At the end of most nights people do nuetral ev flips dealing each other 12 cards for $200 each and who ever has the best 5 card hand wins.

Who is the favorite and by how much if the terms were player A gets 12 cards for $100, player B gets 14 cards for $150.

Secondary question, what is the $ amount to make 12 and 14 equal?

Let’s first simplify the game quite a bit. Suppose Player A got 1 card and Player B got 2. High card wins. Clearly with twice as many cards, B has twice the chance of winning as does A. (It's the same idea that the chance you win a lottery is proportional to the number of lottery tickets you bought.) So, in general, one’s winning chance is directly related to the number of combinations. For just two players, A and B, we have:

A. For 12 cards, poker combos = C(12,5) = 792. Prob A Wins = 792/2794 = 28.3%

B. For 14 cards, poker combos = C(14,5) =2002. Prob B Wins = 100% – 28.3% =71.7%

Therefore B is a 2002/792 = 2.53 to 1 favorite. If A pays 100 and B pays 150 and winner takes all, then

EVa = 0.283*250 – 100 = -29.25

EVb = 0.717 *250 - 150 = 29.25

For break-even, B should pay 2.53 as much as A. So if A pays 100 and B pays 253,

EVa = 0.283 * 353 – 100 = 0

EVb = 0.717*353 – 253 = 0

Statman might be right, but I'm not sure. I think basing the percentages on the ratio C(14,5) to C(12,5) is an approximation rather than the exact answer.

Consider a game where the goal is to make a flush. Bob is dealt 5 cards and Alice is dealt 6. Though Alice can make 6 times as many 5-card combos, she cannot make 6 times as many flushes and her chance of winning (when they play until someone wins) is not 6/7 but a little less.

In full poker the flushes aren't the only complication.

Edit: On second thought, I don't think what I said above is an issue in full poker because each combo is a hand and you just need the better hand.

Edit #2: Nor do I think there's an issue with ties if they play until someone wins. So as of now I'm in agreement with Hal.

I had some of the same concerns as heehaww. For example, if most of the cards you get are bummers, having another added card may not help you very much. However, if there is good potential, say for high flushes or straights or full houses, one or more added cards can be a great help. So, assuming a natural 'balancing', I can't think of any other way than counting the number of 5-card hands even recognizing that the hands have a great deal of card overlap.

Of course what statmanhal did is an approximation. Think about A receiving 5 cards and B the remaining 47. Just counting the combos, A should be C(47,5):1 underdog. Actually, A can never win and can just tie when he is dealt a royal flush (4 times out of C(52,5)). So, his equity is 2/C(52,5) which is slightly different from 1/C(47,5).

However, I think that his approximation is good (just a feel, don't have nothing to support it).

The combination argument was my first thought too. But then I thought about diminishing marginal returns. Then I thought about running all cases in my holdem program. Then I realized that there are more than 10^27 cases. Then I thought about running a simulation. Then I realized that there would never be enough cases in the simulation to provide meaningful results and/or finish running in my lifetime.

P.S. If anyone has a super fast holdem program, maybe they can run through all the 7 vs 5 card situations (or 6 vs 5).

You don't need a lot of simulations, regardless of home many potential outcomes there are, right, because no matter what you're binning into wins and losses. If there's no bias in the selection of your trials there won't be bias in the outcome of the simulations. And you can get within, say, 1% of the true result without that many simulations, worst case is usually something like 20k iterations I think.

(If you don't agree, then tell me an example of a distribution where random sampling won't return an unbiased result)

So is what stat said correct? Or close enough to correct for arguments sake? On the outside before the math it seems like A would be the right choice and A was what most people guessed.

You would definitely want to be Player B paying 150 to win 100. The consensus seems to be a fair line would be about B's 250 to A's 100, although there is a thought that the chances for the "best hand" may not actually be the same as ratio of the quantity of hands.

Intuitively, I think it's correct that the line should be slightly smaller, but I could be wrong, I also think, like Rusty says, that you could get a close enough approximation by running a couple thousand Monte Carlos Simulations.

I'm inclined to believe you will make money over the long haul if you could structure a deal of posting 200 to win 100 and taking the 14 Cards.

Later I changed my mind again but for a less convincing reason than Nick's example so I was going to sleep on it. Nick's example is proof that it's only an approximation, but below is how I understand why it's only that.

In my flush example, the reason it's not 6/7 is that multiplying by C(6,5) overcounts the 6-card flushes. Then in an edit I said that in full poker, any combo is a hand, but that's not true; many combos are the same hand and so the same problem will occur. When dealt 14 cards + community, you have a lot of redundant cards since you're only making a 5-card hand. So just like there are many ways to make the same 5-card flush (many combos of redundant cards), there are multiple ways to make many 5-card hands. For that reason, each extra combo the advantaged player has is not an extra way to win. (This paragraph is more convincing now that I wrote it out than when it was in my head.)

So the exact answer would be a big chore.

P(14-card hand wins) = P(royal vs lower) + P(K-high SF vs lower) + ...

And then we'd also need P(tie).

Way too many cases to bother with unless we were to split the work among like 10 of us each being assigned different cases. (A sim sounds better.)

Yes, I spent about an hour last night starting to write out all the cases but quit since I was still only in the straight flushes. Lots and lots of cases to consider with that many cards.

Okay, I performed a couple of simulations. Unfortunately, I have one of the slowest computers on the face of the earth, so I have to walk before I can run (maybe crawling before walking is more accurate).

In 60,000 deals of 6 cards vs. 5 cards, the 6-card hand won 62.78% of deals. From a combinatorics perspective, of course, 6-card hands have 6 times the number of 5-card hand possibilities than a 5-card hand, so would be expected to win (using combinatorics alone) 6/7 = 85.71% of the deals. The standard error of the estimate from these 60,000 samples is around 0.20%.

In 20,000 deals of 7 cards vs. 5 cards, the 7-card hand won 74.37% of deals. From a combinatorics perspective, of course, 7-card hands have 21 times the number of 5-card hand possibilities than a 5-card hand, so would be expected to win (using combinatorics alone) 21/22 = 95.45% of the deals. The standard error of the estimate from these 20,000 samples is 0.31%.

So it seems, based upon these low-card simulations, that the "overlap" effect is an important factor and that the fact that cards are shared significantly lowers the true winning percentage of hands with more cards compared to a pure combinatorics approximation.

I hope my simulations serve to whet the appetite of people who can perform better/faster simulations so that larger number of cards can be considered.

One more to report ...

In 32,000 deals of 7 cards vs. 6 cards, the 7-card hand won 63.41% of deals. From a combinatorics perspective, of course, 7-card hands have 21:6 times the number of 5-card hand possibilities than a 6-card hand, so would be expected to win (using combinatorics alone) 21/27 = 77.78% of the deals. The standard error of the estimate from these 32,000 samples is around 0.27%.

Ok so you're saying it's always gonna be a number less than the combinations?

I don't want to jump to any numerical conclusions just from the above simulations, but I think most people would have agreed that the combinatorics approximation had to be an upper bound. Nobody knew how much lower the actual percentage was compared to the percentage purely based upon the combinatorics.

I think it is the nature of poker hands that the implicit "independence" assumption inherent in the pure combinatorics approximation is not correct. This is akin to a type of diminishing marginal returns of additional cards I referenced above.

Others may have different views or better ways of saying this.

One more to report ...

In 32,000 deals of 8 cards vs. 5 cards, the 8-card hand won 84.88% of deals. From a combinatorics perspective, of course, 8-card hands have 56 times the number of 5-card hand possibilities than a 5-card hand, so would be expected to win (using combinatorics alone) 56/57 = 98.25% of the deals. The standard error of the estimate from these 32,000 samples is around 0.20%.

One more to report ...

In 24,000 deals of 8 cards vs. 6 cards, the 8-card hand won 76.13% of deals. From a combinatorics perspective, of course, 8-card hands have 56:6 times the number of 5-card hand possibilities than a 6-card hand, so would be expected to win (using combinatorics alone) 56/62 = 90.32% of the deals. The standard error of the estimate from these 24,000 samples is around 0.28%.

Two more to report ...

In 24,000 deals of 9 cards vs. 6 cards, the 9-card hand won 85.64% of deals. From a combinatorics perspective, of course, 9-card hands have 126:6 times the number of 5-card hand possibilities than a 6-card hand, so would be expected to win (using combinatorics alone) 126/132 = 95.45% of the deals. The standard error of the estimate from these 24,000 samples is around 0.23%.

In 24,000 deals of 9 cards vs. 7 cards, the 9-card hand won 75.85% of deals. From a combinatorics perspective, of course, 9-card hands have 126:21 times the number of 5-card hand possibilities than a 7-card hand, so would be expected to win (using combinatorics alone) 126/147 = 85.71% of the deals. The standard error of the estimate from these 24,000 samples is around 0.28%.

One more to report ...

In 24,000 deals of 9 cards vs. 5 cards, the 9-card hand won 92.55% of deals. From a combinatorics perspective, of course, 9-card hands have 126:1 times the number of 5-card hand possibilities than a 5-card hand, so would be expected to win (using combinatorics alone) 126/127 = 99.21% of the deals. The standard error of the estimate from these 24,000 samples is around 0.17%.

Hopefully, we are building up a store of results that can help us estimate the 14 vs. 12 case (which is far beyond my simulation capabilities).

Thanks for all the sims whosnext

No way the combos give the right odds the correlation effect is massive.

I'll maybe try and run a sim tonight but think about the game this we. Both players have 12 cards. Will getting two more cards, when behind, allow you to outdraw your opponent 20% of the time? Seems absurdly high to me.

One more to report (trying to fill in some of the earlier holes) ...

In 24,000 deals of 8 cards vs. 7 cards, the 8-card hand won 63.73% of deals. From a combinatorics perspective, of course, 8-card hands have 56:21 times the number of 5-card hand possibilities than a 7-card hand, so would be expected to win (using combinatorics alone) 56/77 = 72.73% of the deals. The standard error of the estimate from these 24,000 samples is around 0.31%.

Two more to report ...

In 24,000 deals of 10 cards vs. 6 cards, the 10-card hand won 92.17% of deals. From a combinatorics perspective, of course, 10-card hands have 252:6 times the number of 5-card hand possibilities than a 6-card hand, so would be expected to win (using combinatorics alone) 252/258 = 97.67% of the deals. The standard error of the estimate from these 24,000 samples is around 0.17%.

In 24,000 deals of 10 cards vs. 7 cards, the 10-card hand won 85.02% of deals. From a combinatorics perspective, of course, 10-card hands have 252:21 times the number of 5-card hand possibilities than a 7-card hand, so would be expected to win (using combinatorics alone) 252/273 = 92.31% of the deals. The standard error of the estimate from these 24,000 samples is around 0.23%.