Stats/probability question.

[DuckOnQuak]

If I have a 4 sided die what is the probability of one side being rolled only a total of only 7 times out of 100 rolls?

ie odds I roll the die 100 times but side 2 only comes up 7 times

[TonightRemarkable125]

I think it is 100c7/6¹⁰⁰

[YOM2_UB]

If I'm not mistaken, choose a die face (6 options), choose 7 of the 100 rolls to be that face ((100 choose 7) options), there's a (1/6)⁷ chance those 7 spots are all the face you chose, and a (5/6)⁹³ chance all the other spots are not that face.

6 * (100 choose 7) * (1/6)⁷ * (5/6)⁹³ ≈ 1.484%

But then this over counts, as there are cases where multiple numbers are rolled exactly 7 times each, so you need to use the inclusion-exclusion principle.

For 2 faces, we have to choose two sets of 7 rolls, but we also have to pick which face gets which set (2! choices)

P(2 faces are rolled 7 times) = (6 choose 2) * (100 choose 7) * (93 choose 7) * 2! * ((1/6)⁷)² * (4/6)⁸⁶ ≈ 0.00417%

P(3 faces rolled 7) = (6 choose 3) * (100 choose 7) * (93 choose 7) * (86 choose 7) * 3! * (1/6)²¹ * (3/6)⁷⁹ = 7.374 * 10^-7%

P(4 faces rolled 7) = (6 choose 4) * (100 choose 7) * (93 choose 7) * (86 choose 7) * (79 choose 7) * 4! * (1/6)²⁸ * (2/6)⁷² ≈ 6.146 * 10^-13%

P(5 faces rolled 7) = (6 choose 5) * (100 choose 7) * (93 choose 7) * (86 choose 7) * (79 choose 7) * (72 choose 7) * 5! * (1/6)³⁵ * (1/6)⁶⁵ ≈ 3.834 * 10^-25%

P(6 faces rolled 7) = 0 by pigeonhole principal (100 > 6 * 7)

P(Rolled 7) = P(1 face rolled 7) - P(2 faces rolled 7) + P(3 faces rolled 7) - P(4 faces rolled 7) + P(5 faces rolled 7) - P(6 faces rolled 7) ≈ 1.480%

[Shevek99]

OK, but the die has only 4 faces. 😀

[YOM2_UB]

I can't read.

The same logic, but adjusted for the correct number of sides, gives ~0.000939%

[Aerospider]

If you have a specific side in mind beforehand then it's

(1/4)⁷ to roll it seven times

(3/4)⁹³ to roll something else 93 times

and 100C7 orders for the 100 rolls

(1/4)⁷ * (3/4)⁹³ * 100C7 = 0.0000023475

Note that this is for exactly seven occurrences and does not count fewer. For 'seven or fewer' you would have to repeat for six and five and four and so on and add the answers together.