Value Of Pocket Pairs

[nicksite 0]

So this morning I was pondering the value of pocket pairs. Take a look at what I found.Hands that will be coinflip or better against...AANoneKKAAQQAAKKAKJJAAKKQQAKAQKQetc...Turns out there's a formula to find out how many hands are coinflip or better.(0.5)x^2+(0.5)xx is...for AA- 0for KK- 1for QQ- 2etc.I'm wondering if this is a new finding... or if it's helpful or anything. Feedback?

[No_Neck 0]

EDIT. misread OP's post.

[LongLiveYorke 38]

I like the fact that you're thinking. Here's a derivation for those who don't like formulas that appear out of thin air:add the number of over pairs which, my your system, is equal to xadd the combinations of overcards, which in your system would be x choose 2so, our formula is:= x + x choose 2= x + x!/ (2!(x-2)!)the above is by the binomial formula (choose formula)now simplify:= x + x(x-1)/2= (x)(x+1)/2= .5 x^2 + x/2QEDBut, again, I think this is all meaningless. Let me know if I'm wrong, but wouldn't it just be easier to take the total amount of cards above ours (as in the x in your system multiplied by 4 to include suits) and just do 4x choose 2?For our considerations, it really doesn't matter if the above cards are pocket pairs or overcards. This way, we get the total number of hands possible in a more meaningful way. In other words, my formula would be:4x choose 2= (4x)!/ (2! (4x-2)!)= 2x (4x-1) = 8x^2 - 2xI think this is right. Correct me if I'm wrong.Now, there are 52 choose 2 hands in hold'em= 1,326So, against a random hand, our pocket pair of strength x would have the following probability of being a coinflip or worse:2x (4x-1)/1326= x (4x-1) / 663= .006033 x (x-.25)Using the above formula, we get the following chart:Hand "x" Probability AA - 0 - 0.0%KK - 1 - 0.5%QQ - 2 - 2.1%JJ - 3 - 5.0%TT - 4 - 9.0%99 - 5 - 14.3%88 - 6 - 20.8%77 - 7 - 28.5%66 - 8 - 37.4%55 - 9 - 47.5%44 - 10 - 58.8%33 - 11 - 71.3%22 - 12 - 85.1%Again, the probability on the right is the probability that against one hand we are a coinflip or worse. I hope this is right.

[nicksite 0]

I like the fact that you're thinking. Here's a derivation for those who don't like formulas that appear out of thin air:add the number of over pairs which, my your system, is equal to xadd the combinations of overcards, which in your system would be x choose 2so, our formula is:= x + x choose 2= x + x!/ (2!(x-2)!)the above is by the binomial formula (choose formula)now simplify:= x + x(x-1)/2= (x)(x+1)/2= .5 x^2 + x/2QED

Yeah, I just used a calculator and used the quadratic regression function. That logic makes sense to me though.

But, again, I think this is all meaningless. Let me know if I'm wrong, but wouldn't it just be easier to take the total amount of cards above ours (as in the x in your system multiplied by 4 to include suits) and just do 4x choose 2?For our considerations, it really doesn't matter if the above cards are pocket pairs or overcards. This way, we get the total number of hands possible in a more meaningful way. In other words, my formula would be:4x choose 2= (4x)!/ (2! (4x-2)!)= 2x (4x-1) = 8x^2 - 2xI think this is right. Correct me if I'm wrong.Now, there are 52 choose 2 hands in hold'em= 1,326So, against a random hand, our pocket pair of strength x would have the following probability of being a coinflip or worse:2x (4x-1)/1326= x (4x-1) / 663= .006033 x (x-.25)Using the above formula, we get the following chart:Hand "x" Probability AA - 0 - 0.0%KK - 1 - 0.5%QQ - 2 - 2.1%JJ - 3 - 5.0%TT - 4 - 9.0%99 - 5 - 14.3%88 - 6 - 20.8%77 - 7 - 28.5%66 - 8 - 37.4%55 - 9 - 47.5%44 - 10 - 58.8%33 - 11 - 71.3%22 - 12 - 85.1%Again, the probability on the right is the probability that against one hand we are a coinflip or worse. I hope this is right.

Ahh... very nice point. I didn't really check the math but the formula makes sense to me.

[LongLiveYorke 38]

Yeah, I just used a calculator and used the quadratic regression function. That logic makes sense to me though.

I guess you were lucky then to come up with the exact formula.

[Zach6668 513]

You guys are dorks

[Actuary 3]

we need a math's geek sub forum for nerds like you two.my goodness!

[AZWildcat 0]

nerd alert! j/k...... I wish I was that good at math..... Long live the "# of outs times 2" rule!