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10-17-2005, 10:25 PM
I'm new to the forum and fairly new to poker. I have a bunch of basic questions that I'll be asking over the next few months and I hope you guys can help learn.

I read Caro's site that the odds of being dealt any wired pair is 220-1. I was wondering if somebody could "show me the math" on how you get there.

Thanks,

Drew
PS I'm not doubting the accuracy, just wanna know how to do the work.
10-17-2005, 10:28 PM
This belongs in the probability section.

That said:
there are 52 chose 2 = 26*51 = 1326 possible pairs of hole cards.
there are 4 chose 2 = 2* 3 = 6 possible pairs of a particular value
6 / 1326 = 1 / 221

Which corresponds to 1 to 220 odds.
Xhad
10-17-2005, 10:44 PM
You can also view it as two independent events (think of the dealer tossing you each of your cards one at a time, it doesn't matter that other people are getting dealt in because their cards are still unknown)...

Assuming you want, say, aces you first need one of four cards out of the deck, so the probability of drawing one of your cards is 4/52.

Then you need to draw one of the three aces remaining, so multiply by 3/51.

(4/52)(3/51) = 12/2652, which is 1/221.
10-17-2005, 11:17 PM
Thanks very much! That's perfect.
Golden
10-28-2005, 02:55 AM
It seems like I get pockets a lot more than 1 in 220. Am I missing something?
10-28-2005, 03:13 AM
He means any particular wired pair.

So it's 220-1 to get dealt Aces, 220-1 for Kings, 220-1 for deuces etc.
MikeBandy
10-28-2005, 04:35 AM
[ QUOTE ]
Golden: It seems like I get pockets a lot more than 1 in 220. Am I missing something

[/ QUOTE ]
[ QUOTE ]
FeNeF: He means any particular wired pair.

So it's 220-1 to get dealt Aces, 220-1 for Kings, 220-1 for deuces etc.

[/ QUOTE ]
Right. Perhaps the following is too simplistic, but we may take it one step further. The odds against getting a pair are 16:1.

As Rufus demonstrated, there is one chance in 221 of getting a pair of a certain rank. There are thirteen ranks, so there is one chance in seventeen (221/13) of getting a pair.

We can arrive at the same conclusion using the algorithm Rufus provided:

There are 52 choose 2 = (52*51) / 2 = 1326 possible pairs of hole cards.
There are 4 choose 2 = (4*3) / 2 = 6 possible pairs of a particular value.
There are 6*13 = 78 possible pairs.
1326 / 78 = 17

Which corresponds to odds of 16:1 against.
Precept2
10-28-2005, 06:17 PM
You're still making it harder than it needs to be.

Suppose you look at your first card. There are 3 more like it out of the 51 unseen cards. Which means, 48 of them won't make a pair. 48:3 is 16:1.
MikeBandy
10-28-2005, 10:40 PM
[ QUOTE ]
You're still making it harder than it needs to be.

[/ QUOTE ]
Thanks, Precept.