Don't you consider opponents cards in estimating odds?

[Wozza]

Rank newbie here - hell, it's only my second post...

Been reading up on calculating, for example, the odds of hittng a flush on the turn if you hold 2 spades and the flop drops another 2. Most of what I've read says that there are then 9 further spades available and you start the calcs based on that assumption.

But say there are 7 players, when the turn hits there are 20 cards out. On average, that means that 5 spades have been dealt (20/53 x 13). You've got 2, the flop's got 2 which means that one of the bad guys has also got one and the number of outs should be 8 not 9.

OK, it's not a huge difference, but shouldn't you consider what everyone else is holding when considering outs?

[PairTheBoard]

No.

PairTheBoard

[Mike Haven]

But say there are 7 players, when the turn hits there are 20 cards out. On average, that means that 5 spades have been dealt (20/53 x 13). You've got 2, the flop's got 2 which means that one of the bad guys has also got one and the number of outs should be 8 not 9.

I'll just rewrite this slightly, to help me with my explanation:

"But say there are 8 players, when the flop hits there are 19 cards out. On average, that means that 4.75 spades have been dealt (19/52 x 13). You've got 2, the flop's got 2 which means that one of the bad guys has also got one and the number of outs should be 8 not 9."

No.

Yes: before the 19 cards are dealt the chances are that (19/52) x 13 = 4.75 of them will be spades.

However, once they are dealt, just because you have seen five cards and four of them happen to be spades doesn't mean 0.75 spades are in the 14 unseen cards.

On the other hand, what you could say is that in the other players' 14 unseen cards there are [14/(52-5)] x 9 = (14/47) x 9 = 2.681 spades.

Then you could say that the chances of a spade falling on the turn are
[33 - (9-2.681)]/ (9-2.681) = ( 33 - 6.319)/6.319 = 26.681/6.319 = 4.22/1.

Or.

You could say there are 9 spades in the 47 unknown cards, so your chances are (47-9)/9 = 38/9 = 4.22/1.

[gamble4pro]

Every probability can be calculated, not only odds for own hand. Odds formulas for your opponents' hands are generated by heavier formulas, especially "at least one opponent to..." type events, where the parameter n=number of your opponents is involved. As probability-based strategy, you can compare the odds for an expected own card formation with odds for possible higher formations of opponents.

[d10]

If you knew exactly what everyone held, you would, but you don't. The best you can do is estimate what is most likely. But if you're estimating based on normal distribution, that means you also must account for the fact that close to 3/4 of the cards dealt to the other opponents are not of your suit, and then the resulting mix in the deck is still the same as it would have been if you had just ignored what your opponents might have. So, if you don't know what your opponents hold, don't try to figure it into your calculations.

[PairTheBoard]

"OK, it's not a huge difference, but shouldn't you consider what everyone else is holding when considering outs?"

No.

There are 7 other players. Their 14 cards will Not be the next card dealt. But that is also true of the card burned before the flop, the card to be burned before the Turn, and the 30 remaining cards in the deck which will not be dealt on the Turn. You know NOTHING about any of these cards except that they won't be dealt on the Turn. If you're going to Pretend you know something about the 14 cards in the other players hands you may as well Pretend to know the same thing about the remaining 32 cards not coming on the Turn. Or better yet, don't pretend to know something about any of them because you don't.

PairTheBoard

[Wozza]

Mike,

Thanks for taking the time to post those figures. They certainly helped to clarify the situation. I guess if you play enough (and read enough), these things become a little more obvious.