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 02-21-2019, 09:11 PM #1 ScotchOnDaRocks veteran   Join Date: May 2015 Location: SealsWithClubs Posts: 2,037 2-7 Razz Probability of Making Straight 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!
02-21-2019, 10:10 PM   #2
heehaww
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Re: 2-7 Razz Probability of Making Straight

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

Quote:
 What are the odds that 234 ends up hitting 56 in the four cards remaining that you will dealt?
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

 02-21-2019, 11:56 PM #3 ScotchOnDaRocks veteran   Join Date: May 2015 Location: SealsWithClubs Posts: 2,037 Re: 2-7 Razz Probability of Making Straight 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?
 02-22-2019, 01:07 AM #4 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,554 Re: 2-7 Razz Probability of Making Straight 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%
02-22-2019, 01:11 AM   #5
ScotchOnDaRocks
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Re: 2-7 Razz Probability of Making Straight

Quote:
 Originally Posted by heehaww 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!

02-22-2019, 10:09 AM   #6
heehaww
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Re: 2-7 Razz Probability of Making Straight

Quote:
 Originally Posted by ScotchOnDaRocks Then a follow-up is how often 237 will collect 456 over the next four cards?
[(4^3)*37 + 3*C(4,2)*4^2] / C(49,4)

Quote:
 Last follow-up would be is how often will the 247 collect either a 356 or a 568 among the next four cards?
[4^4 + 2(4^3)*33 + 2*3*C(4,2)*4^2] / C(49,4)

02-23-2019, 03:17 AM   #7
ScotchOnDaRocks
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Re: 2-7 Razz Probability of Making Straight

Quote:
 Originally Posted by heehaww [(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%..

 02-23-2019, 02:16 PM #8 ScotchOnDaRocks veteran   Join Date: May 2015 Location: SealsWithClubs Posts: 2,037 Re: 2-7 Razz Probability of Making Straight 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
03-03-2019, 01:30 PM   #9
heehaww
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Join Date: Aug 2011
Location: Tacooos!!!!
Posts: 4,554
Re: 2-7 Razz Probability of Making Straight

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

 03-03-2019, 02:30 PM #10 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 5,343 Re: 2-7 Razz Probability of Making Straight 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%
 03-05-2019, 04:19 PM #11 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 5,343 Re: 2-7 Razz Probability of Making Straight 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. Last edited by whosnext; 03-05-2019 at 04:34 PM.

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