Say we had a 5 player game with each player having a 20% chance of winning (calculated based on their last 500 games for example), then obviously player 1s probability of winning this particular game is 20%.
But if these numbers were changed, say these 5 players had a really good track record (against other players who are not present at this game) and they had the following probabilities respectively: [20%, 30%, 40%, 50%, 60%] How would I calculate the probability of player 1 winning this particular game?
To be more clear, the [20%, 30%, 40%, 50%, 60%] probabilities would mean that of the last 500 games, player 1 won 100, player 2 won 150, player 3 won 200, player 4 won 250 and player 5 won 300.
Best Answer
The problem can not be solved. To see this, suppose that you simplify the situation.
Player-A plays Player-B.
Player-A wins $70\%$ of the time against everyone.
Player-B wins $60\%$ of the time against everyone.
There is no telling how Player-A and Player-B will do in head to head competition. You can reasonably infer that since Player-A's winning percentage against everyone else is higher than Player-B's winning percentage, that Player-A is the stronger player.
This allows the reasonable conclusion that Player-A's chance against Player-B is $> 50\%$. However, there is no telling how much greater.
Contrast the above problem, with the following (somewhat classic) artificial problem that is often used to introduce Probability students to Bayes Theorem.
There are $2$ weather forecasters A and B.
A is accurate $90\%$ of the time.
B is accurate $80\%$ of the time.
One morning, A predicts that there will be rain that day.
At the same time, B predicts that there will not be rain that day.
You are given the highly artificial premise that the scientific factors used by A and B are generally independent of each other.
What are the chances of rain, that day?
The only way to solve this, is to (somewhat) trip the light fantastic, with respect to the artificial premise.
That is, you have to presume that normally, given the independence of the scientific factors, the following $4$ events are possible:
Then, you are supposed to infer that since A and B disagree, it is impossible for them to both be right, or both be wrong. Then, you conclude that the probability that A is right and B is wrong must therefore be
$$\frac{.18}{.18 + .08} = \frac{9}{13}.$$
The difficulty in trying to graft the weather forecast approach on to the Player-A vs Player-B situation, is that the weather forecast problem can not be solved unless you are given the head-to-head probabilities of the pertinent events.
In the Player-A vs Player-B problem, this information is missing.