Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

I once got rolled up 5's back to back in 7 stud, what're the odds of that?

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

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I once got rolled up 5's back to back in 7 stud, what're the odds of that?

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Somewhat worse than 7-1

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

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DS,

We have a book forum.

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I'm more likely to get involved here.

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

What about subscripts? Can you program this forum to accept subscripts?

Xn


Doesn't seem to work.

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

David,

This question isn't that great, but it's probably less disappointing than most of the others. I ordered the book from amazon almost a month ago, and it has been delayed a few times. Do you know anything about this? problems with supply, shipping, whatev?

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

craziest thing that happened to me is a guy getting kings four straight times in a live game. He got it 3rd time...they changed the deck and he got it again. He literally busted the entire table...at least it broke up aftewards.

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

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[ QUOTE ]
DS,

We have a book forum.

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I'm more likely to get involved here.

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LOL. Oh so desperate...

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

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You were only off about 7 orders of magnitude. That's not so much, in the grand scheme of things.

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Although an accurate estimate, properly it's this:

I'll use the notation (x y) for "x over y", which is x!/((x-y)!*y!), or combinations.

First, you need to be dealt 2 deuces: (4 2) / (52 2). That is, 2 cards out of 4 good cards, out of 52 total.

Second, you need the complement chance of the flop not showing a deuce: 1 - ((48 3) / (50 3)). That is 3 cards out of 48 good cards (50 remaining minus 2 deuces) out of 50 total.

Then you can multiply all this and raise it to the 3rd power for it to happen 3 times in a row.

(4 2) = 4!/2!*2! = 4*3/2 = 6
(52 2) = 52*51/2 = 1326
(48 3) = (48*47*46)/(3*2) = 17296
(50 3) = (50*49*48)/6 = 19600

Gives: (6/1326) * (1 - (17296/19600)) = 0.0045248 * (1-0.88244) = 0.000531935.

That's the probability of it happening once, 3 times it's: 0.000531935^3 = 1.51*10^-10. Or about one in 6,643,900,265. (that's rounded)

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

1. Have you read the book yet?

2. How do you rate it on a 1-10 scale (rounded to nearest .25 multiple)

3. Will the book prove itself immediately useful for limit or NL hold em?

Re: Special Thread For Chen-Ankenman Mathematics of Poker
 

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You were only off about 7 orders of magnitude. That's not so much, in the grand scheme of things.

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lol, are you [censored] high? go read the "verizon puts a 71$ beat on someone" thread in BBV, and realize that your level of stupidity is up there with the average verizon service rep.