You roll a die until you get a $5$, what is the expected value of the minimum value rolled?

I have found a different way to think about to this problem for those still a little stuck. Let m = the minimum number found. E(m) = 1*P(m=1) + 2*P(m=2) + ... + 5*P(m=5) Notice we do not include the probability that m=6 as we know we have rolled a 5 (this is when the game stops) so we cannot have our smallest number to be 6.

Lets have a look at P(m=1): This is effectively the event that we get a 1 before we get a 5. Whats more likely, getting a 1 before a 5 or a 5 before a 1? Naturally they are both equally likely so P(m=1) = 0.5

Now lets look at P(m=2). We need two things here. To get a 2 before a 5 and to get a 5 before a 1. I.e we have 2 as our minimum but the game has stopped before we had a chance to change that minimum to 1. We now have 6 possible choices in the race of 1,2 and 5. We could have 512 or 125 or 251 or so on. Think of it as a race between 3 horses, all outcomes equally likely. (in essence we are also "racing horse 4 5 and 6" but we dont care how they perfom in the race just the relative positions of 1 2 and 5) We have 3 choices for First place, 2 for second and 1 for third. 3! = 6. Of these 6 choices the ONLY way in which m=2 is if we have 251. 1 choice out of 6 thus P(m=2) = 1/6.

Now onto P(m=3) again we are looking at the positions of horses 1,2,3 and 5. With a total of 4! possibilities we must have either 3512 or 3521. P(m=3)=2/4! = 1/12.

P(m=4). we have 5 horses now racing: 1,2,3,4 and 5. with a total of 5!, 120 outcomes and m=4 only occuring on the outcomes: 45xyz. with x,y,z taking any choice in 123. there are 3! choices for xyz thus 3!/5! = 1/20 = P(m=4)

P(m=5). Just look the step above we are looking at the relative positions between horses 1,2,3,4 and 5. This time m=5 if and only if 5 wins the race! So successful outcomes are of the form 5xyzt. with xyzt in 1234. There are 4! ways to arrange horses 1234 thus P(m=5) = 4!/5! = 1/5.

Can you spot a pattern for P(m=i)?

Putting this all in gives E(m) = 1*1/2 + ... + 5*1/5 = 137/60.

:) Oskar

I seem to get $\dfrac{137}{60}$, slightly more than $2$, which seems as plausible to me as your similar answer

More precisely: $$1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15$$

I think you are saying that if only $6$s are thrown before the first $5$ then the minimum is $6$ to calculate

$$1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac16+6 \times \frac1{30}$$

I am saying the minimum is $5$, since a $5$ is thrown to stop the game. This is what is causing the difference between $\frac{139}{60}$ and $\frac{137}{60}$

I wanted to point out that we don't need infinite series or fancy formulas to solve this problem.

Any infinite sequence of die rolls yields a permutation of $\{1,2,3,4,5,6\}$ by simply retaining all of the first appearances of each number. By symmetry, all $6!$ permutations are equally likely. Therefore to calculate the desired expectation, one can simply enumerate the $6!$ permutations, find the minimum values before the $5$ in each case, and average the results.