12th January 2021, 3:16 PM
(This post was last modified: 12th January 2021, 3:16 PM by Atari4600. Edited 1 time in total.)
Well it has been about a week with some highly active feedback
(Bi3liu also sent me a pm with his solution which fit nicely in my pr2 inbox)
Here is the solution:
Question 1: What is the probability that you would have seen a 5-of-a-kind at least once in 1296 rolls.
The phrasing of this question is worded very particularly, because even scoring a 5-of-a-kind twice or three times, or all 1296 are all valid samples that the question. The phrasing "at least once" allows for every possible variation. Because of that wording, it makes seeing a five-of-a-kind at least once and not seeing at all mutually exclusive (if you don't have A then you must have B). Which means you can apply the law of P(A) + P(B) = 1 (or 100%). so P(A) = 1 - P(B).
So now, it is very easy to determine P(B) (The probability of NOT seeing A) which is (1295/1296)^n (there are 1295 ways you will not see a 5-of-a-kind) and because these tests are run in succession you need to multiply each time this happens up to 'n' times.
so now the solution to question 1 can be written as P(A) = 1 - (1295/1296)^n, where n is equal to the denominator r 1296. Which results in the number of 0.6323 or around 62.23%
Question 2: Once you work out the solution to question 1, determine how many rolls will give you a 50% chance of seeing a 5-of-a-kind at least once. A 75% chance? a 99% chance?
From this question we are now working backwards to solve for 'n'. In other words, we are given P(A), so we can rewrite the expression from P(A) = 1 - P(B) to P(A) = 1 - (1295/1296)^n and we are solving for 'n'.
So we rewrite the expression as 1 - P(A) = (1295/1296)^n, to isolate 'n', we need to log both sides of the equation. So, log(1-P(A))=n log (1295/1296). Solving for 'n' yields: n = log(1-P(A))/log(1295/1296), where now we are given P(A) to be equal to a number of values: 0.5, 0.75, and 0.99.
For P(A) = 0.5, n = 897.97 or 898 rolls.
For P(A) = 0.75, n = 1795.94 or 1796 rolls
for P(A) = 0.99, n = 5969.997 or 5970 rolls
Question 3: Lets consider you have a 1 in 'n' chance of a desired outcome (where n is an arbitrary whole number) and you had an 'n' amount of attempts to achieve the desired result. What happens as n gets really large (That would imply that the odds of a given outcome is getting smaller as 1/n goes to zero, but also, you are giving that event more chances at showing up with that larger number."
The solution to this is very similar to the first question, lets recall that P(A) = 1 - P(B),
where P(B) instead = ((n-1)/n)^n (rewriting question three algebraically). Simplifying yields: P(B) = (1-1/n)^n, where if you take the limit as n goes towards infinity you achieve the result of 1/e, therefore P(A) = 1 - 1/e as n goes towards infinity.
This is a very interesting result as this value (1-1/e) is the time constant found in step responses and many other fields of math/science/engineering fields.
(Bi3liu also sent me a pm with his solution which fit nicely in my pr2 inbox)
Here is the solution:
Question 1: What is the probability that you would have seen a 5-of-a-kind at least once in 1296 rolls.
The phrasing of this question is worded very particularly, because even scoring a 5-of-a-kind twice or three times, or all 1296 are all valid samples that the question. The phrasing "at least once" allows for every possible variation. Because of that wording, it makes seeing a five-of-a-kind at least once and not seeing at all mutually exclusive (if you don't have A then you must have B). Which means you can apply the law of P(A) + P(B) = 1 (or 100%). so P(A) = 1 - P(B).
So now, it is very easy to determine P(B) (The probability of NOT seeing A) which is (1295/1296)^n (there are 1295 ways you will not see a 5-of-a-kind) and because these tests are run in succession you need to multiply each time this happens up to 'n' times.
so now the solution to question 1 can be written as P(A) = 1 - (1295/1296)^n, where n is equal to the denominator r 1296. Which results in the number of 0.6323 or around 62.23%
Question 2: Once you work out the solution to question 1, determine how many rolls will give you a 50% chance of seeing a 5-of-a-kind at least once. A 75% chance? a 99% chance?
From this question we are now working backwards to solve for 'n'. In other words, we are given P(A), so we can rewrite the expression from P(A) = 1 - P(B) to P(A) = 1 - (1295/1296)^n and we are solving for 'n'.
So we rewrite the expression as 1 - P(A) = (1295/1296)^n, to isolate 'n', we need to log both sides of the equation. So, log(1-P(A))=n log (1295/1296). Solving for 'n' yields: n = log(1-P(A))/log(1295/1296), where now we are given P(A) to be equal to a number of values: 0.5, 0.75, and 0.99.
For P(A) = 0.5, n = 897.97 or 898 rolls.
For P(A) = 0.75, n = 1795.94 or 1796 rolls
for P(A) = 0.99, n = 5969.997 or 5970 rolls
Question 3: Lets consider you have a 1 in 'n' chance of a desired outcome (where n is an arbitrary whole number) and you had an 'n' amount of attempts to achieve the desired result. What happens as n gets really large (That would imply that the odds of a given outcome is getting smaller as 1/n goes to zero, but also, you are giving that event more chances at showing up with that larger number."
The solution to this is very similar to the first question, lets recall that P(A) = 1 - P(B),
where P(B) instead = ((n-1)/n)^n (rewriting question three algebraically). Simplifying yields: P(B) = (1-1/n)^n, where if you take the limit as n goes towards infinity you achieve the result of 1/e, therefore P(A) = 1 - 1/e as n goes towards infinity.
This is a very interesting result as this value (1-1/e) is the time constant found in step responses and many other fields of math/science/engineering fields.
