I roll a dice, every time I get a 6, I give you X amount of money.
Every time I do NOT get a 6, I increase X by 1
Example in 5 rolls:
begin X = £1, Money = £0
1- 1 (X = 2)
2- 3 (X = 3)
3- 6 (Money = Money + 3 = 3) (X = 3)
4- 4 (X = 4)
5- 6 (Money = Money + 4 = 7) (X = 4)
So in total you win:
0
0
3
0
4
so total money you won = £7
Now what I'm trying to calculate is:
- The average money you get awarded in each of the 5 rolls.
- The average overall money you win in the 5 rolls
For the very first roll, the average is easy to calculate and equal to (1/6 * £1)
For the second roll, you can calculate manually as:
chance win on 1 * chance win on 1 * £1 + chance no win on 1 * chance win on 2 * £2 =
1/6 * 1/6 * £1 +
5/6 * 1/6 * £2 = 1/36 + 10/36 = 11/36 * £1
Also a slightly more complicated version of this problem is:
If you get a 6 OR 5 I increase the number of available rolls by 1, so another example:
Money = £0, X = £1, N = 5
1- 5 (X = 2) (N = N + 1 = 6)
2- 3 (X = 3)
3- 6 (Money = Money + 3 = 3) (X = 3) (N = N + 1 = 7)
4- 4 (X = 4)
5- 2 (X = 5)
6- 6 (Money = Money + 5 = 8) (X = 5) (N = N + 1 = 8)
7- 2 (X = 6)
8- 3 (X = 7)
Total rolls = 8, total Money you won = £8
Predicted answer (corrected):
For the simple version of the problem, I wrote a simulation and the correct answer for the average win in 5 rolls is:
2.221261
Let $X_n$ denote the value of $X$ at the start of dice roll $n$.
Then, you have the recurrence relation $X_n=\begin{cases} 0 & \text{w.p. }\frac{1}{6} \\ X_{n-1}+1 & \text{w.p. }\frac{5}{6} \end{cases} $ Therefore you have the recurrence relation $E[X_n]=\frac{5}{6}\left(E[X_{n-1}]+1\right)$.
Solve this recurrence for $E[X_n]$ using your favorite approach. You can just do it by hand if you only want it for up to $n=5$ dice rolls:
$E[X_1]=1$
$E[X_2]=5/3$
$E[X_3]=20/9$
$E[X_4]=145/54$
$E[X_5]=995/324$
The expected money received after dice roll $n$ is $M_n=\frac{1}{6}E[X_n]$. If you compute the expected money after 5 rolls, you get 1.77
Does this help?
By the way, I wrote a python code to simulate this experiment and the complex experiment as well for 10000 trials. The average I obtained agree with my calculations above. You must make sure you have enough trials, otherwise the randomness will skew your calculations.
Here is the result:
Simple Experiment: 1.7867
Complex Experiment: 2.8905