Thought up this question (which has no practical value I'm aware of) and was curious because I don't even know how to start setting it up.

10 players are playing hold'em. I deal them two random cards.

I then let the first Nine choose to discard and take two new cards.

Then I deal out the board.

Questions:
--What hands should the first nine players discard?
--Player 10, who did not get a chance to re-draw, what's the lowest hand he can have and still be a favorite to win the hand?

If it makes the math easier we can assume that all ten hands were dealt from a fresh deck because if you start factoring in card replacement I'm sure the only reasonable answer is to simulate it.

I just thought it was interesting because technically all ten people still have two "random" cards, yet clearly player 10's chances to win have gone down dramatically.

The deal is now biased. We can assume the 9 made rational decisions to keep better than average hands and discard worse than average hands, replacing them with an expectation of getting average hands. Which is better than "worse than average" hands. Repeating this trial many times will result in the 9 having a better average value hand than the 1. So yes random, but biased to be better than average.

It doesn't really matter which ranking system is used to determine the average hand. But for the 1 player, he needs a MUCH better than average hand to expect to have the best hand now. Something pretty close to AA probably.

If you define “favorite to win” as greater than 50% equity, then (assuming no folding) your player who doesn’t redraw could never be a favorite. AA vs 9 random hands has (IIRC) about 30% equity. Allowing redraws would only worsen that equity.

Now, in that situation, you could define “favorite” as +EV, which would be a more interesting definition. AA vs 9 random hands clearly is a favorite by that definition. You get 9:1 odds, so any hand with more than 10% equity is +EV. I’m not sure how high a hand has to be vs 9 random hands to be +EV; maybe play with some equity calculators and figure that out, then adjust to a tighter range for your question.

Neither can any of the other 9. So that wouldn't be a useful definition. Whoever has the highest equity is the favorite to win, even if that equity is only 15%. He will win more often than the others.

Maybe start easier and consider all hands have value on the [0,1] interval.

Then what should the discard cutoffs for P1 - P9 be? (Theoretically they should depend upon prior action. But maybe simplify further to simultaneous decisions.)

How about we make it even easier? What if there are 2 players? P1 can discard but P2 cannot. I assume that problem should be easy.

Then keep adding players who can discard.

Since I had some time I pursued the I idea I posted above. Rather than dealing with real poker hands, I simplified to the standard [0,1] poker hands. Players are "dealt" a random number between 0 and 1 and high value wins.

In our experiment, if there are N total players at the table, all but one of them can discard their initial hand-value and draw another random hand-value if they so choose. The Nth player does not have the ability to discard/redraw. To simplify things further, we will assume that the N-1 players who can discard/redraw make their decisions simultaneously.

When N is small, it is possible to derive analytically the optimal equilibrium cutoff each of the first N-1 players should utilize (each of these N-1 players utilizes the same cutoff in equilibrium).

Then it is straightforward to derive the probability each of the N-1 players will win a hand (of course each of the N-1 has the same probability) and the probability that the Nth player will win a hand.

I derived the analytical solutions for N=2, 3, and 4. Beyond that is too much to work through the calculations manually so I programmed a simulation.

Before I present the numerical results, I want to present an interesting finding which, I think, generalizes to higher N.

It is straightforward to show that:

P1 = [1 + c - c^2]/2 for N=2, where P1 is the prob that Player 1 (the player who can discard/redraw) wins and c is the optimal cutoff Player 1 utilizes.

P1 = [2 + c - c^2 + c^3 - c^4]/6 for N=3.

P1 = [3 + c - c^2 + c^3 - c^4 + c^5 - c^6]/12 for N=4.

If this pattern holds, then:

P1 = [4 + c - c^2 + c^3 - c^4 + c^5 - c^6 + c^7 - c^8]/20 for N=5.

P1 = [5 + c - c^2 + c^3 - c^4 + c^5 - c^6 + c^7 - c^8 + c^9 - c^10]/30 for N=6.

Etc.

At this point these higher-N formulas are conjectures, but they do seem consistent with my simulations. If anybody has any thoughts on those formulas, feel free to share them. I wonder if this is a "well-known" result that I stumbled upon (assuming that the formulas are even correct).

The table below assumes those formulas are correct and continue all the way through the N=10 case.

Total Number of Players	Optimal Cutoff	_____ P1 _____	_____PN _____
2	.50000	.62500	.37500
3	.60583	.38774	.22452
4	.67033	.28041	.15877
5	.71454	.21942	.12231
6	.74705	.18015	.09926
7	.77214	.15277	.08341
8	.79218	.13259	.07187
9	.80860	.11711	.06309
10	.82235	.10487	.05621

Of course, the optimal cutoff naturally rises as the number of players increases. The lone player who cannot discard/redraw naturally is at a significant disadvantage in this game.

Hmm finally came back to this, dang good work dude!

I wonder if there is some practicality in the fact that ...well other than me having a deja vu which is super weird on such a highly technical question I almost certainly was never dealing with before...

where was I

..oh yah it's interesting that people should be chucking an 82 if there's ten people at the table...you mean that sort of implies that at poker as the number of players goes up you really need to have a top % hand to figure to win at showdown.

Also in the OP I asked what's the lowest # that the person that can't redraw can have and still expect to win...hrmm is it the optimal cutoff for the other players? Seems like it would need to be higher as there's a pretty good chance 1 of the 9 opponents had an 85+ or re-drew to one.

I think you are correct.

In the N=2 case (Player 1 can discard, Player 2 cannot), it is clear that Player 1 should use a cutoff of .625 [which is 5/8].

The math is pretty straightforward in this case and if I did the math correctly, the hand-value where Player 2 has a 50% chance of winning is .69231 [which is 9/13].

The N=3 case is a bit more tedious to work through and I don't have the time right now to do it.


eta: I forgot I have a simulator so I ran a few cases.

For N=3, optimal cutoff is .60583 (found above), and value where PN = 1/N is approx .738

For N=10, optimal cutoff is .82235 (found above), and value where PN = 1/N is approx .876

.

what if we were playing heads-up, and i can redealt 23 times (assumes "no burn" card). is my hand still "random"?

a bit like the Monty Hall problem. look it up if you don't know it....


basically there's big prize behind one of 3 doors.... you choose door, then they take away one of the other 2 doors. they give you a choice to switch to the lone remaining other door.. should you do it? yes, but many people (including myself) initially can't see why. and some math profs (LOL) debated the wrong answer.

it's easy to visualize if there are 100 doors, you choose one... they take away 98 doors and let you change to the other doors........ you had 1% chance of being right at start. and you still do. the last remaining door basically represents 99 doors. 1% it is still there randomly (you have the prize behind your door) and 99% chance it didn't disappear because it has prize behind it.... and to be clear, they didn't take the other doors away randomly, if they chose the door with prize behind it, it stayed.