Hey, I need some help for 2-7 Razz where aces are high and A2345 is not a straight

What are the odds that 234 ends up hitting 56 in the four cards remaining that you will dealt? This game is unlike TD in that if you make a straight on 5th or 6th you can't get rid of it.

C(49,4) = 211,876 is as far as I got lol

Then a follow-up is how often 237 will collect 456 over the next four cards? Then what about 89TJ?


Last follow-up would be is how often will the 247 collect either a 356 or a 568 among the next four cards?

Good vibes to anyone that can help!

I don't know Razz and hopefully I'm not missing a detail.

A slick way to tackle this is 1-P(miss), using inclusion-exclusion for P(miss):

1 - [2*C(45,4) - C(41,4)]/C(49,4) = 7.153%

You were probably attempting the additive approach which would be this:

[4*4*C(41,2) + 2*C(4,2)*4*41 + 2*4*4 + C(4,2)^2] / C(49,4) = same answer

Thanks heehaw! You gave me a lil too much credit. I was able to obtain the denominator C(49,4).

The 7% was a little higher than I thought it would be but certainly seems possible. Thought it would be lower because even when you get the first 5 or 6 you still just have a gutshot.

I'm good with the concept of some things being able easier to calculate as 1-P(miss) though.

So the 345 would need either the 26 or the 67. This does not seem like it would be double the probability because the 6 is needed for both?

N(miss) = N(no 6's) + N(no 7's or 2's) - N(no 6's or 7's or 2's)
→ P(miss) = [C(45,4) + C(41,4) - C(37,4)] / C(49,4)
→→ P(hit) = 13.053%

Thanks!

[(4^3)*37 + 3*C(4,2)*4^2] / C(49,4)

[4^4 + 2(4^3)*33 + 2*3*C(4,2)*4^2] / C(49,4)

Thanks heehaw!, assuming I did the ordering of operations right I get 1.3% and 2.4% respectively. Definitely appears reasonable.

Now 456 is a hand that can be trouble because it can make more straights. Can I just get one more?, I'm not sure how to extrapolate from the above.

What is the probability that 456 will collect 23, 37, or 78 among the next four cards? I'm guessing around 18%..

Actually I'm somewhat of an idiot. Having a straight will not "foul" your hand. For example, if you hold (23)456T(J) you will just simply have 2345T as your final low.


That said, I can still use everything that was calculated in here in terms of knowing how often we make hands.


7.1% is prob that 234 will make 56 over the next four cards, but its also the prob of making 57 for a wheel

Sorry, when you asked this I got sick for a couple days. Had the fever lasted a little longer, I might have been writing this post as a sheep because my thermometer kept trying to tell me I was a sheep.

[4^4 + 3*C(4,2)^2 + 3*16*(2+C(33,2)) + 4!*C(4,2)*33 + 2*9*16*C(4,2)] / C(49,4) = 15.237%

That sounds a bit low, so I likely messed up.

Of course, this is very easy to brute force on a computer. Which I did.

Computer finds that a player holding 456 gets 23, 37, or 78 in his remaining 4 cards:

= 40,156 / 211,876

= 18.9525949%

Here's what I got. Find cases where next 4 cards contain some number of {2,3,7,8} with {2,3}, or {3,7}, or {7,8}.

Case 1: all 4
= C(4,1)*C(4,1)*C(4,1)*C(4,1)
= 256

Case 2: 3 of them with a dupe
= C(4,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)
= 1,152

Case 3: 3 of them without a dupe
= C(4,3)*C(4,1)*C(4,1)*C(4,1)*C(33,1)
= 8,448

Case 4: 2 of them with trips
= 3*C(2,1)*C(4,3)*C(4,1)
= 96

Case 5: 2 of them with 2 dupes
= 3*C(4,2)*C(4,2)
= 108

Case 6: 2 of them with 1 dupe
= 3*C(2,1)*C(4,2)*C(4,1)*C(33,1)
= 4,752

Case 7: 2 of them without dupes
= 3*C(4,1)*C(4,1)*C(33,2)
= 25,344

TOTAL:
= 40,156

heehaww's expression above is very close to these terms but is different in one or two spots.