One more to report ...

In 24,000 deals of 10 cards vs. 8 cards, the 10-card hand won 75.10% of deals. From a combinatorics perspective, of course, 10-card hands have 252:56 times the number of 5-card hand possibilities than a 8-card hand, so would be expected to win (using combinatorics alone) 252/308 = 81.82% of the deals. The standard error of the estimate from these 24,000 samples is around 0.28%.

Using whosnext's sim results (after adding the 5,5 case), I did a simple regression for predicting the actual result from the combinatorial ratio. A log-log fit gave reasonably good results with an R-square of 83.5% (in log-land, though); the prediction equation follows:

Win % = 26.52%*exp[1.194*(C1/(C1+C2)]

where C1 and and C2 are the number of hand combinations for the higher and lower number of dealt cards respectively (14 and 12 for OP's example).

While well outside the data range for number of combos (not the combo ratio, though), at this point we can predict for OP's case the 14-card holder's win rate would be 62.4%. Incidentally, a semi-log equation did almost as well in R-square terms yielding an estimate of 63.3%.

I'm somewhat hoping that whosnext doesn't get to sim this case for it may show this simple analysis is way off. Okay, do it, I'll take my lumps.

Prompted by statmanhal's post above, let me summarize the simulation results so far, grouping them by the difference in cards dealt to each player:

W(6,5)=62.8
W(7,6)=63.4
W(8,7)=63.7
W(9,8)=tbd

W(7,5)=74.4
W(8,6)=76.1
W(9,7)=75.8
W(10,8)=75.1

W(8,5)=84.9
W(9,6)=85.6
W(10,7)=85.0

W(9,5)=92.6
W(10,6)=92.2

W(10,5)=tbd

A simple relationship seems to fit the data pretty well:

W(H1,H2) = WinPct = 50 + (15.384 * ((H1 - H2) ^ 0.707))

where H1 and H2 are the number of cards dealt to each player (H1>=H2).

That is, a simple non-linear relationship between the difference in number of cards dealt and the win pct seems to fit.

This formula predicts the following:

Diff in Cards WinPct
-------------- --------
0 50.0%
1 65.4%
2 75.1%
3 83.4%
4 91.0%
5 98.0%

This formula (similar to the one statmanhal posted above) can be constrained so that the WinPct tends to 100% as the difference in number of cards dealt gets large, and equals 50% when the difference equals 0. For now, in my formula above I posited that the WinPct equals 98% when the difference equals 5. Of course, other "end-states" can be built into such a formula. Essentially, right now I have capped the difference in cards at 5 in the formula or else it will predict WinPct's greater than 100%.

This is not to say that I believe that the above formula holds over the entire range of number of cards dealt to each player. I think the WinPct will eventually come down (holding H1-H2 constant) as H1 and H2 increase, reflecting a second "dimension" of the "diminishing marginal returns" argument people have discussed in this thread. But it has not shown up yet in the values of H1 and H2 that have been simulated so far.

This was the impetus of my estimation exercise. Think of it as a companion to the one posted by statmanhal above. If we eventually can know the true formula over the entire range of values, I am confident that it will look like a mixture of a formula like statmanhal's and the one I posted here.

I think I have really reached the limits of my PC's abilities to perform these simulations. However, I will attempt to push these limits so that we can learn a little more about the WinPct function. Again, if anyone else has a fast PC I encourage them to perform some simulations into this question.

Thanks much.

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

In 24,000 deals of 9 cards vs. 8 cards, the 9-card hand won 63.53% of deals. From a combinatorics perspective, of course, 9-card hands have 126:56 times the number of 5-card hand possibilities than a 8-card hand, so would be expected to win (using combinatorics alone) 126/182 = 69.23% of the deals. The standard error of the estimate from these 24,000 samples is around 0.31%.

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

In 32,000 deals of 10 cards vs. 5 cards, the 10-card hand won 96.57% of deals. From a combinatorics perspective, of course, 10-card hands have 252:1 times the number of 5-card hand possibilities than a 5-card hand, so would be expected to win (using combinatorics alone) 252/253 = 99.60% of the deals. The standard error of the estimate from these 32,000 samples is around 0.10%.

Here's one more important one to report ...

In 32,000 deals of 11 cards vs. 9 cards, the 11-card hand won 72.49% of deals. From a combinatorics perspective, of course, 11-card hands have 462:126 times the number of 5-card hand possibilities than a 9-card hand, so would be expected to win (using combinatorics alone) 462/588 = 78.57% of the deals. The standard error of the estimate from these 32,000 samples is around 0.25%.

Okay, here's the next in the series when one person receives two more cards than the other person:

In 32,000 deals of 12 cards vs. 10 cards, the 12-card hand won 70.28% of deals. From a combinatorics perspective, 12-card hands have 792:252 times the number of 5-card hand possibilities than a 10-card hand, so would be expected to win (using combinatorics alone) 792/1044 = 75.86% of the deals. The standard error of the estimate from these 32,000 samples is around 0.26%.

Adding this result to the previous results of 2-card differences:

W(7,5)=74.4
W(8,6)=76.1
W(9,7)=75.8
W(10,8)=75.1
W(11,9)=72.5
W(12,10)=70.3

So we are now seeing the 2-card difference win pct start to decline as predicted by several posters in this thread.

I plan on simulating 13 vs. 11 cards, but it will take a few days before I expect to get any results due to the exponential increase in the number of hands that must be considered.

Huzzah, here's the next (and maybe final) in the series when one person receives two more cards than the other person:

In 30,000 deals of 13 cards vs. 11 cards, the 13-card hand won 67.28% of deals. From a combinatorics perspective, 13-card hands have 1287:462 times the number of 5-card hand possibilities than a 11-card hand, so would be expected to win (using combinatorics alone) 1287/1749= 73.58% of the deals. The standard error of the estimate from these 30,000 samples is around 0.27%.

Adding this result to the previous results of 2-card differences:

W(7,5)=74.4
W(8,6)=76.1
W(9,7)=75.8
W(10,8)=75.1
W(11,9)=72.5
W(12,10)=70.3
W(13,11)=67.3
W(14,12)=???

So we are now seeing the 2-card difference win pct start to decline as predicted by several posters in this thread.

I am not sure if I will get anywhere on a 14 vs. 12 simulation since the hand combinations continue to grow exponentially, considering it took around a week to complete the 13 vs. 11 simulation.

Based upon the results posted above, it seems that a reasonable guess would be that a 14-card hand would probably win around 65% of the time versus a 12-card hand.

Last post (yea) in the series when one person receives two more cards than the other person:

In 50,000 deals of 14 cards vs. 12 cards, the 14-card hand won 65.26% of deals. From a combinatorics perspective, 14-card hands have 2002:792 times the number of 5-card hand possibilities than a 12-card hand, so would be expected to win (using combinatorics alone) 2002/2794 = 71.65% of the deals. The standard error of the estimate from these 50,000 samples is around 0.21%.

Adding this result to the previous results of 2-card differences:

W(7,5)=74.4
W(8,6)=76.1
W(9,7)=75.8
W(10,8)=75.1
W(11,9)=72.5
W(12,10)=70.3
W(13,11)=67.3
W(14,12)=65.3

Remember OP's original question concerned Player A getting 12 cards for $100 and Player B getting 14 cards for $150. Clearly, this would be a fair game if Player B were expected to win 150/(150+100) = 150/250 = 60%.

Since Player B is now seen to be expected to win around 65.3% of the hands, a fair price for Player B, given Player A pays $100, would be (100*.653)/(1-.653) = $188.

All the best.

Great work whosnext

thanks alot whosnext. pretty interesting