The odds for being dealt pair (2-2, 3-3, ..., K-K, A-A) is 16 against one: because if the second card is dealt, there are 51 cards left, of which 3 can make a pair to the first card, So: Wills = 3; Willnots = Total - Wills = 51 - 3 = 48 --> 48:3 or 16:1
This is fairly simple, but I'm wondering the following:
Let's say there are 9 people at a table:
--> What are the odds that at least 1 person gets dealt a pair?
--> What are the odds that exact two people get a pair?
--> What are the odds that at least two people get a pair?
Let's say the 4 people in front of me let me know they have not a pair... I have a pair (6-6) and there are still 4 people behind me:
--> What are the odds that at least 1 of the 4 remaining people has a pair?
--> What are the odds that 1 of the 4 people has a higher pair (7-7, 8-8, ... K-K, A-A)?
--> What are the odds that 1 of the 4 people has a lower pair (5-5, 4-4, 3-3, 2-2)?
--> What are the odds that 2 of the 4 remaining people get dealt a pair?
I'm not looking for just an answer: more I'd like to figure out how I can calculate this for all handgroups... for example, the odds of getting any ace (A-2, A3, ... AA) is about 6:1. I'm looking to get an insight in how likely it is that at a table of 9 people, 2 of these same cardgroups ('any ace' or 'pocket pair', ...) occur at the same time.
That's why I'd appreciate it if you give the steps in between, so I can do this myself for all groups that I'd like to figure out.
Thanks in advance!
Andy
This is fairly simple, but I'm wondering the following:
Let's say there are 9 people at a table:
--> What are the odds that at least 1 person gets dealt a pair?
--> What are the odds that exact two people get a pair?
--> What are the odds that at least two people get a pair?
Let's say the 4 people in front of me let me know they have not a pair... I have a pair (6-6) and there are still 4 people behind me:
--> What are the odds that at least 1 of the 4 remaining people has a pair?
--> What are the odds that 1 of the 4 people has a higher pair (7-7, 8-8, ... K-K, A-A)?
--> What are the odds that 1 of the 4 people has a lower pair (5-5, 4-4, 3-3, 2-2)?
--> What are the odds that 2 of the 4 remaining people get dealt a pair?
I'm not looking for just an answer: more I'd like to figure out how I can calculate this for all handgroups... for example, the odds of getting any ace (A-2, A3, ... AA) is about 6:1. I'm looking to get an insight in how likely it is that at a table of 9 people, 2 of these same cardgroups ('any ace' or 'pocket pair', ...) occur at the same time.
That's why I'd appreciate it if you give the steps in between, so I can do this myself for all groups that I'd like to figure out.
Thanks in advance!
Andy