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View Full Version : Math Mistake in THE THEORY OF POKER?

CrisBrown
11-01-2003, 01:59 PM
Hi All,

This has probably been discussed to death before, so feel free to direct me to the appropriate thread if it has.

On two occasions, David uses the following example to illustrate Bayes' Theorem: The game is hold'em, you hold QQ, and an opponent raises. You know this opponent would not raise without AA, KK, or AK. How do you decide which he has? David says that the probability for AA is 0.45% (that's correct) with the same probability for KK, for a total probability of 0.9% that he has one or the other. So far so good.

David goes on to say that the probability of AK is 1.2%, so the odds are 4:3 that he has AK rather than either AA or KK. Ummmmmmmm......................

There are 12 possible ways to catch AA (one of four possible suits for the first Ace, one of three remaining suits for the second Ace), and 12 possible ways to catch KK. There are 16 possible ways to catch AK (one of four Aces, one of four Kings). Thus, there are 24 possible combinations he could hold for AA or KK, and only 16 possible combinations he could hold for AK.

Rather than 4:3 odds that he holds AK, it's 3:2 that he holds either AA or KK.

Am I wrong?

Cris

P.S. David's conclusion that you should fold is still correct. Two-thirds of the time you're a huge underdog, and the other third of the time you're only a marginal favorite. Easy decision. Muck the QQ.
1800GAMBLER
11-01-2003, 02:15 PM
[ QUOTE ]
David says that the probability for AA is 0.45% (that's correct)

[/ QUOTE ]

[ QUOTE ]
There are 12 possible ways to catch AA

[/ QUOTE ]

---

[ QUOTE ]
one of four possible suits for the first Ace, one of three remaining suits for the second Ace

[/ QUOTE ]

You're double counting. A /images/graemlins/spade.gif A /images/graemlins/club.gif is the same as A /images/graemlins/club.gif A /images/graemlins/spade.gif
Nottom
11-01-2003, 02:18 PM
[ QUOTE ]
There are 12 possible ways to catch AA (one of four possible suits for the first Ace, one of three remaining suits for the second Ace), and 12 possible ways to catch KK.

[/ QUOTE ]

You are counting A /images/graemlins/spade.gifA /images/graemlins/club.gif and A /images/graemlins/club.gif A /images/graemlins/spade.gif as different hands. There are 6 ways to be dealt AA.
1800GAMBLER
11-01-2003, 02:23 PM
Strange how we used the same suits.
Nottom
11-01-2003, 02:37 PM
I just think its funny that your post kept growing everytime i looked at it.
CrisBrown
11-01-2003, 04:01 PM
Jay and nottom,

Thanks. Doh ... difference between permutations and combinations.

Cris
Louie Landale
11-01-2003, 05:35 PM
When you disagree with someone who knows what they are talking about, there is a lot more "EV" in presuming you are wrong. Say something like "this is how I see it, where am I wrong".

- Louie

Unless of course its me that's disagreeing.
1800GAMBLER
11-01-2003, 07:59 PM
Hah. I edit a lot. I gave my first reply and thought it wasn't much of a reply and looked bad, then i thought that about my 2nd edit. 3rd time lucky, i figured that was enough info.
CrisBrown
11-01-2003, 09:49 PM
Louie,

Thus the title "Math Mistake in THE THEORY OF POKER?" (note the question mark), and my last sentence "Am I wrong?"

Cris
Copernicus
11-02-2003, 04:26 AM
Even though its been shown how you got the math wrong, I'm not sure why you were even doing the math...again.

Early in the post you agree that the prob. of AA is .45%, and KK another .45%. How could you agree there, but then go on to miscalculate the same number?
johnnyhearts
11-02-2003, 07:04 AM
Cris,

You have been punished for your impudence! From this day forward never shall you question nor reveal a lack of understanding of a 2+2 author's published word and/or actions! If you do, you will be shouted down more brutally than you can possibly imagine!

Johnny Hearts
BruceZ
11-02-2003, 08:24 AM
Early in the post you agree that the prob. of AA is .45%, and KK another .45%. How could you agree there, but then go on to miscalculate the same number?

That's easy, he just counts the total number of hands the same way. 12/(52*51) = 0.45%. Or 4/52 * 3/51 = 0.45%.

You can use either permutations or combinations, as long as you use the same one in both the numerator and denominator. Fractions are the same as permutations as you can see above. This works for probability, but the error appears when you want to just count the hands. Then you have to use combinations so you don't double count. C(4,2) = 4*3/2! = 6.
rayrns
11-02-2003, 08:27 AM
"There are 12 possible ways to catch AA (one of four possible suits for the first Ace, one of three remaining suits for the second Ace), and 12 possible ways to catch KK. There are 16 possible ways to catch AK (one of four Aces, one of four Kings). Thus, there are 24 possible combinations he could hold for AA or KK, and only 16 possible combinations he could hold for AK." (I don't know how to work that quote thingy.)

Your figures could still have worked ok, except for the way you figured for the A-K. The way you were working on it there would be 16 ways to make A-K. (4 Aces-4 Kings) but the King could come first and the A second. Therefore 16 ways (4 Aces-4 Kings) plus 16 ways(4 Kings-4 Aces) for 32. 32-24 would still equal 4:3 which is the number you are looking for. Just a longer math way of getting there.
CrisBrown
11-02-2003, 01:11 PM
Yup, that's what she did.

Cris