There is a game where, rolls between 1 and 8 let you roll again (up to a total of 6 rolls), if you roll a 1 or 8, or have threw the die 6 times in total, the game is over. The player's overall score is determined by the number of times they managed to throw the die multiplied by a trivial constant a.
The task is to find the expected winnings. To solve this problem, I started off by defining V as the amount of turns the player gets through
v = {1 if only rolls once, 2 if rolls twice, 3 if rolls 3 times,..., 6 if rolls 6 times}
To construct the probability mass function for this, I have considered the fact that when the player throws the die, in essence two things can happen:
- They got 1 or 8, in which case this was their last throw
- They got a value between 1 and 8 so they get to throw again (unless this is the 6th throw)
To calculate {1}, I did 2/8 since there are two possibilities out of 8 which makes the player end up in a scenario where they only threw the die once. The rest of this is where I am stuck. I have tried doing
$$\frac{6}{8} * \frac{2}{8} + (\frac{6}{8})^2$$ for the probability of the second one (two turns) since the player can either have rolled 2-7 first and 1 or 8 just now, OR (which is where my plus sign comes in) they got a value 2-7 in the first turn, and just now, in the second turn.
If I do this for all of them however the total probability within my pmf is not 1 (2.25 something) which tells me it is wrong. Where am i going wrong with my calculation for the probability or how could I approach this problem differently? Thanks
The probability they get $n$ for $n \lt 6$ is $(\frac 68)^{n-1}\cdot \frac 28$. The probability they get $6$ is $(\frac 68)^5$ because they just have to not throw $1$ or $8$ five times and it doesn't matter what happens the sixth throw. You should not have the $(\frac 68)^2$ term because if they roll two other than $1,8$ they roll again.