How to find n average number of trials before criteria are met with differing probabilities per outc-CodePudding

I've spent a few days trying to figure this out and looking up tutorials, but everything I've found so far seems like it's close to what I need but don't give the results I need.

I have a device that produces a single letter, A-F. For simplicity's sake, you can think of it like a die with letters. It will always produce one and only one letter each time it is used. However it has one major difference: each letter can have a differing known probability of being picked:

A: 25%
B: 5%
C: 20%
D: 15%
E: 20%
F: 15%

These probabilities remain constant throughout all attempts.

Additionally, I have a specific combination I must accrue before I am "successful":

As needed: 1
Bs needed: 3
Cs needed: 0
Ds needed: 1
Es needed: 2
Fs needed: 3

I need find the average number of letter picks (i.e. rolls/trials/attempts) that have to happen for this combination of letters to be accrued. It's completely fine for any individual outcome to have more than the required number of letters, but success is only counted when each letter has been chosen at least its minimum amount of times.

I've looked at plenty of tutorials for multinomial probability distribution and similar things, but I haven't found anything that explains how to find average number of trials for a scenario like this. Please kindly explain answers clearly as I'm not a wiz with statistics.

CodePudding user response：

In addition to Severin's answer that logically looks good to me but might be costly to evaluate (i.e. infinite sum of factorials). Let me provide some intuition that should give a good approximation.

Considering each category at a time. Refer this math stackexchange question/ answer. Expected number of tosses in which you would get the k number of successes for each category (i) can be calculated as k(i)/ P(i):

Given,
p(A): 25% ; Expected number of tosses to get 1 A = 1/ 0.25 = 4
p(B): 5% ; Expected number of tosses to get 3 B's = 3/ 0.05 = 60
p(C): 20% ; Expected number of tosses to get 0 C = 0/ 0.20 = 0
p(D): 15% ; Expected number of tosses to get 1 D = 1/ 0.15 = 6.67 ~ 7
p(E): 20% ; Expected number of tosses to get 2 E's = 2/ 0.20 = 10
p(F): 15% ; Expected number of tosses to get 3 F's = 3/ 0.15 = 20

you get an idea that getting 3 B's is your bottleneck, you can expect on average 60 tosses for your scenario to play out.

CodePudding user response：

Well, minimum number of throws is 10. Average would be infinite sum

A=10•P(done in 10) 11•P(done in 11) 12•P(done in 12) ...

For P(done in 10) we could use multinomial

P(10)=Pm(1,3,0,1,2,3|probs), where probs=[.25, .05, .20, .15, .20, .15]

For P(11) you have one more throw which you could distribute like this

For P(12) you have to distribute 2 more throws. Note, that there are combinations of throws which are impossible to get, like Pm(2,3,0,2,2,3|probs), because you have to stop earlier

And so on and so forth