Hello,
A fair, six-sided die is rolled repeatedly.
If any side except 1 turns up it is summed.
If a 1 turns up the sum becomes zero. And the next series of rolls begins.
If a sum becomes larger than 20 (i.e. 21, 22, 23, 24, 25, 26) the next series of rolls begins.
What are the probabilities of the sums larger than 20 (i.e. the sum is 21, 22, 23, 24, 25, or 26). In other
words, what is the probability of starting a new series of rolls after 21? And after 22?... And after 26?
(Also what is the probability of starting a new series of rolls if a 1 turns up)
Examples:
2, 3, 4, 1 - a series of 4 rolls, the sum of this series is 0 (as there is 1 at the end);
5, 5, 2, 3, 1 - another series where the sum is 0;
6, 6, 6, 3 - a series of 4 rolls, the sum is 21 (6 + 6 + 6 + 3);
4, 3, 2, 2, 6, 5 - a series of 6 rolls, the sum is 23 (4 + 3 + 2 + 2 + 6 + 5)
The number of rolls can vary from 4 (3 * 6 + 3) to 11 (10 * 2 + 2). Also what matters is the sum of rolls
not their number. So I can't treat these series as permutations or combinations.
I have written a little computer program to simulate the rolls and got the following numbers:
0: 64.137%
21: 9.5194%
22: 9.0316%
23: 7.0866%
24: 5.2019%
25: 3.3639%
26: 1.6596%
I know my numbers are correct because the book I have taken this problem from contains a table with the
probabilities:
21: 9,49%
22: 9,08%
23: 7,08%
24: 5,17%
25: 3,38%
26: 1,66%
Now, my idea was to count the number of series which give 21, 22, ..., 26 (N(21), N(22), ..., N(26)). Then
count total number of all possible series which is number of series having 1 at the end plus number of
series having the sum of 21 plus number of series having the sum of 22 and so on.
N_total = N1 + N(21) + N(22) + ... + N(26)
So, for example, P(21) = N(21)/N_total
Obviously:
N(n) = 0, where n < 0;
N(n) = 1, where n = 0 (there is only one way to get the sum = 0, do not roll the die at all);
N(n) = N(n - 2) + N(n - 3) + ... + N(n - 6); (There is no 1 because if a 1 turns up the sum is 0)
I guess I am on the right track, but I can not be sure because I do not know how to find out the value of
N1 (the number of series having a 1 at the end).
I have tried:
N1 = N(2) + N(3) + N(4) + ... + N(20) + 1 (the idea behind this is at any point before the sum is larger
than 20 a 1 can turn up; if the sum is already, say, 4, the next roll can give a 1, and so there are N(4)
ways to get 1). BUT if I calculate N1 in this way the probabilities do not much with the number I have in the book nor with my the numbers my own program gives.
Also, is it possible to calculate, say, N(21) without previously having to calculate N(20), N(19), ..., N(1)?
Thank you!
A fair, six-sided die is rolled repeatedly.
If any side except 1 turns up it is summed.
If a 1 turns up the sum becomes zero. And the next series of rolls begins.
If a sum becomes larger than 20 (i.e. 21, 22, 23, 24, 25, 26) the next series of rolls begins.
What are the probabilities of the sums larger than 20 (i.e. the sum is 21, 22, 23, 24, 25, or 26). In other
words, what is the probability of starting a new series of rolls after 21? And after 22?... And after 26?
(Also what is the probability of starting a new series of rolls if a 1 turns up)
Examples:
2, 3, 4, 1 - a series of 4 rolls, the sum of this series is 0 (as there is 1 at the end);
5, 5, 2, 3, 1 - another series where the sum is 0;
6, 6, 6, 3 - a series of 4 rolls, the sum is 21 (6 + 6 + 6 + 3);
4, 3, 2, 2, 6, 5 - a series of 6 rolls, the sum is 23 (4 + 3 + 2 + 2 + 6 + 5)
The number of rolls can vary from 4 (3 * 6 + 3) to 11 (10 * 2 + 2). Also what matters is the sum of rolls
not their number. So I can't treat these series as permutations or combinations.
I have written a little computer program to simulate the rolls and got the following numbers:
0: 64.137%
21: 9.5194%
22: 9.0316%
23: 7.0866%
24: 5.2019%
25: 3.3639%
26: 1.6596%
I know my numbers are correct because the book I have taken this problem from contains a table with the
probabilities:
21: 9,49%
22: 9,08%
23: 7,08%
24: 5,17%
25: 3,38%
26: 1,66%
Now, my idea was to count the number of series which give 21, 22, ..., 26 (N(21), N(22), ..., N(26)). Then
count total number of all possible series which is number of series having 1 at the end plus number of
series having the sum of 21 plus number of series having the sum of 22 and so on.
N_total = N1 + N(21) + N(22) + ... + N(26)
So, for example, P(21) = N(21)/N_total
Obviously:
N(n) = 0, where n < 0;
N(n) = 1, where n = 0 (there is only one way to get the sum = 0, do not roll the die at all);
N(n) = N(n - 2) + N(n - 3) + ... + N(n - 6); (There is no 1 because if a 1 turns up the sum is 0)
I guess I am on the right track, but I can not be sure because I do not know how to find out the value of
N1 (the number of series having a 1 at the end).
I have tried:
N1 = N(2) + N(3) + N(4) + ... + N(20) + 1 (the idea behind this is at any point before the sum is larger
than 20 a 1 can turn up; if the sum is already, say, 4, the next roll can give a 1, and so there are N(4)
ways to get 1). BUT if I calculate N1 in this way the probabilities do not much with the number I have in the book nor with my the numbers my own program gives.
Also, is it possible to calculate, say, N(21) without previously having to calculate N(20), N(19), ..., N(1)?
Thank you!