Probability of the same thing happening 16 out of 19 times?

You may or not know the formula for probability is:

P(x=r) = (n!/(r! * (n-r)!)) * (P^r) * (Q^(n-r))

P(x=r) = nCr * (P^r) * (Q^(n-r))

where n is the total number of attempts (19 in your case), P is the probability of getting the outcome (1/9 in your case), Q is the probability of not getting the outcome (8/9 in your case), and r is the amount of times out of the total your looking for the probability of.

You need to calculate the value of:

P(X=16) <-- meaning the probability (P) of the outcome (X) happening 16 times

P(X=16) = 19C16 * (1/9)^16 * (8/9)^3

P(X=16) ≈ 3.6727 * 10^(-13)

If every side has a different number on it, and each side has the same probability of being rolled, the probability of EXACTLY 16 rolls the same would be:

9 * 19! / (16! * 3!) * (1/9)^16 * (8/9)^3 = 3.3* 10^-12: about 1 in 300 blion

However - I don't know of any evenly weighted 9-sided dice. Are you sure this isn't a TEN-sided die, with numbers 0-9? The odds would be a little worse then - about 1 in 2 trillion.

Each time you roll the die, there is a 1/9 chance of landing on any number.

Landing on that number on the second throw is the same, 1/9.

What you have to do is multiply 1/(9^16)*(8^3)/(9^3)

which gives you roughly 8^3 over 9^19.

Which is roughly 1 over 2640000000000000.

This isn't a particularly result startling actually.

Repeating it would be.

You would have been better off choosing an even amount of times to roll like 20, because 19 doesn't go into 100, meaning if you wanted a percent, that's impossible. If you want a fraction, the only one possible is16/19 because 19 is prime, and it can't be simplified.

Please reference the site below for Probability.