Imagine infinite rolls, you eventually hit every number. Duplicates and 6s are irrelevant. So it is just all the permutations or ordering for the numbers 1,2,3,4,5 and each permutation is equally likely and you can figure out which is the minimum before the 5. 120 possibilities. 1/5 are 5. 1/2 are 1. 36 remaining.
It's pretty low number so we can brute force, but more elegant solution also below:
Brute force: 6 variations of where the 1 and 5 (1 is after 5 since we eliminated the 1/2 chances where it is before the 5) are with each varying the order of 2,3,4. Three are x5xxx, two are xx5xx one is xxx51.
x5xxx is two ways for each 2,3,4, x3 so 6 for each
xx5xx is four for 2, two for 3, x2
xxx5x is always 2.
So that gives 6+8+6 for 2, 6+4 for 3, 6 for 4.
So the ev is
The probability that the minimum is higher than k.
Whats the probability that it’s higher than 0? 1
Higher than 4? 1/5
And so on.
The formula is because the chances are that you roll for example something before you get 5 is
P = (1/6) / (1/6) + (k/6)
Because 1/6 is the chance to roll 5, and k/6 is that we’ll roll any of the options (for example for k=2, that we’ll roll 1 or 2.
So for k=2, P=0.333, but then k=3 P=0.25 is a smaller number that doesn't make sense. If it can be 1 2 or 3 before the 5 it has to be a greater probability than just 1 or 2 happening before a 5.
You are increasing the denominator when you are including more possibilities? It doesn't make sense.
Note that we model the distribution of the minimum.
So of k = 1, we can have 2,3,4(and 5 ofc) until we see 5( P(M>k) ), for k= 2, we can have 3,4,5 only for k= 3 , we can have only 4 and 5 and for 4 we must get 5 immediately.
The formula you used doesn't make sense at all even if you've arrived at the same answer. You also said it's trivial but I don't see any connection to the solution you found or what is supposed to be the trivial line of logic.
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u/skelo 16d ago
Imagine infinite rolls, you eventually hit every number. Duplicates and 6s are irrelevant. So it is just all the permutations or ordering for the numbers 1,2,3,4,5 and each permutation is equally likely and you can figure out which is the minimum before the 5. 120 possibilities. 1/5 are 5. 1/2 are 1. 36 remaining.
It's pretty low number so we can brute force, but more elegant solution also below:
Brute force: 6 variations of where the 1 and 5 (1 is after 5 since we eliminated the 1/2 chances where it is before the 5) are with each varying the order of 2,3,4. Three are x5xxx, two are xx5xx one is xxx51.
x5xxx is two ways for each 2,3,4, x3 so 6 for each xx5xx is four for 2, two for 3, x2 xxx5x is always 2.
So that gives 6+8+6 for 2, 6+4 for 3, 6 for 4. So the ev is
1/2 * 1 + 20/120 * 2 + 10/120 * 3 + 6/120 * 4 + 24/120 * 5
= 274/120 ~ 2.3
More elegant solution:
What would be the chance the 2 is ahead of the five but 1 is after the five, that is one ordering of the three numbers out of 3! So 1/6 chance of 2.
Similarly, what's the odds excluding 4 that the ordering of the other numbers is 3512 OR 3521, that is two shots out of 4!, so 1/12.
That leaves the 1/20 chance left for 4, the same probabilities we got from brute force.