I open UTG with TT. What are the odds I'm beat?

[Moose747]

I hate to be asking such a simple question, but I don't really know how to solve this one.

Say I raise with TT utg in a 10-handed game. What's the probability someone has JJ+ ? What about 11-handed? 12-handed? Is there an easy formula?

[BruceZ]

[ QUOTE ]
I hate to be asking such a simple question, but I don't really know how to solve this one.

Say I raise with TT utg in a 10-handed game. What's the probability someone has JJ+ ? What about 11-handed? 12-handed? Is there an easy formula?

[/ QUOTE ]

For the exact solution, you need the inclusion-exclusion principle as shown in this post for some other pairs. For the probability that your TT is against JJ,QQ,KK or AA, the first 3 terms of inclusion-exclusion get you to within 0.1% as follows:


10-handed:

9*24/C(50,2) -
C(9,2)*24*19/C(50,2)/C(48,2) +
C(9,3)*24*(18*14+1*18)/C(50,2)/C(48,2)/C(46,2)

=~ 16.5%.


11-handed:

10*24/C(50,2) -
C(10,2)*24*19/C(50,2)/C(48,2) +
C(10,3)*24*(18*14+1*18)/C(50,2)/C(48,2)/C(46,2)

=~ 18.2%.



12-handed:

11*24/C(50,2) -
C(11,2)*24*19/C(50,2)/C(48,2) +
C(11,3)*24*(18*14+1*18)/C(50,2)/C(48,2)/C(46,2)

=~ 19.8%.


You can also use the following approximations which are much simpler, and as you can see by comparing to the above, they are sufficiently accurate for most purposes. There are 24 higher pairs.

10-handed: 1 - [1 - 24/C(50,2)]^9 =~ 16.3%

11-handed: 1 - [1 - 24/C(50,2)]^10 =~ 18.0%

12-handed: 1 - [1 - 24/C(50,2)]^11 =~ 19.6%


See the series of articles by Brian Alspach titled "I'm In...No I'm Out" for more information about the inclusion-exclusion principle, and some tabulated probabilities for higher pairs.

[Moose747]

Thanks Bruce. I think I have a decent grasp on this now. Or at least I know where to look should I need it.

[smbruin22]

good subject... i'm always wanting to do calculations like this.

what does the C(x,y) mean??... is it something i have to look up on a table??

[BruceZ]

[ QUOTE ]
good subject... i'm always wanting to do calculations like this.

what does the C(x,y) mean??... is it something i have to look up on a table??

[/ QUOTE ]

The number of ways to choose y things out of x things without regard to the order they are chosen.

C(x,y) = x!/(x-y)!/y! = x*(x-1)*(x-2)*...*(x-y+1) / [y*(y-1)*(y-2)*...*1].

Note that the numerator has y factors, since the lower x-y factors of x! are canceled by (x-y)!

E.g., C(9,3) = 9*8*7/(3*2*1). Also, C(52,2) = 52*51/2 is the number of hold'em hands. C(52,5) = 52*51*50*49*48/(5*4*3*2*1) would be the number of hands in 5 card draw.

We can use this last example from draw poker to see the rationale for the expression. Note that there are 52 ways to choose the first card, leaving 51 ways to choose the 2nd card, leaving 50 ways to choose the 3rd card, etc. Then we divide by 5! = 5*4*3*2*1 because this is the number of rearrangements or "permutations" of each 5-card hand, and we want to count each unique 5-card hand only once, without regard to the order of the cards. To see why there are 5! ways to order 5 cards, we note that there are 5 ways to choose the first card, leaving 4 ways to choose the 2nd card, leaving 3 ways to choose the 3rd card, etc. In general, there are n! permutations of n things.

You can evaluate C(x,y) in Excel using the =COMBIN function, or even in Google by typing in, for example, 52 choose 5.