August 7th, 2015 at 9:20:28 AM
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At a 6-handed table, what is the probability that at least one pocket pair is dealt from exactly 2 different ranks from of a set of 3 ranks?

For instance, suppose our "set of 3 ranks" is {A,K,Q} meaning our focus is on QQ+. What is the probability that exactly two different QQ+ pairs occur this hand?

For further clarification, if it goes AA|AA|xx|xx|xx|xx that doesn't count (which makes this different than the other pairs question asked recently).

AA|KK|... counts

AA|AA|KK|... counts

AA|KK|QQ|... doesn't count

I've tried 2 different methods and got slightly different answers. My slightly more confident answer is .0019453214 and my other answer is .0019522663.

Brute force (adding up a bunch of stuff) is a dreadful option for this problem.

For instance, suppose our "set of 3 ranks" is {A,K,Q} meaning our focus is on QQ+. What is the probability that exactly two different QQ+ pairs occur this hand?

For further clarification, if it goes AA|AA|xx|xx|xx|xx that doesn't count (which makes this different than the other pairs question asked recently).

AA|KK|... counts

AA|AA|KK|... counts

AA|KK|QQ|... doesn't count

I've tried 2 different methods and got slightly different answers. My slightly more confident answer is .0019453214 and my other answer is .0019522663.

Brute force (adding up a bunch of stuff) is a dreadful option for this problem.

August 7th, 2015 at 5:00:42 PM
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I get 0.00194532102035809 which differs from your first answer in the 10th decimal place.

In the following, dfact(n) is the double factorial n!! = n*(n-2)*(n-4)*...*1 for n odd which is the number of ways to partition n+1 cards into 2 card hands. By inclusion-exclusion

P(AA and KK) =

C(6,2)*C(4,2)^2/C(52,4)/dfact(3) -

C(6,3)*C(4,2)*dfact(3)*2/C(52,6)/dfact(5) +

C(6,4)*dfact(3)^2/C(52,8)/dfact(7)

≈ 0.000662525310415919

P(AA and KK and QQ) =

C(6,3)*C(4,2)^3/C(52,6)/dfact(5) -

C(6,4)*dfact(3)*C(4,2)^2*3/C(52,8)/dfact(7) +

C(6,5)*dfact(3)^2*C(4,2)*3/C(52,10)/dfact(9) -

C(6,6)*dfact(3)^3/C(52,12)/dfact(11)

≈ 1.40849702965542e-05

We want

P( (AA and KK) or (AA and QQ) or (KK and QQ) ) - P(AA and KK and QQ)

= 3*P(AA and KK) - 3*P(AA and KK and QQ)

≈ 0.00194532102035809

In the following, dfact(n) is the double factorial n!! = n*(n-2)*(n-4)*...*1 for n odd which is the number of ways to partition n+1 cards into 2 card hands. By inclusion-exclusion

P(AA and KK) =

C(6,2)*C(4,2)^2/C(52,4)/dfact(3) -

C(6,3)*C(4,2)*dfact(3)*2/C(52,6)/dfact(5) +

C(6,4)*dfact(3)^2/C(52,8)/dfact(7)

≈ 0.000662525310415919

P(AA and KK and QQ) =

C(6,3)*C(4,2)^3/C(52,6)/dfact(5) -

C(6,4)*dfact(3)*C(4,2)^2*3/C(52,8)/dfact(7) +

C(6,5)*dfact(3)^2*C(4,2)*3/C(52,10)/dfact(9) -

C(6,6)*dfact(3)^3/C(52,12)/dfact(11)

≈ 1.40849702965542e-05

We want

P( (AA and KK) or (AA and QQ) or (KK and QQ) ) - P(AA and KK and QQ)

= 3*P(AA and KK) - 3*P(AA and KK and QQ)

≈ 0.00194532102035809

August 7th, 2015 at 6:42:23 PM
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This was my first thought, but then I looked at the individual things being over-counted by 3*P(AA & KK) and to me it looked like they needed different coefficients than all +/- 3. I'll show what I did and maybe you can point out where I went wrong. Actually, while writing this post, I noticed a few screwups I made before. My new answer is closer to yours, with all but my final coefficient matching yours.Quote:BruceZ3*P(AA and KK) - 3*P(AA and KK and QQ)

Our P(AA & KK) is exactly the same. I see you've even started using double-factorials like me, or was that because you knew I like them?

After those lines, we still have to eliminate the cases of three-pair, quads+2p, pair+2quads, and triple-quads.

Our first line counted the two-pair cases, which in turn:

- count the three-pair cases 3 times
- count the quads+2p cases 5 times
- count the pair+2quads cases 8 times
- count the triple-quads cases 12 times

We can just look at an arbitrary quads+2p, so let's look at AAKKQQQQ

AAKK counted it once

AAQQ counted it twice

KKQQ counted it twice

Total times AAKKQQQQ counted = 5

AAKK counts AAKKKKQQQQ twice

AAQQ counts it twice

KKQQ counts it 4x

total = 8

Our second line subtracted the pair+quads cases, which in turn:

- subtracted the quads+2p cases 2x
- subtracted the pair+2quads cases 6x
- subtracted the triple-quads cases 12x

Our third line added back the double-qauds cases, which in turn:

- counted the pair+2quads cases once
- counted the triple-quads cases 3x

So after all three lines, the three-rank tallies are as follows:

- The three-pair cases are counted 3 times.
- The quads+2p cases are counted 5 - 2 = 3 times
- The pair+2quads cases are counted 8 - 6 + 1 = 3 times
- The triple-quads cases are counted 12 - 12 + 3 = 3 times

By doing that, one is also:

- subtracting the quads+2p cases 6x
- subtracting the pair+2quads cases 12x
- subtracting the triple-quads cases 24x

But adding those back in also meant:

- adding the pair+2quads cases 12x
- adding the triple-quads cases 36x

Next I subtract those 3 times which also means subtracting the triple-quads cases 6 times.

Then the triple-quads cases are counted 3-24+36-6 = 9 times, so my final step is to subtract them 9 times. (The -9 sticks out like a sore thumb and must be my mistake, especially since our answers would be the same if it were a +3 as the pattern would suggest.)

My new, less-wrong answer is .001945321

Edit: I don't know why I originally was unsure the quoted would work. It makes perfect sense. I guess I wanted the reassurance of seeing exactly what was happening with all the counts, but then my problem was messing up some counts. After the weekend I'll still try to find the mistake that led to the -9, since I like being able to derive the counts despite it being unnecessary for this problem.

August 7th, 2015 at 11:51:06 PM
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Here's the mistake. It's not 6, it's 18.Quote:AlmondBreadAfter that, my pair+2quads cases were counted 3-12+12 = 3 times

Next I subtract those 3 times which also means subtracting the triple-quads cases 6 times.

AAKKKKQQQQ counts triple-counts twice

So do the 2 other pair+2quads.

For a total of 6, but that gets multiplied by 3 so 18.

Then triple-quads are counted 3-24+36-18 = -3 times so we add them back in 3 times.