Probably very easy question

[w_raedy99]

I have just started reading Ken Warren's 'Winnder's Guide to Texas Hold'Em Poker' (someone gave me a copy for free) and ran into something that got my attention.

He states "When you have an Ace in the pocket there will be an Ace dealt to at least one other player 74.7% of the time." Now this is obviously ten-handed.

For some reason this number seems high to me, perhaps that is naiive on my part. I would be very interested in hearing if this number is correct, and also how to calculate it.

I've also ran into other statements that I definitely did not agree with, e.g. "Play people with tattoos because anybody stupid enough to get a tattoo will not be a good poker player" and "Play against young people, because even with the internet, etc. you will likely have more experience still." I know I'd love to exploit these two tips when people play against me who have read this book, having tattoos and being young [img]/images/graemlins/smile.gif[/img].

Will

[umdpoker]

ken warren's book sucks. don't take anything he says too seriously.

[jimymat]

That was one of the first books I read. Its good for someone who has to where velcro shoes. Once you learn what the blinds are you can sell this book on ebay for a pony ride and a stick of gum. Thers a reason why you got the book for free. Good luck.

[TripleH68]

[ QUOTE ]
"When you have an Ace in the pocket there will be an Ace dealt to at least one other player 74.7% of the time."

[/ QUOTE ]

Thinking simple...
You have AJ.
50 cards remain. 3 of them are aces.
18 cards are in other players hands.

50/18=3/x. x being the number of aces, on average, in the 18 cards in other players hands. x = 1.08.

[BruceZ]

[ QUOTE ]
I have just started reading Ken Warren's 'Winnder's Guide to Texas Hold'Em Poker' (someone gave me a copy for free) and ran into something that got my attention.

He states "When you have an Ace in the pocket there will be an Ace dealt to at least one other player 74.7% of the time." Now this is obviously ten-handed.

For some reason this number seems high to me, perhaps that is naiive on my part. I would be very interested in hearing if this number is correct, and also how to calculate it.

[/ QUOTE ]

This is correct. If you understand combinatorics, you can calculate it as

1 - C(47,18) / C(50,18) = 74.7%.

If you don't understand combinatorics, you can calculate it as follows:

Compute 1 minus the probability that your opponents hold no aces. There are 50 cards remaining, and 47 are non-aces, so the number of ways to deal 18 non-aces to your 9 opponents is

47*46*45*...*33*32*31*30.

The total number of ways to deal 18 cards to your opponents is

50*49*48*47*46*45*...*33.

So to get the probability of no aces, just divide these

47*46*45*...*33*32*31*30 / (50*49*48*47*46*45*...*33).

After cancelling like terms, this reduces to:

32*31*30 / (50*49*48) = 25.3%.

One minus this is the probability that at least one ace is out. 1 - 25.3% = 74.7%.

[jedi]

[ QUOTE ]
I have just started reading Ken Warren's 'Winnder's Guide to Texas Hold'Em Poker' (someone gave me a copy for free)

[/ QUOTE ]

I think you got a bad deal.

[LetsRock]

My personal theory on dealt cards distribution is this. There are 20 cards dealt ten handed. A very informal form of probablity tells me that there will be about 1.5 of each rank dealt. Using that very loose "formula" (it's not scientific by any means), I usually assume that there 2 Aces out in every deal. If I have one, then someone else likely has exactly one. If I have two, there's a pretty good chance that noone else has one.

Now, I don't live and die under this theory - I just use it as a rough guide to help me put people on hands......

[submariner]

I like that. A quick and easy to use rule of thumb. Thanks.

[slickpoppa]

I have found several statistical mistakes in that book. I suggest throwing it out and never looking at it again so you do not corrupt your mind.

[Earthy Tones]

ken warren is a retarded monkey.