Difficult Probability question

[theredpill]

Forget about Poker for a second or just put it in the back of your mind. Let's just think about math.
Suppose you knew what your win % was for a certain hand against one player on the table, call him player A, then you knew what your win % against Player B was, and finally Player C.

So you have 3 win percentages, one on Player A, one on Player B, and one on Player C, how would you figure out what your win % would be for a given hand when all 3 players are in the flop against you at the same time?

Like if it was 10% that you win against Player A
25 % that you win against Player B
14 % that you win against Player C

But these were all computed exclusively heads up but now you are playing them all at the same time. Is there a mathematical way of figuring these up ? My first "dumb" answer was to add all of the percentages and divide by 3 and find an average but this seems to simple. Any geniuses on here that can help me?

[pokerclacker]

with the information you've provided there's no way of figuring out your % chance to win. the calculations would be a bit more complicated than adding and dividing by 3.

lets say you're in a simpler situation, playing a fictional dice game with a 100 sided. the point of the game is to role a # that no one else has selected. for every $ you throw in you get a random # and thus wager =payoff, if winner takews all and ties are split pots. player A pays $90 selects 90 random #'s. You have a 10% chance to beat player A and win the $90 and you have a 90% to loose to player A and loose your ante. Player B pays $75 and recieves 75 random #'s (many of which are the same as player A's). You have a 25% chance to beat player B. Player C pays $86 to get 86 random #'s. you have a 14% chance to beat player C. You role the die. You have a 10% chance to beat player A, but 75% of the time you beat player A, you will still lose to player B thus .10x.25= .025, or a 2.5% chance to beat both players A and B simultaniously. you only have a 14% chance on top of that to beat both players A, B and C, so .025x.14=.0035, thus you have a .35% chance to beat all three players at once, not so good is it?

this .35% chance isn't entirly realistic though. when looking for specific cards it may become much more complicated. lets look at another situation where you're drawing 'almost' dead, while still having outs against 2 apponents individualy. lets say you hold J /images/graemlins/diamond.gif 10 /images/graemlins/diamond.gif, Player A holds J /images/graemlins/spade.gif J /images/graemlins/heart.gif, and player B holds A /images/graemlins/diamond.gif K /images/graemlins/diamond.gif, and the flop comes: J /images/graemlins/club.gif 5 /images/graemlins/diamond.gif 3 /images/graemlins/diamond.gif.
you have a 28.18% chance to win against player A alone, and a 51.31% chance to win against player B. we would expect you to have a 14.45 shot to win the hand, however due to killed outs and cards that will improve your apponents more than you, you really have a 0 percent chance to win the hand with both players in!! don't believe me? go to www.twodimes.net (http://www.twodimes.net) and see for yourself.

basically the moral is, if you're outs are killed, get out, and if you're drawing against a large field be very carefull to have the odds necessary to make it worthwhile.

[theredpill]

Thanks for your reply, clacker. I went to twodimes but I can't figure out how to use it. Where do I put my hand in, and what are these "dead cards" all about. Aren't dead cards the ones you accidentally see when someone tosses their cards in and they are flipped over? Can you tell me where to put my hand at ? And other players hands? Or direct me to a manual ?

Thanks for the example. I understand what you mean. I guess it is the multiplication of the probabilities. However, lets talk about the long run. If it were the long run then, I would just multiply all of my probabilities and not really be worrying about the outs and in a perfect world and perfect distribution, it would work out. I understand that it is never perfect, though.

Anyone else have a second opinion?

[theredpill]

Nevermind I figured it out but I still don't understand what the dead cards are for.

[pokerclacker]

dead cards are exposed cards no longer involved in the pot. for instance if some one folds in 7stud with a Q /images/graemlins/diamond.gif 7 /images/graemlins/spade.gif, showing. you would have to account for these cards in whatever probability calculation you may be doing. Basically all dead cards are, are cards no longer in the deck and no longer in an active player's hand. (if some1 accidently flips a card, it's a dead card also)

[theredpill]

Another question.
In general, with a flop like

10 /images/graemlins/diamond.gif 6 /images/graemlins/diamond.gif K /images/graemlins/club.gif

and you holding A /images/graemlins/spade.gif 9 /images/graemlins/diamond.gif

Would you expect to win more hands against a late position player or an early position player? Assuming that

1) You play each one heads up
2) They hit something on the flop : top pair, straight draw, flush draw, flush, straight, 3 of a kind, two pair, full house, overpair etc...
3) we are excluding crappy hands like bottom pair and overcards

That means it excludes a pair of 6's and a pair of 10's.

Thanks.