I teach AP Stats, and one of the other math teachers came to me with a game he and his buddies came up with. I have a feeling it was a drinking game, but whatever. I like math and I'm a nerd (and he knows that).

To win the game, you need to roll three 1s before you roll a 6, on a fair die. All non 1s and 6s result in a reroll.

1a) What's the probability of rolling three 1s before you roll a 6?
1b) What's the expected number of trials before you win?

2) What's the expected number of total rolls before you win (including 2-5s)?


----------------------------My Thoughts-------------------------------------------

1a) Since P(1) = P(6) = 0.5, the probability of winning this game is 1/8.

1b) Since this game would be a Geometric Distribution, 1/(1/8) = 8 trials expected to win this game.

2) To figure out the expected number of rolls to win, I ran a simulation, but I have no idea how to calculate this one.

I fiddled with this tonight.

1) Yes, those are straightforward to understand and demonstrate

2) If you write out the formula for the probability of winning on roll T (making sure to normalize across only winning sequences), you arrive at the following formula for the expected number of rolls to win (where I have substituted Q=T-3).

E = [Sum((Q^3)*p^Q) + 6*Sum((Q^2)*p^Q) + 11*Sum(Q*p^Q) + 6*Sum(p^Q)] / [Sum((Q^2)*p^Q) + 3*Sum(Q*p^Q) + 2*Sum(p^Q)]

where the Sums all go from Q = 0 to infinity and p=2/3.

The solutions of all of these power series in terms of p are all well-known and available in books and on the internet.

Solving yields:

E = 486 / 54

E = 9

Hope this helps and matches the results of your simulations.

P.S. This is probably obvious, but the way you win on roll T is having 2 1's and T-3 2-5's in the first T-1 rolls and then rolling a 1 on the T-th roll.


Edit: Others in this forum are much better at solving these types of questions in a much more clever way than my solution approach which was to laboriously write out all the formulas, simplify, and then solve the resulting power series.

This problem is interesting since at first sight one might say that E = 18 (since you are waiting for 3 ones, each of them has a 1/6 chance of hitting) or E = 15 (since you don't want to see a 6, and there are 5 other numbers and so one might say that, given that no 6 has occurred, the chance of having a 1 is 1/5).

Actually, the correct result is 9 (as whosenext showed) and the correct reasoning is the following. Say that you want just one 1. The game stops whether you see a 6 (you lose) or a 1 (you win). The combined chance of having a 6 or a 1 is 1/3. So, the game lasts on average 3 rounds and the expected number of rolls to have a 1 given that no 6 occurred is 3.

Once you got a 1 (and you need 3 rolls on average to reach it), you of course need other 3 rolls to get another 1 if you don't get a six and another 3 to have 3 1s. So, the total is 9.

good enough for me.
This game and Qs
cries out for a Markov chain solution ( I do them in Excel 1st) but also made a simple recursion for this too. Easy to get the distributions.
and see what goes on.
simple math too

Here is my Excel in Google link for those interested
https://goo.gl/kh8vtm

looks like Google sheets likes to change the default Excel precision too

I too get 9 rolls given a win for the mean
mean loss given a loss: 4.23 rolls
either:5.25

p(win) = 1/8
*****
only winning 1 in 8 tries
makes me want to drink!

enjoy
thanks for the nice question too

Because English is not your first language you said "correct reasoning" rather than "easiest reasoning", not realizing that your words could be construed as impolite.

But your words were essentially true. Markov chains? Power series? Cmon.

hey, some like heavy artillery. nothing wrong with that!
Oh, I see now
being funny
LOL
*****
the basic math
*****
1s and 6s (as the other faces are not relevant)
the probability of one or the other would be 1/6 + (1/6 + 1/6) = 1/2 = p
the craps principle there

the winning sequences
111 = p^3 = 1/8
6 = p^1 = 1/2
16 = p^2 = 1/4
116 = p^3 = 1/8

for the mean # of total rolls for either face1 wins or face6 wins
3* 1/8 + 1* 1/2 + 2* 1/4 + 3* 1/8 = 3/8+4/8+4/8+3/8 = 14/8=1.75
the average # of rolls to see either face1 or face6 = 6/2 = 3
3*1.75 = 5.25

for the mean # of total rolls given face1 wins =
3* (6/2) = 9

for the mean # of total rolls given face6 wins =
we have to normalize here (left to the reader as a fun exercise)
1.571428571*3=4.714285714 (rounded)

fun is fun
math is fun!