I have been having trouble with the following problem. There are three slots, each randomly picking one of the elements in their set. R, M, and V can only occur once in all three slots, but N can occur any number of times across the three slots. The first slot picks an element first, then the second slot, then the third slot.
The first slot: {R, M, N, N, N}.
The second slot: {V, N}.
The third slot's set can change depending on what was rolled in the first and second slot, but its initial set is: {R, M, V, N, N, N, N, N, N, N, N, N, N}.
In case the above is not clear with the randomization process, the first slot is chosen first, then the second slot is chosen second. Both of these slots are independent, since the second slot's set will not change as result of the first slot's choice. The third slot is dependent on the first two, as the number of elements can be 13 if the first and second slots are N, 12 if one of them are not N, and 11 if both of them are not N.
What is the probability that n slots did not choose N?
I have found 17 valid configurations, given as first-second-third: NNN, RNN, MNN, NVN, RVN, MVN, NNR, MNR, NVR, MVR, NNM, RNM, NVM, RVM, NNV, RNV, and MNV. After calculating the probabilities for each and adding them together, I get around 95.3928%, almost 5% unaccounted for. Please include how you get the answer in your answer, as I have similar issues with another problem. Thank you.
I did something like this to calculate the probability of the last slot chose N: $$\sum_{n=0}^2 (P(n)*\frac{10}{13-n})$$
where $P(n)$ is the probability that n slots before the third slot did not choose N, which is trivial enough to calculate. For R in the third slot, I did the same thing, but $P(n)$ is the probability that n slot did not choose N or R. I did the same for M and V. After these calculations, for the third slot, I got that N has about an 82.9254079% chance of occurring, R and M have about a 6.5501166% chance each, and V has a 3.974359% chance.
I multiplied the probabilities for each slot as necessary to get the following probabilities, which add up to around 95%:
NNN: $0.6*0.5*0.829254079$
RNN: $0.2*0.5*0.829254079$
MNN: $0.2*0.5*0.829254079$
NVN: $0.6*0.5*0.829254079$
RVN: $0.2*0.5*0.829254079$
MVN: $0.2*0.5*0.829254079$
NNR: $0.6*0.5*0.065501166$
MNR: $0.2*0.5*0.065501166$
NVR: $0.6*0.5*0.065501166$
MVR: $0.2*0.5*0.065501166$
NNM: $0.6*0.5*0.065501166$
RNM: $0.2*0.5*0.065501166$
NVM: $0.6*0.5*0.065501166$
RVM: $0.2*0.5*0.065501166$
NNV: $0.6*0.5*0.03974359$
RNV: $0.2*0.5*0.03974359$
MNV: $0.2*0.5*0.03974359$
Edit: I believe I solved part of this. Taking Ross Millikan's advice, I added together the probabilities where the last slot was N, where it was R, where it was M, and where it was V. After doing so, I found that the sums were not equal to the probabilities in the third slot, but rather, the following:
The probability of N was accurate, the probabilities of R and M were off by a factor of 0.8, and the probability of V was off by a factor of 0.5. With this, I surmised that I must've multiplied by the probability that the element was not chosen before twice. Once was explicit in the probabilities right above. The other must've been implicit in how I calculated the probabilities in the third slot. To account for this, I changed MNR, MVR, RNM, and RVM into (M/R)NR, (M/R)VR, (R/M)NM, and (R/M)VM; and I changed NNV, RNV, and MNV into N(N/V)V, R(N/V)V, and M(N/V)V. This led the sum to be 1.
As result, the probability that 0 slots did not chose N is 24.8776224%, the probability that 1 slot did not choose N is 47.7773893%, the probability that 2 slots did not choose N is 24.7249417%, and the probability that 3 slots did not choose N is 2.6200466%. Is this correct, and was the method I used valid?