How many time do I need to throw a dice to get 3 times the same number with 99% probability?

Question

I throw a dice with 6 faces (1,2,3,4,5,6) n times.

Each time I record the number obtained.

I want to find how many throw n is needed to get 3 times the same number with 99% probability.

Answer 1

nC3 *(1/6)^6 = 0.99

=> nC3 = (0.99) * (6)^6

=> n(n-1)(n-2) = (0.99) * (6)^7 = 277137

Trial and error process to find n

n should be close to (277137)^(1/3) = 65

n = 64 => n(n-1)(n-2) = 274560

n = 65 => n(n-1)(n-2) = 287430

Thus, n = 65 will ensure the same number 3 times with 99% probability.

Answer 2

You want to know

" how many times you need to throw a dice to get the same number 3 times straight"

the possibility of getting the same number 3 times straight

is  1/6 x 1/6 x 1/6 = 1/216

(1/6 means  , any one number among  1, 2, 3, 4, 5, 6)

it means, If you throw a dice 216 times

at least one time you will get the same number 3 times

so you need to throw a dice at least 216 times

216 times

Answer 3

Assuming you are asking the question how many roles of a 6 sided dice you need to get the any number to come up 3 times in a row (three of a kind) with a 99% probability, the answer is as follows,

For rolling the dice three times, the number of possible number combinations that you could get (1,1,1), (1,1,2), (1,1,3), etc.. is equal to 6 * 6 * 6 = 216.

Out of those 216 combinations, there are only 6 that represent the same number coming up in a row. e.g., (1,1,1), (2,2,2), (3,3,3)....(6,6,6)

Therefore, the probability you will get 3 of a kind with the first three rolls is then 6 / 216, or 2.78%.

This leaves you with the number of combinations that fail (not getting three in a row) as 216 - 6 = 210.

So your failure rate for the first three rolls is then, (210 / 216), or 97.22%

Not counting the first 2 throws, since you need to throw at least 3 times to be successful, the probability that the first n throws fail to give you a three of a kind is then ( 210 / 216 ) ^ n (raised to the power of n)

Therefore the probability that you will roll three of a kind is represented by the equation 1 - ( 210 / 216 ) ^ n, and the chance you will get a 99% confidence is,

1 - ( 210 / 216 ) ^ n = 0.99

Let's solve this for n, isolating the exponent and taking the log of both sides,

1 - 0.99 = (210/216) ^ n

log (1 - 0.99) = log (210/216) ^n

log (0.01) = n * log (0.9722)

n = log (0.01) / log (0.9722)

n = 163

Therefore you will need at least 163 rolls of the dice to get any number to come up three times in a row with 99% confidence. And if you want to add in the first 2 rolls (since you need to roll at least 3 times to be successful) this makes it 165 rolls.

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Answer 4

If you don't care which number comes up 3 times, then you would need to roll the dice 13 times.  One of the 6 numbers will have been repeated 3 times by then.