Quick probability question... |
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Posted Thu Aug 12, 2004 1:16 pm GMT by Johnny T
If I hold an A 4, what's the chances of getting two pairs on the flop (ie. a flop like A 4 x) ?
How do you work out the probabilities of something occurring.. ? Is there a quick shortcut method that you can use to work them out as you play ?
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Posted Thu Aug 12, 2004 8:41 pm GMT by wEbMaStEr
I'm sure someone will correct me if i am wrong here, i always said, numbers are not my strong point.
ok so you have A4 you want to know what chances are of pairing both....
there are 4 of each card so as you already hold 1 of each there are 3 left of each.
now it depends on how many others are playing and what they have, but lets suppose for ease you are heads up.
There are 52 cards in the deck you have 2 cards and your opp has 2 cards thats 4 from 52 which is 48
out of this 48 you have a chance to hit 6 so, 48/6 or to simplify 8/1
so you have an 8 to 1 probability of the 1st card out being any 1 of the cards you want.
the chances of hitting any 1 of those 6 increases ever so slightly on each subsequent card to come out because you are only 48/6 on the 1st card, the second card out is 47/6, the 3rd 46/6 and so on as the amount of cards avaliable decreases.
Now, this is where i get lost myself, the above are the chances of any
1 card coming out.
I just had to delete this whole next part and start over because the math was wrong,  should be right now...
to get the probability of both cards coming you have to multiply the odds for each event.
So for the 1st 2 cards out to both be 1 of your cards you would multiply 48/6 by 47/3 (this was the part i got wrong, the number of cards you can now hit goes down because we are presuming you already hit 1 so you now only have 3 of the other card to hit)
so for cards 1 & 2 being an A and 4 probability would be 48/6 X 47/3 which is 125.333333/1
This is specific for cards 1 & 2 tho, not your flop, to get that you would have to work out the chances of your cards hitting in positions 1&3 and 2&3 also
which would be 48/6 X 46/3 = 122.666667/1
and 47/6 X 46/3 = 120.111111/1
and then add these together and then divide by 3 to find your average
(120.111111 + 122.666667 + 125.333333) / 3 = 122.703704
So your probability of holding an A4 and hitting both on the flop would be
122.703704/1
I think?
nah im pretty sure thats right 8)
that took me the best part of an hour to work out, yet when playing i can work out outs in a second 
Don't dare ask about the full house!!!
lol nah j/k you can work that out also using the above method
Posted Thu Aug 12, 2004 8:59 pm GMT by wEbMaStEr
sorry you wanted a quick way? 
yeah sure whatcha do is very roughly, unless you can do that sorta thing instantly in your head,
for my 1st card i have 6 outs from 48 = 8/1
for my 2nd i then have 3 outs from 48 (cos its easier than 47 and this is very roughly) = 16/1
multiply...
128/1
how exact you are doesn't really matter cos in poker its all relative...
there are 2 cards left to come and you need an A and a 4 ...
odds = about 120/1
relative odds of it being a 6 and a 9 also = 120/1
see what i mean? odds are relative cos the next card could pretty much be anything
Posted Fri Aug 13, 2004 12:46 pm GMT by Johnny T
wEbMaStEr,
A big thanks for your reply mate. I read it a couple of times... then I got the calculator.. then I read it again and I understand where your coming from. It is all relative I suppose because your opp could be holding A 4 himself?
Still, it gives me something to go on. I've not managed to get into the calculating probability or odds in my head as I play yet. But I've realised that to advance my game I'm going to have to start doing it sooner or later. At the moment I just do it by 'feel'. This method is very hit and miss as you can imagine but generally if I'm in doubt I fold.
The short way you descibed sounds ideal.
Again, a big thanks for your post, its appreciated
:D
Posted Fri Aug 13, 2004 1:55 pm GMT by robbieg
The mistake you are making is disregarding the cards your opponents hold from the pool of possibilities. Because you haven't seen these they are just as unknown as the cards left in the deck. Thus, for the first card you would have six outs out of 50 (not 48 ), or .12 chance to get one pair. Then you are left with only three outs because you here want not trips but 2 pair. But you have seen one more card so now your chances are 3 out of 49, or .061 percent. If you didn't make it on the second card, the odds on the third are 3 (still 3 outs) out of 48 (because you have eliminated one more bad card), or .0625 percent. But you have two shots on getting the second pair on the second or third card, so I would combine those odds (.061 + .0625 = .1235). Now, you need both of these things to happen (1 AND (2 or 3), so you multiply the two chances by each other: .12 X .1235, which gets you about 1 .015 or 1.5% chance of getting exactly 2 pair on the flop. This could be totally whack but it's what I remember from high school.
Posted Fri Aug 13, 2004 2:52 pm GMT by Johnny T
robbieg,
Thanks for that. I've since found this link through this site:
http://www.cardplayer.com/poker_odds/#
There's a card odds calculator on there that just shows the odd for any given hand. Its quite useful.
Thanks again for your post. Much appreciated.
:D
Posted Mon Aug 16, 2004 3:43 pm GMT by vegasholdem
The odds of you flopping a 2pair on the flop is 28-1 against!!
click on my signature to learn more...
Posted Mon Aug 16, 2004 4:44 pm GMT by Johnny T
| vegasholdem wrote: | The odds of you flopping a 2pair on the flop is 28-1 against!!
click on my signature to learn more... |
First glance at your site and it looks good. Will have a closer inspection tomorrow when I have more time. Cheers.
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