How do I calculate the probability of x result occurring y times in z rolls?

Well let's take your example of a 20 on the D20 and 7 or less on the D10, the probability of that event (call it E) on one roll is p=(1/20)*(7/10)=7/200.

To calculate the probability of E occurring a specific number of times let's define the complement event E' which is *not* rolling 20 on the D20 and 7 or less on the D10, this event has the probability q= 1 - 7/200 = 193/200.

So now we want to calculate the probability of E occurring on 6 of the rolls out of 168 rolls, say. This means that E' must occur on the other 168-6=162 rolls.

So we could get a string of results

|--- 6 Es, 162 E's -------|

E' E' E' E E E' E' .... E E'

the probability of this *particular* result is p^6*q^162.

Unfortunately this is not the whole story because the 6 successful events E can occur anywhere in the 168 rolls, we must add the above probability for each possible arrangement of 6 Es and 162 E's. So all we need to know is how many arrangements there are. Well lucky for us this is trivial and given by the binomial coefficient C(168,6)=168!/(6!162!) hence the required probability is

prob(6 Es) = C(168,6)*p^6*q^162 = 0.16 (approx.)

similarly for 10 Es we get

prob(10 Es) = C(168,10)*p^10*q^158 = 0.04 (approx.)

This is referred to as a binomial distribution, we can also use it to calculate ranges of occurrences of E by adding up probabilities E.g.

prob(6, 7, 8, 9 or 10 Es) = prob(6 Es) + prob(7 Es) + prob(8 Es) + prob(9 Es) + prob(10 Es)

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Note carefully that these are *probabilities* which are the *fraction* of successes to total outcomes (prob = (successes)/(total outcomes)).

*Odds* are related but expressed differently, the odds of an event are the ratio of successes to failures ((successes):(failures)), note also that (successes) + (failures) = (total outcomes).