quick help for a disagreement :)

[TxSteve]

ok..i'm mixed up in this discussion about odds of being dealt pocket aces..

from WPT etc..i'm of the impression that the odds are 220-1
can anyone give me a quick run down of how to support this?

thanks

[uuDevil]

There are 52 cards in a deck, 4 of these are aces.

The probability the 1st card you are dealt is an A: 4/52.

Given you have 1 ace, there are 51 cards left in the deck, 3 of them aces.

Probability you will now get one of 3 remaining aces: 3/51.

Probability of AA is the product: (4/52)*(3/51)=(1/13)*(1/17)=1/221.

Translated to odds, 1/221 is 220:1.

QED /images/graemlins/laugh.gif

[tech]

Probability of being dealt AA = 4/52 * 3/51 = .004525 = .4525%

Odds = (1-.004525) / .004525 = 220

Thus the odds are 220:1.

[TxSteve]

thank you..i'm with you on all of that..

but this person in the argument with me thinks that the # of players in the hand is a factor...

he is saying that the as each successive card is dealt...your odds aren't 3/51 the next time your card is dealt..they are 3/46 or 0/46 depending on what cards were dealt..

sheesh this is frustrating since im no math whiz

[tech]

Unless you know what those cards are, it is irrelevant whether they are dealt or not.

[uuDevil]

Ah, ok.

Tech is right, but if you want more points of view, TomCollins gave a succint answer in this post. (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=672999&page=&view=&sb =5&o=&vc=1)

If you want a full discussion, BruceZ and others go through it in detail in this thread. (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=565745&page=&view=&sb =5&o=&vc=1)