Math Puzzle #1

[Camer the Dragon]

Qu 1:

Spoiler

The probability of seeing at least 1 5-of-a-kind is the same as not seeing none
not getting a 5-of-a-kind is probability (1295/1296) and so not seeing it for 1296 rolls is (1295/1296)^1296 since each roll is independent of the others
so the probability of seeing at least 1 5-of-a-kind is 1 - (1295/1296)^1296 which is roughly 63.23%

Qu 2:

Spoiler

50% chance of seeing it is the same as 50% chance of not seeing it so (1295/1296)^x = 0.5
x*ln(1295/1296) = ln(0.5)
x = ln(0.5) / ln(1295/1296)
x = 897.97 so roughly 898 rolls

75% chance of seeing it is the same as 25% chance of not seeing it so (1295/1296)^x = 0.25
x*ln(1295/1296) = ln(0.25)
x = ln(0.25) / ln(1295/1296)
x = 1795.9 so roughly 1796 rolls

99% chance of seeing it is the same as 1% chance of not seeing it so (1295/1296)^x = 0.01
x*ln(1295/1296) = ln(0.01)
x = ln(0.01) / ln(1295/1296)
x = 5966.0 so roughly 5966 rolls

(all x to 5 s.f.)

Qu 3:

Spoiler

The chance of seeing the result after n trials is 1 - (1 - 1/n)^n (i.e. not having all of the trials be "failures")
The limit of this as n tends to infinity is a known formula for e^x (about 2.71828) with x in this case being -1
Thus the chance of seeing the result after n trials for large n is approximately 1 - 1/e which is about 63.21%

Bit rusty on this so hopefully didn't oof any of it lol