View Full Version : How would you calc probability of opponents hands?

Jon Poker
12-14-2004, 04:43 PM
If the board showed:
10h, Jc, 8h, 4h

If you are against 3 opponents, how would you calculate the probability that someone has:
- a pair of jacks
- 2 pair
- a straight
- a flush

In general, I am trying to figure out what equation or process one could use to mathematically determine the probability of certain hands based on the cards on the board.

Any help would greatly be appreciated.

12-14-2004, 07:49 PM
there is an essay in the archives about hand reading...almost every other thread on here has some mention meathods of putting your opponents on certain hands given their image, betting patterns, etc...

12-14-2004, 10:28 PM
In general, I am trying to figure out what equation or process one could use to mathematically determine the probability of certain hands based on the cards on the board.

[/ QUOTE ]


The only way to be mathematical about it is to assume that your opponents are dealt random hands out of the unseen cards. This method is not too helpful, though, as it ignores important facts, like the meaning of your oppo's previous bets and the fact that even poor players throw many hands away (like 72o) preflop.

However, if you are interested in how to solve the problem assuming your opponents cards are dealt randomly out of the unseen cards, I solved a similar problem here:

link (http://forumserver.twoplustwo.com/showflat.php?Cat=&Board=probability&Number=1379771 &Forum=f11&Words=&Searchpage=0&Limit=25&Main=13769 50&Search=true&where=bodysub&Name=20330&daterange= 1&newerval=4&newertype=d&olderval=&oldertype=&body prev=#Post1379771)

See my last response, especially, for the kind of methods you use. I urge you not to try to learn this stuff to improve your poker game, though -- there are much better ways.

If you are still really interested in seeing the specific problem posted solved, let me know and I'll do it. Again, the only real use of this exercise is to improve your probability skills in a way that will not translate directly to the poker table.


Jon Poker
12-15-2004, 11:52 AM
I am still looking for someone to answer my questions. I am in need of a mathematical approach or formula to determine the probability of the outcomes I listed above. If anyone is mathmatically smart enough to solve it please do so. It would be much appreciated.

As far as using it to improve my poker game, ignore that. That is not my goal. I am just curious because I saw somewhere were someone was using the probability (not taking into account betting patterns, image, etc) of opponents holding certain hands.


12-15-2004, 03:03 PM
I don't know about a formula, but here is the thought process:

Opponents holds:
Jacks: 3 ways / c(50,2)
2pair: 54 ways / c(50,2) ie JT,J8,J4,T8,T4,84
Str: 32 ways / c(50,2) ie Q9 and 97
Flush: 120 ways / c(50,2)

So you would sum all of these if you wanted the probablity that he has any one of the hands, but make sure you aren't double counting the straights and the flushes...


12-15-2004, 03:46 PM
3 ways / c(50,2)

[/ QUOTE ]

This is completely wrong. Given the flop (and not knowing our own cards, since the OP did not give them) there are 48 unseen cards left. 45 of them are non-jack cards. We'll calculate the probability that no oppo has a J, and then subtract it from 1 to determine the chance that at least one oppo has either a pair or a set of jacks:

1 - ((45 choose 2)/(48 choose 2))*((43 choose 2)/(46 choose 2))*((41 choose 2)/(44 choose 2))= .3362

So there is about a 34% chance that at least one of your oppo's has at least one jack.

Maybe I'll do the others later.