ReadOnlyAccnt posted...Divide 1 by the number of heroes in the game then multiply that probability by its self 3 times. This is assuming you we're always the first to random and all hero choices were still in the pool. Is that right?

What this is is the chance of randoming Lina three times in a row. You might random her in the last three games, or one of the many variations of it, which I believe you would get by doing:

10!/(7!*3!)

(ten factorial over seven multiplied by three factorial)

Which comes out at 120 possible variations

In the case of a single, specific hero such as Lina determined before the first random it would be:

10C3*(1/95^3)*)(94/95)^7)

Which comes out at 0.00013 (aprox).

I also assumed the question was getting

*any hero* three times in the ten games, which is much more complicated.

For example, it doesn't matter what the first hero randomed is. So to figure out the chance of getting the first hero you random would be the same but r=2 n=9

But that doesn't take into account that you might random a different hero the second time, and the chance of them being chosen three times, and the third time etc.

If it was twice it'd be a lot easier; that way you can assume that you'll either: A get the same hero twice and thus you're good, or B you'll get a different hero, which we can use nCr to figure out the prob' of. With three times you can get the same hero twice and this means you have to find all the probabilities regarding that possibility.

Of course, this all assumes the question is what I think it is: that after ten randoms you'll have three of a single hero. Even if it's not what he originally intended I'm going to have to work it out now >.>

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