? from ssnl

[tarheeljks]

someone posted this in ssnl and being the lazy sob that i am, i'm reposting it:

[ QUOTE ]
Any math buffs out there, help me figure this out. I'm too lazy to look up the correct method of solving this.

The probability of a player having AA or KK on any random shuffle is 1/114 (approx - so give or take a few on that). Anyway...

If there are 6 players at a table, all dealt a random hand, what is the probability that ONE of the players has AA or KK? And how do you figure this?

[/ QUOTE ]

[BruceZ]

[ QUOTE ]
someone posted this in ssnl and being the lazy sob that i am, i'm reposting it:

[ QUOTE ]
Any math buffs out there, help me figure this out. I'm too lazy to look up the correct method of solving this.

The probability of a player having AA or KK on any random shuffle is 1/114 (approx - so give or take a few on that). Anyway...

If there are 6 players at a table, all dealt a random hand, what is the probability that ONE of the players has AA or KK? And how do you figure this?

[/ QUOTE ]

[/ QUOTE ]

Do you want EXACTLY 1, or AT LEAST 1? For at least 1, we can get the exact answer from the inclusion-exclusion principle similar to these problems to get:

6*12/C(52,2) -
C(6,2)*12*7/C(52,2)/C(50,2) +
C(6,3)*12*(6*2+1*6)/C(52,2)/C(50,2)/C(48,2) -
C(6,4)*12*(6*2+1*6)/C(52,2)/C(50,2)/C(48,2)/C(46,2)

=~5.35%


For exactly 1, we make the following modification:

6*12/C(52,2) -
2*C(6,2)*12*7/C(52,2)/C(50,2) +
3*C(6,3)*12*(6*2+1*6)/C(52,2)/C(50,2)/C(48,2) -
4*C(6,4)*12*(6*2+1*6)/C(52,2)/C(50,2)/C(48,2)/C(46,2)

=~ 5.28%


As always, this much simpler approximation is very accurate:

1 - (1314/1326)^6 =~ 5.31%

[tarheeljks]

he asked for exactly 1, but i'm sure he'll be happy to see both. anyway, thanks for doing the calcs; the last approximation is what was offered in ssnl b/c people were too lazy [img]/images/graemlins/smile.gif[/img]

[jay_shark]

P(AAorKK) = 12/1326 = 1/110.5

A quick approximation of the probability that AT LEAST one of the players has AA or KK is 1-[(1326-12)/1326]^6 ~ 5.308%

Likewise , the approximate probability that exactly one of the players has AA or KK is 6*12/52c2*(50c2-7)/50c2*(48c2-7)/48c2*(46c2-7)/46c2*(44c2-7)/44c2*(42c2-7)/42c2 ~ 5.2466%

[BruceZ]

[ QUOTE ]
Likewise , the approximate probability that exactly one of the players has AA or KK is 6*12/52c2*(50c2-7)/50c2*(48c2-7)/48c2*(46c2-7)/46c2*(44c2-7)/44c2*(42c2-7)/42c2 ~ 5.2466%

[/ QUOTE ]

This is not exact, and while close in this case, this kind of approximation will not perform well in more complicated problems, and the independence approximation will perform much better. The -7 implies that there are 7 remaining AA or KK, but this is only true if no one else has a single A or K.

[jay_shark]

It's true that the exact solution isn't much more complicated at all .

It's interesting to note that this method is not bad even when we calculate the probability that at exactly one player holds a pocket pair at a 6 handed table . If you can tolerate being off by about 3% [img]/images/graemlins/smile.gif[/img]

6c1*78/1326*[(1326-78)/1326]^5 ~ 26.06%

Using the other method , we get:

6*78/1326*1147/1225*1050/1128*957/1035*868/946*783/861 ~23.73%

[BruceZ]

[ QUOTE ]
It's true that the exact solution isn't much more complicated at all .

It's interesting to note that this method is not bad even when we calculate the probability that at exactly one player holds a pocket pair at a 6 handed table . If you can tolerate being off by about 3% [img]/images/graemlins/smile.gif[/img]

6c1*78/1326*[(1326-78)/1326]^5 ~ 26.06%

Using the other method , we get:

6*78/1326*1147/1225*1050/1128*957/1035*868/946*783/861 ~23.73%

[/ QUOTE ]

Actually, for exactly 1 pair, the method of independence is more accurate when done like this:

6*78/1326*[(1225-73)/1225]^5 =~ 25.959%

I computed the exact answer to lie between 25.969% and 25.972%, so this independence approximation is within about 0.01%.

Also, if we use this independence approximation for the AA/KK problem, we get

6*12/1326*(1218/1225)^5 =~ 5.2765%

The exact answer lies between 5.2754326% and 5.2754319%, so this independence approximation is within about 0.001%.

The reason yours is less accurate is actually because you attempt to account for the changing number of cards remaining in the denominator with each term, without a corresponding change in the numerator. It is actually more accurate to keep both of these the same, as the effects then tend to largely cancel out.

[jay_shark]

Cool , I will use that method from now on .