Probability to win against more than one opponent?

[Peter-23]

Lets say I know my exact probability to win against one opponent is 35%.

If I somehow know that my probability to win a specific hand is the same against any opponent what is my probability to win against two opponents? Or three?

Or another situation:

Lets say I estimate my probability to win against:
Opponent 1 is 35%
Opponent 2 is 25%
Opponent 3 is 45%
What would be my combined probability?

I would greatful if someone could give me a formula for this kind of calculations.

[pzhon]

There is no such formula. There isn't enough information.

Suppose hand A beats hand B and C 51% of the time individually.

Hand A could beat both 53% of the time: A2o versus T9s and T9s in different suits. http://twodimes.net/h/?z=787316

Hand A could beat both 30% of the time: 33 versus AKs and QJo. http://twodimes.net/h/?z=787314

You have to think about how a hand wins. How often does it make huge hands that will be best against many opponents? How often does it make a weak hand that wins only if the opponents miss?

[Peter-23]

Doesn't that have to do with how I came up with my probabilitys to win? Im looking for somthing when that work is done.

Lets use the the first situation but change the probability to 55% to win against one opponent.

I am at the river and I have two opponents left. Exactly 45% of the possible twocard combinations not accounted for will beat me, the rest will give me the pot (for simplicity we can say there is no possible split here).

For simplicity we can say that i know for sure that these two always play everyting to the end. So the only information I lack is what they have at this instance.

What is the probability for both of them to have a hand among the 55% that I beat? And is there a formula for calculating this?

Do I really lack information here?