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 Medium-High Stakes Full Ring Discussion of \$400+ pot-limit and no-limit and 5/10 live texas hold'em full ring games, situations and strategies

 03-05-2018, 08:19 PM #1 pokerslut321 journeyman   Join Date: Jul 2008 Posts: 226 Probability question Random question...at a full ring poker table, what’s the chances that the players directly to your left and right share one of your whole cards? For example...I’m in seat 5 and my hand is Q10 offsuit. How often or what percentage of the time will seat 4 OR seat 6 have either a Q or a 10? What’s the equation to quantify this?
 03-06-2018, 09:17 AM #2 diskoteque Joey Local     Join Date: Aug 2010 Posts: 30,318 Re: Probability question Not sure but I think it’s 50%, either they have it or they don’t
 03-06-2018, 04:14 PM #3 iamallin centurion   Join Date: Feb 2016 Location: Lonely planet Posts: 198 Re: Probability question About 41% I tried to calculate it. May have made some mistakes. 50% was actually a very good guess disko
 03-06-2018, 10:27 PM #4 grant2 Carpal \'Tunnel     Join Date: Nov 2010 Posts: 8,871 Re: Probability question i think it's ~39.6% When you have a pocket pair, there is ~84.5% chance that none of your neighbours' 4 cards will be either of the 2 out of 50 that are same rank as your hole cards. When you have a non-pair, there is a ~58.9% chance that none of the your neighbours' 4 cards will be any of the 6 out of the 50 that are the same ranks as your hole cards. You get a pocket pair ~5.9% of the time, unpaired ~94.1% of the time. (5.9% x 84.5%) + (94.1% x 58.9%) = 60.4% chance of missing 1 - 60.4% = 39.6% chance of hitting.
 03-07-2018, 03:42 PM #5 shorn7 veteran   Join Date: Aug 2007 Posts: 2,645 Re: Probability question Isn't this as simple as 1- ((6/50) + (6/48)) or 75.5%? I guess not sure why you really care as there are 6 other players at the table too.
 03-07-2018, 07:03 PM #6 RoyalRumble journeyman   Join Date: Jun 2012 Posts: 278 Re: Probability question Assuming you want to count the times they have more than 1 of your cards between them: Chance they have none of your cards is 44/50 * 43/49 * 42/48 * 41/47 = 58.9% Chance they have one or more is 41.1%
03-13-2018, 03:16 AM   #7
grant2
Carpal \'Tunnel

Join Date: Nov 2010
Posts: 8,871
Re: Probability question

Quote:
 Originally Posted by RoyalRumble Chance they have one or more is 41.1%
Unless you have a pocket pair... which happens 6% of the time... see above!

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