairs, trips, full houses and quads.

There are smaller effects like knowing someone has four of a kind makes it slightly easier to make pairs and other rank matches, slightly harder to make a flush.

But my guess is none of this is strong enough to displace the ten high straight as the median winning hand. I think we'll need a simulation to settle it for sure.

Yes, I think I understand. My disagreement is that I don't think you can forget the buckets.

Take an extreme case. 49% of the time the hand is won by a hand better than a straight, and 51% of the time it's won by a straight. If only one player gets that straight, you still expect it to be the median straight. The same thing is true if 49% of the time the hand is won by a hand weaker than a straight, and 51% of the time it is won by a straight.

You're thinking looking for the hand that has a 50% chance of winning if you get it. Then ranking the hands without buckets makes sense, although you then get severe computational difficulties due to card conflicts between hands.

I'm looking for the hand such that 50% of the time a better hand wins, and 50% of the time a worse hand wins. That's an easier question to answer, if you accept my bucket simplification.

Despite your lengthy (and good) post, you haven't responded to my post in the least.

---

Cross-posts.

No, I know what a median is, believe me.

You are simply not understanding my (simple) point.

Forget it, one of us (or both) is being obtuse.

I'm the one being obtuse, sorry. You're right.

I was thinking of the question, given that a straight wins the hand, what is the median rank of the straight? That only depends on the number of people who get straights, not on the probabilities of hands better or worse than straights.

But you're correct that the question should be, on all the deals, not just the ones won by straights, what is the median winning straight? As you say, that should be roughly the 20th percentile straight, or the 7 high. Accounting for the fact that there might be more than one straight might bring you up to an 8 high.

Thanks Aaron, your response has made my day (I thought I was losing my marbles)!

P.S. And I always enjoy your contributions to this website.

Because I disagreed with you? If that were a symptom of insanity, few people would be sane.

Hi Aaron (can't believe you disagreed with my solid PQL post)

If you look at winning hands (including ties) it's a 7 high str8 and if you take ties out (i.e. ties do not count as a winning hand then it's actually a 10 high str8.

You shouldn't be calculating the average str8 but the given percentile (as mentioned in above posts).

Straight makes a lot of sense intuitively since that hand gains the most benefit from 7 cards, I think.

This seems like such an easy Monty Carlo just parsing a random deck into 7 card arrays, but I've been sidetracked so much, lately I really shouldn't spend the time.

Since I still need to create code to read final hand strength anyway, I may try to squeeze it in but I can't make any promises.

Why would you code a Monte Carlo simulation when it can be done for free with PQL? See my first post.

I have never seen or used PQL before. In your first post, is the "75,704" the number of trials of the simulation? If so, that seems like a very small sample of the zillions of ways 7 7-card hands could be dealt. (I am not be critical, I am just asking.)