Someone double check my math, please.

I'm converting a board game from using D10 to multiple D6 and I want to make sure I'm calculating the probabilities right.

So if a 'hit' occurs on a 5 or 6, the odds of a single hit is obviously 1/3.

So for two dice, the odds of a single hit is:  P = 1 - (2/3)^2  ??

And then the odds of at least one hit on N dice is P = 1 - (2/3)^N ?

Do I just multiply those probabilities to get the odds of, 2 or more hits?

So the odds of M hits on N dices is P^M?

Seems like I'm missing something, but I don't have my textbooks with me and I'm having to rely on web based math sites.

I know I can trust TR.  (<--- memescape material). 

ETA: Fixed basic prob cuz I can't read my own equations. (I did that part right in my spreadsheet).

I think your probability statement is wrong since it doesn't result in the correct answer for only one die.

P = 1 - (1/3)^1 = 2/3 which is wrong for your set-up.

For N dice, the probability of getting exactly M hits (no more, no less) is N!/(M!(N-M)!) (1/3)^M (2/3)^(N-M)

You can see that that works out to (2/3)^N in the case M=0 (no hits), or to (1/3)^N in the case M=N (all hits).

So probability of exactly one hit is N (1/3) (2/3)^(N-1).

Probability of at least one hit is, as you said, 1 - (2/3)^N.

Probability of at least two hits is 1 - (2/3)^N - N (1/3) (2/3)^(N-1).

I don't know of any neat and tidy formula for the probability of at least K hits.

I would just add up the probability of exactly M hits for M running from K to N.

Or else add up the probability of exactly M hits for M running from 0 to K-1, and subtract that sum from 1.

Use an app