Given a regular dice game in which each side has an equal probability of $1/6$, you roll the dice. If you get an even number, you roll two dice now and if the sum of the new values is again even, you roll three dice now and this continues until you get an odd sum. This odd sum is your score. What's your average score per turn?
This problem really confuses me as I've been learning probability. What if you continuously get even sums?
On
I am pretty sure the answer is 6 2/3 instead of 6,75
I took off where the previous posts started and wrote everything down and discovered a pattern. I used excel to do the math and will include the image.
Solution:
Half the time you hit an odd number...
So there is 50% chance you hit an odd number throwing 1 die (1+3+5) which is an average of 3. So half the time you can throw again with 2 dice and 1/4 of the time you can throw 3 dice and so forward.
With 2 dice you can throw: odd-even, odd-odd, even-even. Only odd-even ends up in an odd number. the average for an odd die is 3 and the average for an even die is (2+4+6) is 4. So you have 25% of an average of 7 (3+4)
With 3 dice only odd-odd-odd and odd-even-even end up in an odd number for 3+3+3=9 and 3+4+4=11. So you have 12.5% of hitting an average of 10 (9+11/2)
4 dice gives 6,25% of an average of 14 - 5 dice gives 3,125% of an average of 17 - 6 dice gives .... average of 21
The pattern is 3-7-10-14-17-21 (3+4+3+4+3+4+3+4) etc..
Now take a look at the image
I simple added 50% x 3, 25% x 7, 12.5% x 10 .. etc and devided by total %% used which is close to 1.
These type of questions can usually be tackled using recursion. In this case consider the following:
As Peter well pointed out, the probability of rolling an odd sum is always $\frac{1}{2}$, and you have probability 1 of the game ending at some point. So let's say that $E$ is the expected value of the game, while $E_{n}$ is the expected value of the game given that your starting turn is throwing $n$ dice. First time rolling, game will finish if you roll an odd number. This happens with probability $\frac{1}{2}$ and $\mathbb{E}(\text{result}|\text{roll was odd})=3$, while with probability $\frac{1}{2}$ we will roll an even number and now the expected result will be the expected result of a game with two dice (what we called $E_{2}$).
So we write down the following:
$$E = \frac{1}{2}.3+\frac{1}{2}.E_{2}$$
Next step is to work out $E_{2}$, which following our reasoning (you can run the cases), renders:
$$E_{2} = \frac{1}{2}.7+\frac{1}{2}.E_{3}$$
which we can replace in our previous expression for $E$ resulting in: $$E = \frac{1}{2}.3+\frac{1}{2}.\frac{1}{2}.7+\frac{1}{2}.\frac{1}{2}.E_{3}$$
It's easy to see from here that:
$$E = \frac{1}{2}.3+\sum_{n=2}^{\infty}\frac{\mathbb{E}\left( \text{sum of n dice}|\text{sum is odd}\right)}{2^{n}}$$
Now I haven't found an analytical solution for this, but upon running the cases it looks as if for $n\geq2$, $\mathbb{E}\left( \text{sum of n dice}|\text{sum is odd}\right)=3.5n$
Now, if this is true then the answer would be:
$$E = \frac{1}{2}.3+3.5\sum_{n=2}^{\infty}\frac{n}{2^{n}}$$ $$E = \frac{27}{4}$$