Legit way of calculating odds? |
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Posted Wed Aug 20, 2003 5:37 pm GMT by mrfloppy
Standard formula for figuring odds...take the number of unknown cards after the flop 47(3 card flop and the 2 cards you're holding of course for 52) and divide it by your number of outs. So if you're looking at a standard flush equation...you've got pocket clubs 2 clubs come on the flop...that leaves 9 outs(13 clubs minus the 2 on the flop and the 2 in your hand) so you take 9/47 and get about 19.1% chance of hitting your flush.
Now I wanna look at some things I thought up in my head to see if it's justifiable...Let's figure a 10 person table...that means 20 cards are dealt. There are 4 suits so 20/4...you figure there's a total of 5 of each suit on the table. Let's say you have those pocket suited clubs again. Flop comes and you grab 2 more clubs of course. So you've dumped 5 clubs on the deal, and 2 more on the flop for a total of 7. 13-7 that leaves you 6 outs to hit your flush. 20 cards dealt on the deal plus the 3 card flop 52-23=29 so you divide the 6 remaining clubs into 29 and that takes your percentage to 20.6% of hitting that flush.
So my question is does that 1.5% variation mean much? Is this a legit way of figuring odds? I haven't checked this theory out on other odds draws to see if it affects the percentage greatly or not. Just looking for opinions, critcism or otherwise.
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Posted Mon Aug 25, 2003 3:30 pm GMT by Adam Marshall
Good question 
The derivation comes from the fact that you assumed that 5 clubs are dealt out, and YOU got two of them. Since you are already assuming that you are getting dealt two clubs, you have to refigure the distrubution among the other players. There's only 11 clubs left outta 50 cards. So the other 9 players are going to be dealt 18 of those 50 cards. Chances of getting a club is not typical anymore. The chance of one player getting dealt one or two clubs is 1 - ( 39/50 )( 38/49 ) = 39.51%
instead of the usual 1 - ( 39/52 )( 38/51 ) = 44.12%
that's where the inequaliy comes from. I'd get into more detail, but I hope you get the picture as it would destroy my brain to figure out the equation.
Posted Mon Aug 25, 2003 7:10 pm GMT by kluCAR
no no no. it doesn matter what the other player has got. when you are lookin for a card u just devide number of outs with UNSEEN cards (and that includes cards in others players hands). They are randomly delt so u dont need to calculate it in your odds.
Posted Wed Aug 27, 2003 9:18 pm GMT by Adam Marshall
I know!
We're talking about the derivation that occurs if you consider the statistical probability that other players hold those cards. Totally different kinda thinking.
For instance, the chance of getting dealt pocket aces is 1 in 221 or so. If you get dealt those two magical aces the chance of someone else having two aces also wouldn't be 1 in 221. It'd be 1 in 1225 or something. By the same type of thinking we're applying, if you didn't get dealt any aces, the chance of someone else holding pocket aces jumps up to about 1 in 204. Now assign that value to j...
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