ten rolls of a ten side die, 90 or higher?

Question

What are the odds on rolling a 90 or higher, given ten rolls of a ten sided die?

Answer 1

There is no easy way to do this. All the preceding answers are wrong.

There are 10 dice, and 10 possible results on each, so the total number of possible results (counting each die as distinct) is 10^10. The probability of getting 90 or higher
equals
P(100)+P(99)+P(98)+.... +P(90)

There's only one way to get 100: every die must be a 10.
P(100) = 1/(10^10)

There are ten ways to get 99: the first die can be a 9 and the other nine dice 10's, or the second die can be a 9 and the other nine 10's, etc.
P(99) = 10/(10^10)

There are 55 possible ways to get 98. There are 10 ways to roll an 8 and nine 10's, and another 45 ways to roll 2 nines and 8 10's. (45 = 10! /(8!*2!))
P(98) = 55/(10^10)

To get 97, you can roll a 7 and nine 10's (10 ways) or an 8, a 9, and eight 10's (90 ways, 10!/(8!1!1!), or three 9's and seven 10's (120 ways, 10!/(7!*3!))
P(97) = 220/(10^10)

I'll let you grind through the numbers if you really want to. It gets to be progressively more work, the further you get from 100, and I have no great desire to crank out more numbers past P(97). I've given you enough to see the pattern.

It is possible to estimate the answer by calculating the standard deviation and figuring out how many standard deviations 90 is away from the mean, and then consulting tables. The answer won't be precise, but it'll be good enough for most practical purposes.

Answer 2

since u have 10 rolls and 90 is the minimum sum, each roll has to be 9 or 10
probability= 2/10=1/5

P of getting a min. sum of 90 = (1/5)^10
= 1.02400 × (10)^-7

Answer 3

To get 90 or higher you will have to get 9 or 10 on every roll.

Roll 1 = 2/10
Roll 2 = 2/10
etc.
Roll 10 = 2/10

P >90 = (0.20)^10 = 0.0000001024

Answer 4

Ten percent, if you mean you're adding up the scores from ech roll. If you want to roll over 90 on a ten sided die, you're out of luck