Probability Dealt Wheel (23457) with six cards

That's the probability of a wheel or a six-card straight to the 6.
If you want specifically a wheel, replace 47 with 46.
If you want the probability of a wheel or 6-high straight (with or without the A), that's (4^5)(47+46) / C(52,6)

Thanks heehaww

Heehaww, just so on same page a wheel in 2-7 lowball is 23457 as it’s the best low hand as an ace is high and straights count against you

So 23457 is the 4^5 part and since the 6th card can be anything I just multiply it by 47. (But a six is not a better hand)

At any rate the .22% I come up with is around 5x more likely than with five cards. That’s slightly counterintuitive but explained by the great #combinations that C(52,6) is over C(52,5)? The more I think about it, the more it does make sense

My bad I ignored the 23457 part, also my math was wrong anyway. I only subtracted one card from 47 when I should have subtracted 4, also I double-counted the wheel+pair combos.

So now I'll redo it but first let's address your remark about the intuition. We get 6 cards to make a 5-card hand, so we have 6C5 = 6x as many chances to make it. In a vacuum we'd be 6x as likely to make it, and indeed if the numerator were simply 4^5 * 47 that would be exactly 6x the probability.

However, we don't want a 6 in the hand, so change the 47 to a 43.

But that's not all. A hand like 223457 gets double-counted if we leave it at that. There are a few ways to avoid this. One is to split it into no-pairs and pairs:

N(wheel w/o pair) = 4^5 * (52−4*6)
N(wheel+pair) = 5(4C2)4^4

Adding those and dividing by 52C6, we get .0017856

Another way to get the numerator is: (4^5)43 − 5(4C2)4^4, ie start with the naive number and then subtract the double-counted combos.

That's inclusion-exclusion; another way to use inclusion-exclusion is to count the ways to miss the wheel and subtract from 52C6.

52C6 − 5(48C6) + (5C2)(44C6) − (5C3)(40C6) + 5(36C6) − 32C6

We'd then need to subtract the 234567 combos, of which there are 4^6.

That 3rd method is overkill here, but for other problems it can be the far superior approach.

Thanks heehaww, I thought I might have been double counting with pairs

Anyhow we don’t have to fear the 6 as 6th card. It can just be ignored. If you are dealt 234567 your hand is simply 23457

Therefor the no pair is 4^5*(52-4*5) I believe ending up in a result of .1987% I believe.

Would you agree?

Ah then yes, .1987%

Thanks bud, that does help as it would be probability for 23467 as the 5 on the 6th card would make it 23457