Let's say I have the cards 5 of hearts and 6 of hearts. The 5 cards on the table are 5 of spades, 5 diamonds, 5 of clubs, 6 of diamonds, 7 of diamonds. So now, lets say the only way an opponent will beat me is with a straight flush. I have calculated that there are 3 ways to do this. Opponent can have these following cards to win:
- 3 of diamonds, 4 of diamonds
- 8 of diamonds, 9 of diamonds
- 4 of diamonds and 8 of diamonds
Now, in this case, I am having calculating the conditional probability I win given this information. And how would it change if I had more than one opponent? How would I approach this?
My initial approach was to calculate the probability that the other person would not win: which would be $45\choose2 $ (which stands for all the cards minus the ones on the table and in my hand ) minus $3$. Thus, $({45\choose2}-3)/ {45\choose2} $. However, this was incorrect.
Your answer seems to be correct. Here's another way of calculating it:
In each of the three scenarios your opponent has to have two specific cards, so the probability of this is simply $$3\times 2 \times \frac{1}{45}\times \frac{1}{44}=0.00\overline{30}.$$
The factor of $2$ comes from it not mattering in which order the cards in his/her cards are drawn.
If you have several opponents, nothing significantly changes w.r.t the methodology if only a few (here three) hands can beat you. However, the problem with more opponents is that more hands are possible, so you're having to calculate a lot more.
Say you have two opponents and the same three hands being the only ones to beat you. Now your probability of winning is one minus that of one of the players having one of the three hands plus the probability of both of the players having one of the three winning hands. That is,
$$P(\text{winning})=1-\left[2P(O_1\overline{O_2})+P(O_1O_2)\right],$$ where $O_i$ means that opponent $i$ has a winning hand and where the factor $2$ comes from us not caring whether or not it is opponent one or two winning, as long as it is one of them. As the events are independent, we have $P(AB)=P(A)P(B).$ Can you take it from here? If not, please do let me know.