Probabilities of what the strongest hand is in N-player holdem poker?

I’m looking for a table/chart of the odds on what the strongest hand will be in N-player poker,
for N=2..10. Assume no folding (/everyone stays in till the end).
i.e. the respective odds that the strongest hand among ALL N players at the table is
StraightFlush/ 4OfaKind/ FullHouse/ Flush/ Straight/ 3ofAKind/ 2Pair/Pair/HighCard
tabulated for all the values N=2..10

Ideally want to see a stacked bar-chart which shows how they sum to 100% for each value of N;

I’m curious that as N increases, higher hands become more probable, so the odds of it being won by a single pair or even a highcard diminish. (This was motivated by a hand yesterday with 5 players where everyone saw the river, and the highest hands were only 2 pairs of Queens, even though there was a backdoor straight draw.)
Remember, I said assume everyone stays in and no folding, so ignore that aspect.

(This is a funky little problem to code and visualize on a rainy-day.)
@TMP: title line states "holdem poker".

No folding => this is poker without ANY gambling, bluffing, or indeed betting or a pot.
("the odds of a higher hand taking the pot is far lower than in a normal poker hand" – what are you talking about? If everyone stays in, the highest hand will always take the pot. But here we are not even concerned with the pot – just the probabilities of hands.)

I only want to see how the higher hands become more probable as N increases, and things like 2Pair, Pair, HighCard become weaker.

Here’s some groundwork:
for N players, we deal 5 community cards (CCs) + 2N hole cards (HCs). Thus M=(5+2N) total cards. When we exclude betting and folding, all 5 CCs are essentially dealt all in one go and become interchangeable, thus there are 5! permutations for any given set of 5 CCs and 2! permutations for each player’s HCs.
Hence there are 52*51*50…*(52-M)/ ( 5! (2!)^N )
distinct permutations of hands.
@TheMadProfessor. Now see the ambiguity: assume noone folds EVER (not even on the river). Then the highest hand ALWAYS wins, like I said. We’re purely talking about raw statistics on poker hands – no betting at all.
Correction: There are 52*51*50…*(52-M+1)/ ( 5! (2!)^N ) distinct permutations of hands.
(In response to some private correspondence,
do not overcount equivalent PERMUTATIONS of the same hands; only count COMBINATIONS:
- the 5! permutations of the community cards are equivalent (since there is no betting: flop,turn,river all dealt together)
- the (2!)^N different permutations of each player’s hand are also uninteresting. So only consider each player’s hand ranked so that the higher card is first (and some arbitrary suit order e.g. C<D<S<H)
And only consider the 5 CCs ranked highest-to-lowest and C<D<S<H.

This consideration grows more important as N grows, because otherwise you generate 2^N times as much unnecessary work, i.e. up to 1024 equivalent permutations.

Actually, given your stipulation that no one folds, the odds of a higher hand taking the pot is far lower than in a normal poker hand, since individual hand strength and revealed cards (assuming hold’em, stud or a similar type game) are irrelevant to a given player’s decision whether to stay in or fold.

For computations, you need to stipulate exactly what you mean by ‘poker’ – the answer will be different depending on whether you mean hold’em, stud, draw, Omaha, etc. For example, if hold’em, obviously only one player could possibly have a Royal Flush and at most 2 could have a straight flush, whereas in stud or draw as many as 4 could have Royal Flushes and all could posibly have straight flushes regardless of N.

Edit: by ‘higher hand’, I mean that in a normal game of poker, a no-pair or one-pair hand is far less likely to be a winner that in your scenario, since someone holding such a hand is far less likely to remain to a showdown. Essentially, what is being played here is hold’em Showdown, where betting is limited just to the initial bet.

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