Probability question - seven dice?

It's not 1/k^n because you can get the same result different ways.

That's the probability of getting the same number on each die individually.

Let's start with a much simpler example: 4 two-way dice (aka coin flips).

There are 16 ways it can come out:

1111 - 1 way

1112 - 4 ways

1122 - 6

1222 - 4

2222 - 1

So your chances are either 1/16, 4/16 or 6/16 depending on what you rolled the first time.

Squaring each of those we get

1/16^2 * 2 + (4/16)^2 * 2 + (6/16)^2 =

2/256 + 32/256 + 36/256 =

70/256 chance of a repeat = 27.3%

Now let's look at a somewhat more complex case:

three 3-way dice.

111 - 1 way

112 - 3 ways

113 - 3 ways

122 - 3 ways

123 - 6 ways

133 - 3 ways

222 - 1 way

223 - 3 ways

233 - 3 ways

333 - 1 way

Total 27 outcomes.

So we have the sum:

3 * 1/27^2 + (3/27)^2 * 6 + (6/27)^2 =

3 / 729 + 54 / 729 + 36 / 729 =

93 / 729 = 12.76 %

Both of the above were confirmed by simulations.

For 7 three-sided dice, there would be quite a few patterns to deal with!

Total of 3^7 = 2187 in various batches:

Using 0, 1, 2 instead of 1, 2, 3 ... here's the list:

0000000 1111111 2222222 ... 3 * 1

0000001 0000002 0111111 0222222 1111112 1222222 ... 6 * 7

0000011 0000022 0011111 0022222 1111122 1122222 ... 6 * 21

0000111 0000222 0001111 0002222 1111222 1112222 ... 6 * 35

0000012 0111112 0122222 ... 3 * 42

0000112 0000122 0011112 0012222 0111122 0112222 ... 6 * 105

0001222 0001112 0111222 .. 3 * 140

0001122 0011122 0011222 ... 3 * 210

So it would be quite a task to work out all the details.

The general case of N k-sided dice ... well that's fairly hopeless, I would say.

As we saw for 7 x 3, it's already pretty complex (although feasible in theory),

but for the general case ... well as you say

"not something you can just write down and solve in one step"

(or, I might add, 10 or perhaps even 100 steps!).

It's the sum of the squares of the probabilities of various combinations,

each multiplied by their frequency, as in the examples above.

But for any but the simplest cases, the number of elements in that

computation quickly grows beyond practical limits.

Step (a million): Pick any 2 cube of the 3 cube (two): These 2 dies have to have 7 became on while the opposite die must now not have 7 became on Number of methods you'll be able to pick two cube out of three cube = 3C2 = three Probability of seven on a cube = a million/6 possibility of no 7 on a cube = five/6 Probability of having precisely 2 sevens = three*(a million/6)*(a million/6)*(five/6) =15/216 = five/seventy two

For n dice that are k-sided you have k^n equal probability combinations. Hence, the probility that you win is 1/k^n = k^-n.