WPT Post-Producer seeks help

[BruceZ]

I take it back about the evaluation of your (incorrect) expression, it does come out to 1 in 22.76, but it is incorrect nonetheless.

[irchans]

J'adoube,
I used your method to generate Bruce's answer.

<pre><font class="small">code:</font><hr>
P(no one has A's)
= P(no other A's are out) +
P(1 A is out) +
P(2 A's are out)*((no one has AA (given only two A's are out) ))+
P(3 A's are out)(no one has AA)+
P(4 A's are out)(no one has AA)
</pre><hr>

Let d = c[50,16] and c[x,y] = x! /y! /(x-y)! .

Then, for a 9 handed game (8 opponents), I get

<pre><font class="small">code:</font><hr>
P(no one has A's) = c[46,16] / d = .2014
P(1 A is out) = c[4,1] c[46,15] / d = .4157
P(2 A's are out) = c[4,3] c[46,13] / d = .2923
P(3 A's are out) = c[4,3] c[46,13] / d = .0827
P(4 A's are out) = c[4,4] c[46,12] / d = .0079
</pre><hr>

and

<pre><font class="small">code:</font><hr>
P(no one has AA given only 2 A's are out) = 14/15
P(no one has AA given only 3 A's are out) = 4/5
P(no one has AA given only 4 A's are out) = 8/13.
</pre><hr>

Plugging those values into your formula gives

<pre><font class="small">code:</font><hr>
P(no one has A's) = .2014 + .4157 + .2923 * 14/15 +
.0827*4/5 + .0079* 8/13 = .960938

P(someone has A's) = 1 - .960938 = .039062

1/.039062 = 25.6

</pre><hr>

That matches Bruce's answer.

[J'adoube]

"This approach will work, but your evaluation of all the terms is wrong. I have evaluated your terms correctly, and I get exactly the same
answer I reported earlier. "

I did notice one problem I assumed there was 9 *other players* not 8.



" For example, the probability of no aces being dealt is C(46,16)/C(50,16) = 0.201. That is, there are C(50,16) ways to choose the remaining
cards for the 8 players, and C(46,16) of these have no ace. It is not C(32,4)/C(50,4) = 0.156 as you said. Apparently you are counting the
number of ways the 4 aces can be arranged in the undealt cards (which would be C(34,4) by the way) divided by the number of ways they
can be in 50 cards. This doesn't work since it doesn't count all the combinations generated by the other cards. You can't just focus on aces.
If you want to focus on the undealt cards, you can say there are C(50,34) total ways to choose them, and there are C(46,30) ways to make
sure they have all the aces, so C(46,30)/C(50,36) = 0.201 as before. Your other terms have similar problems. "

I am almost certain what I said is correct (which you say is incorrect) that is If one takes out two K's,

the probability no A's are out if two cards are dealt to 9 other players is the probability no A's are out is the probability the 4 A's are distribulted amoung the other 32 ((52-20)) cards. And thats 32C4/(50C4). Though this should equal (as you said) 46C18/ 50C18.

Ill look at it again.

[J'adoube]

Poster: BruceZ
Subject: Re: WPT Post-Producer seeks help

" I take it back about the evaluation of your (incorrect) expression, it does come out to 1 in 22.76, but it is incorrect nonetheless. "

I don't think so. I noted before that this is for a ten handed game and I think everyone else did it for a ninehanded game.


I did redo it for a nine handed game and got 24.60 to 1, not the number
you stated.

[BruceZ]

Ahhh, if you were evaluating a 10-handed game, then I agree with your answer completely. I get the same answer you got for a 10-handed game using both my method and your method. The evaluation of all your terms is correct for a 10-handed game. The original poster was asking about a 9-handed game.

I did redo it for a nine handed game and got 24.60 to 1, not the number
you stated.

That is the number I've been stating all along. I said 1 in 25.6, which is 24.6 to 1.

All the exact answers I posted for WPTDan are accurate. These anwers have been obtained 2 different ways by 2 people,and now 3 different ways and 3 different people for the KK vs. AA problem.

[BruceZ]

I said:
" For example, the probability of no aces being dealt is C(46,16)/C(50,16) = 0.201. That is, there are C(50,16) ways to choose the remaining
cards for the 8 players, and C(46,16) of these have no ace. It is not C(32,4)/C(50,4) = 0.156 as you said. Apparently you are counting the
number of ways the 4 aces can be arranged in the undealt cards (which would be C(34,4) by the way) divided by the number of ways they
can be in 50 cards. This doesn't work since it doesn't count all the combinations generated by the other cards. You can't just focus on aces.
If you want to focus on the undealt cards, you can say there are C(50,34) total ways to choose them, and there are C(46,30) ways to make
sure they have all the aces, so C(46,30)/C(50,36) = 0.201 as before. Your other terms have similar problems. "

You said:

I am almost certain what I said is correct (which you say is incorrect) that is If one takes out two K's,

the probability no A's are out if two cards are dealt to 9 other players is the probability no A's are out is the probability the 4 A's are distribulted amoung the other 32 ((52-20)) cards. And thats 32C4/(50C4). Though this should equal (as you said) 46C18/ 50C18.


Agreed. Disregard my comments about focusing on the aces, your way is fine, that wasn't the problem. The only issue here is 9-handed vs. 10-handed. Your evaluation is correct for 10-handed, and can be easily adjusted for 9-handed. WPTDan was interested in 9-handed, so that is what I answered. I get the same anwer as you for 10-handed by my method. I also get the same answer as I reported originally for WPTDan for 9-handed using your method (as did irchans).

So comparing our methods:

10 handed:
P(no aces out given we have KK) =
C(46,18)/C(50,18) my way focusing on dealt cards
= C(46,28)/C(50,32) my way focusing on undealt cards
= C(32,4)/C(50,4) your way focusing on only aces

P(no AA given 2 aces out) =
1 - 9/C(18,2) my method
C(9,2)*2^2/C(18,2) your method


9 handed:
P(no aces out given we have KK) =
C(46,16)/C(50,16) my way focusing on dealt cards
= C(46,30)/C(50,34) my way focusing on undealt cards
= C(34,4)/C(50,4) your way focusing on only aces

P(no AA given 2 aces out) =
1 - 8/C(16,2) my method
C(8,2)*2^2/C(16,2) your method

And you used your way for a compact evaluation of no one having AA for the other numbers of aces being out as well. This provides an advantage in certain cases. I will add the counting method you used here to my bag of tools.

[irchans]

Bruce wrote
"
P(no aces out given we have KK) =
C(46,18)/C(50,18) my way focusing on dealt cards
= C(46,28)/C(50,32) my way focusing on undealt cards
= C(32,4)/C(50,4) your way focusing on only aces
"

I was surprized that there were 3 very different ways to compute that probability. I noticed that you can also get the same number with

(*)

P(no aces out given we have KK) = C(78,28)/C(78,32).

Questions:

1) I have no idea why (*) is true. Can anyone figure out a probabalistic reason for (*) ?

2) Are there any other positive integers m,n,q, and q such that

P(no aces out given we have KK) = C(m,n)/C(p,q) ?

My guess is that there are no soultions where m &gt; 78.

[Easy E]

or a syndication fee, or sumtin'....