wonky O8 starting hand statistics - do I have the numbers right?
 

Just playing with some numbers, wanted to make sure my assumptions are correct. If anything is wrong, please drop a line.

Chance of being dealt AA2x where x is not an ace
c(4,2) *c(4,1) *c(47,1)/c(52,4)=0.004167

Chance of being dealt any A23x hand (x not an ace)
c(4,1)* c(4,1) * c(4,1) *c(46,1)/c(52,4)=0.010875

Chance of being dealt AAWx, (no deuce or ace in Wx)
c(4,2) *c(12,1) *c(46,1)/c(52,4)=0.012234

Chance of being dealt any A2HH combination (HH=K, Q or J)
c(4,1) * c(4,2) *c(12,2)/c(52,4)=0.003901

Chance of being dealt any A2xx combination (x=any)
c(4,1) * c(4,1) * c(50,2)/c(52,4)=0.072398

Thanks

Re: wonky O8 starting hand statistics - do I have the numbers right?
 

[ QUOTE ]
Chance of being dealt AA2x where x is not an ace
c(4,2) *c(4,1) *c(47,1)/c(52,4)=0.004167

[/ QUOTE ]Hi Frank - I just looked at the first one (quoted above). I'll look at the others later and report if I see something amiss. Don't have time right now.

For the first one (quoted above)
AA22 6*6=36 ways to be dealt this.
AA2X 6*4*44 = 1056 ways to be dealt this (where X is neither an ace nor a deuce).

All in all there are 36+1056 = 1092 ways to be dealt AA2x where x is not an ace.

Then 1092/C(52,4) = 1092/270725 = 0.00403.

There are probably some other ways to set it up. But you can't simply use 4*47. I'm not very good at explaining why something doesn't work but think of it this way: three of the cards in the 47 are also deuces. You probably can find a way to do the calculation without making a two line chart with a separate line for two deuces - but that's how I did it this time. And that works. And that gives you a different answer.

Hope that helps.

Buzz

Make the first line C(4,2)*C(4,2) instead of 6*6, then calculate to get 36 if you like.

Re: wonky O8 starting hand statistics - do I have the numbers right?
 

That does help. I had the feeling I might have slipped some double-counting in while cutting corners. Thanks.

Re: wonky O8 starting hand statistics - do I have the numbers right?
 

[ QUOTE ]
Chance of being dealt any A23x hand (x not an ace)
c(4,1)* c(4,1) * c(4,1) *c(46,1)/c(52,4)=0.010875

[/ QUOTE ]Frank - Not quite. You have some duplications here too.

There are generally a number of ways to approach this problem.

One way is to make a chart. Here would be my chart:<ul type="square">AA23
A223
A233
A23X (where X is neither an ace, a deuce, or a trey).

then
AA23...6*4*4 ways to do this.
A223...4*6*4 ways to do this.
A233...4*4*6 ways to do this.
A23X...4*4*4*40 ways to do this.[/list]Then multiply as indicated and add the four sub totals.

96+96+96+2560 = 2848.

Then it's 2848/C(52,4) = 2848/270725.

And that's the probability. Divide to convert from a fraction to what I'd call a decimal fraction. (0.01052)

In this case, I'm going to call it about 1% anyway. But your fundamental approach is wrong.

I advise you to make a chart such as the one I made if you want to make your probability calculations more correct. Basically I mentally separated the aces, deuces and treys from the rest of the deck. (I don't know if that makes sense to you or not).

I'll leave it to you to correct the remaining ones, if you choose to do so. That's a better way for you to learn than my showing you how to do each one.

Buzz