There is a sequence of events that return a number. Here is a sequence:
For a die with a probability of each side 1, 2, ..., 6 equal to 0.05, 0.1, 0.15, 0.2, 0.25, 0.25 someone throws a it and records a value. Then a person continues to throw it and if the value is not the same as recorded on the previous step, he records a new value. This continues till he recorded 6 values.
For example a person throws 1, 2, 2, 2, 2, 4, 2, 2, 3, 5. He will record 1, 2 than he skips three 2 because it was recorded previously, then records 4, 2, skips 2 and records 3 and 5. So the resulting output is the sum of all the recorded values 1 + 2 + 4 + 2 + 3 + 5 = 17
Knowing the probability of the loaded die, how can I calculate the variance and expected value of the event?
I know how to calculate it when the I can record any value, but the fact of ignoring the skipping the duplicate makes it really hard.
Using brute-force techniques (https://github.com/barrycarter/bcapps/blob/master/MATHEMATICA/bc-solve-math-1615460.m), I got the answer 470474386324001/18866716141824 which is about 24.9367, slightly lower than the 25.5 value you'd get if you did count repeats. This may seem strange since there's a 50% chance of rolling a 5 or a 6, but the fact that we remove double 5s and double 6s seems to reduce the value more than enough to compensate.
Given the "complexity" of this answer, I don't think there's a non-brute-force way to solve this problem, although I could be very wrong.
My approach:
The probability distribution for the first roll is given, and the expected value happens to be 17/4.
If the first roll is a 2 (for example), the next number we record must be 1, 3, 4, 5, or 6, since we disallow repeats. The relative probabilities of these values are (0.05, 0.15, 0.2, 0.25, 0.25), which adds up to 0.9 (which we expect, since the probability of 2 was 0.1). To turn these probabilities into absolute probabilities, we divide by 0.9 to get: $\left\{\frac{1}{18},0,\frac{1}{6},\frac{2}{9},\frac{5}{18},\frac{5}{18}\right\}$
So, the probability of rolling a 6 after you've rolled a 2 (for example) is 5/18.
Note that I've set the probability of rolling another 2 after the first 2 is 0, because you will never record a 2 following another 2.
Brute force computing the probability of recording an 'n' after an 'm':
where the row is the first roll and the column is the second roll.
For example, to find the chance of rolling a 6 after you've rolled a 2, we look in row 2, and column 6, to get 5/18, as we did earlier.
Note that this is quite different from the chance of rolling a 2 after a 6 (row 6, column 2) which is 2/15.
$\{3,5,6,5,3,1\}$
Note that these 46656 tuples include tuples with consecutive duplicates like this:
$\{3,6,6,2,6,1\}$
but I will compensate for those cases below.
$\{3,5,6,5,3,1\}$
for example, we proceed as follows:
The probability of the first roll being 3 is 0.15 (given)
To find the probability that the next roll will be 5, we look at row 3 column 5 in the table to find 5/17.
Similarly, to have a 6 follow the 5, 1/3
5 following the 6, 1/3
3 following the 5, 1/5
1 following the 3, 1/17
Multiplying these, we get 1/17340.
The sum of this tuple is 3+5+6+5+3+1 or 23, so it's contribution to the expected value is 23/17340.
$\{3,6,6,2,6,1\}$
In the course of this calculation, we will find the probability of transitioning from a 6 to a 6, which is 0. Thus, when we multiply the probabilities together, we will get 0, and this tuple will contribute nothing to the expected value as desired.
EDIT: I tried solving this with arbitrary values for the probabilities, but the results were not useful. See https://github.com/barrycarter/bcapps/blob/master/MATHEMATICA/bc-solve-math-1615460-2.m for more details