an interesting probability problem

Rishi wrote:>>> I kinda fixed the counts. Are we supposed to come up with an expression involving K?


1,1,1,1,1,1,1,1 (1 way)
1,1,1,1,1,1,2 (7 ways)
1,1,1,1,1,3 (6 ways)
1,1,1,1,4 (5 ways)
1,1,1,5 (4 ways)
1,1,6 (3 ways)
1,2,5 (3 ways)
1,3,4 (3 ways)
2, 6 (2 ways)
3, 5 (2 ways)
4, 4 (1 way)



If K=8, the total number of ways Tk=8 is 1. The probability of getting Tk=8 for K=8 is 1/(6**.
If K=7, the total number of ways Tk=8 is 7. The probability of getting Tk=8 for K=7 is 7/(6**7).
If K=6, the total number of ways Tk=8 is 6. The probability of getting Tk=6 for K=6 is 6/(6**6).
If K=5, the total number of ways Tk=8 is 5. The probability of getting Tk=8 for K=5 is 5/(6**5).
If K=4, the total number of ways Tk=8 is 4. The probability of getting Tk=8 for K=4 is 4/(6**4).
If K=3, the total number of ways Tk=8 is 9. The probability of getting Tk=8 for K=3 is 9/(6**3).
If K=2, the total number of ways Tk=8 is 5. The probability of getting Tk=8 for K=2 is 5/(6**2).




So you add up these individual probabilities to get the final answer.

bw wrote:my son's solution:

450295/1679616 or 0.26809401672

the solution is based on writing a recurrence relation for p(n) where p(n) is the probability that one of the running totals is exactly n assuming we toss the die an infinite number of times.

for the basis case, p(0) is 1 (because the running total right at the beginning is always 0), and p(i) is 0 if 'i' is negative (because the running total can never be negative).

the recurrence relation for p(n) is (1/6)(p(n-1)+p(n-2)+...+p(n-6))

we can solve the recurrence relation in linear time by first computing p(1), then p(2) and so on, saving the previous results to avoid re-computation (dynamic programming).

MaxEntropy_Man wrote:

Rishi wrote:>>> I kinda fixed the counts. Are we supposed to come up with an expression involving K?


1,1,1,1,1,1,1,1 (1 way)
1,1,1,1,1,1,2 (7 ways)
1,1,1,1,1,3 (6 ways)
1,1,1,1,4 (5 ways)
1,1,1,5 (4 ways)
1,1,6 (3 ways)
1,2,5 (3 ways)
1,3,4 (3 ways)
2, 6 (2 ways)
3, 5 (2 ways)
4, 4 (1 way)



If K=8, the total number of ways Tk=8 is 1. The probability of getting Tk=8 for K=8 is 1/(6**.
If K=7, the total number of ways Tk=8 is 7. The probability of getting Tk=8 for K=7 is 7/(6**7).
If K=6, the total number of ways Tk=8 is 6. The probability of getting Tk=6 for K=6 is 6/(6**6).
If K=5, the total number of ways Tk=8 is 5. The probability of getting Tk=8 for K=5 is 5/(6**5).
If K=4, the total number of ways Tk=8 is 4. The probability of getting Tk=8 for K=4 is 4/(6**4).
If K=3, the total number of ways Tk=8 is 9. The probability of getting Tk=8 for K=3 is 9/(6**3).
If K=2, the total number of ways Tk=8 is 5. The probability of getting Tk=8 for K=2 is 5/(6**2).




So you add up these individual probabilities to get the final answer.

this is from the purdue problem of the week series. i submitted a numerical answer to them yesterday. will post my answer here tomorrow and the official answer when they post it.

bw wrote:my son's solution:

450295/1679616 or 0.26809401672

the solution is based on writing a recurrence relation for p(n) where p(n) is the probability that one of the running totals is exactly n assuming we toss the die an infinite number of times.

for the basis case, p(0) is 1 (because the running total right at the beginning is always 0), and p(i) is 0 if 'i' is negative (because the running total can never be negative).

the recurrence relation for p(n) is (1/6)(p(n-1)+p(n-2)+...+p(n-6))

we can solve the recurrence relation in linear time by first computing p(1), then p(2) and so on, saving the previous results to avoid re-computation (dynamic programming).

Rishi wrote:Max,

As you said, my approach was correct but I left out some other possibilities of counting. Like 2 2 4 etc.

MaxEntropy_Man wrote:

Rishi wrote:Max,

As you said, my approach was correct but I left out some other possibilities of counting. Like 2 2 4 etc.

yes but you also have to properly count the number of distinct permutations for each k, i.e. 2,6 is different from 6,2 and so on. i got the answer but hate my own method. bw's kid's solution which is the same as the other solution on the purdue site is far more elegant.

i have an engineer's mindset unfortunately. break it down by whatever tool is available. not a lot of finesse.

MaxEntropy_Man wrote:

MaxEntropy_Man wrote:

Rishi wrote:Max,

As you said, my approach was correct but I left out some other possibilities of counting. Like 2 2 4 etc.

yes but you also have to properly count the number of distinct permutations for each k, i.e. 2,6 is different from 6,2 and so on. i got the answer but hate my own method. bw's kid's solution which is the same as the other solution on the purdue site is far more elegant.

i have an engineer's mindset unfortunately. break it down by whatever tool is available. not a lot of finesse.

and i think you were aiming to do that, but had some mistakes. for example the number of distinct permutations of 1,3,4 is 3! (=6) and not 3 as you had computed.

Rishi wrote:

MaxEntropy_Man wrote:

MaxEntropy_Man wrote:

Rishi wrote:Max,

As you said, my approach was correct but I left out some other possibilities of counting. Like 2 2 4 etc.

yes but you also have to properly count the number of distinct permutations for each k, i.e. 2,6 is different from 6,2 and so on. i got the answer but hate my own method. bw's kid's solution which is the same as the other solution on the purdue site is far more elegant.

i have an engineer's mindset unfortunately. break it down by whatever tool is available. not a lot of finesse.

and i think you were aiming to do that, but had some mistakes. for example the number of distinct permutations of 1,3,4 is 3! (=6) and not 3 as you had computed.

>>> From the two different ways to solve the problem, what really matters is p(k=2) thru p(k=. Then why was it mentioned that the dice is being rolled forever? Is it to trick the student or to find out whether the student disregards the irrelevant information?