If a fair die is thrown 420 times, what’s the probability of getting exactly 70 of each number (1 thru 6)?**

I’m gonna try this one:

Here goes my logic:

Let’s say we order the 420 rolls and first require that they be in the exact order of 70 1’s first, 70 2’s second, etc.

Then, the probability of getting this exact order is (1/6)^420

Now, let’s say we start flip-flopping these around, with a specific rule: you can only switch with a number later in the sequence that isn’t the same as the one you already have. This takes care of duplicate scenarios. So, for the 1’s, you can flip each one with 420-70 = 350 different positions.

The way to calculate that would be:

(1/6)^70 *(420-1*70) +

(1/6)^70 *(420-2*70) +

…

(1/6)^70 *(420-6*70) = 3.55*10^(-52)

The 70 exponents are because each group of 70 has that chance of being picked that way.

Man I looked at it thinking it was so easy but boy was I wrong.

Alright this is my input (likely wrong but still something):

1.398*10^-8

This is just a generalization of Bernoulli, no?

6*((420c70)(1/6)^70(5/6)^350)

and then that’s just the ways you can get them in one order. Raise that to the sixth? (just adjusting by a huge factor to take into consideration that order does’t matter - still trying to think this part through).

And finally I got: (1.398*10^-8)^6! (basically 0)

This is like 99% wrong though, I’m still working on it. Will report back soon with new discoveries.

Try your computation method for the greatly simplified case of an imaginary fair 4-sided die* being rolled 8 times, and getting exactly 2 of each (1,2,3,4).

Then compare your answer to the actual answer which can easily be brute-force computed with a simple C program to be 2520/65536.

- for a 4-sided die, the value for each roll would be the side facing downward

How about this:

Throwing the die 420 times, there are 6420[/sub] possibilities (no sorting or counting, just a list of 420 digits).

To get exactly 70 ones, this is the number of combinations of 70 items taken from the 420:

```
420! / (70! * (420-70)!) = 420! / (70! * 350!)
```

This tells you which of each roll was a one.

To get exactly 70 twos, this is the number of combinations of 70 items taken from the remaining 350: 350! / (70! * 280!). This tells you which of each roll that was not a one was a two.

Threes: 280! / (70! * 210!)

Fours: 210! / (70! * 140!)

Fives: 140! / (70! * 70!)

Sixes: 70! / (70! * 0!) [yes there is only one way]

So, the probability of getting exactly 70 of each is the product of the combinations of getting exactly 70 of each number, divided by **all **the combinations. Noting that the two 350!'s cancel, as do the 280!s, etc., this leaves:

```
420! / (70![sUp]6 * 6420)
```

Plugging all this into MS Calculator, I get 5.99 x 10-7, or one in a bit under 1,668,000.

I expected it to be long, but not quite that long. Let’s try that for smaller numbers of throws (multiplied by a milllion for simplicity):

6: 15,432

12: 3,438

18: 1,351

60: 74.6

120: 13.5

The numbers make sense and seem to follow a reasonable progression.

It’s relatively straightforward to determine the probability that exactly 70 of the 420 rolls are ones (a simple binomial distribution problem). Next, it’s relatively straightforward to determine the probability that exactly 70 of the remaining 350 are twos, etc. Using this method, I calculated the probability to be 5.9960458e-7.

Ahh whoops I see my error. I just did binomial for all of them without subtracting 70 each time like a fool.

Now I got 5.99*10^-7 as well. Weird - I thought it would need to be much closer to 0.

Nice work guys.

Was there a certain reason for this question? Just wondering - seems a bit random.

Just a summer diversion.

Hopefully there’s a student or two out there in CD land who was inspired by this discussion to do a bit of summer reading about statistics and/or probability.

Probably me now. I’ve got a Cartoon Guide to Stats that I should pick up this summer.

If you’re interested in looking for other quizzes, just search for threads with Quiz in the title (on advanced search). Ether has a bunch of them, ranging over Math, Physics, and even Geekdom, and there are a bunch of others out there as well.

Alright, not looking at any of the posts yet, this question looks like it’s essentially “how many ways can you arrange 70 each of 6 options”

The formula for permutations with repetition is

```
n!/(n1! * n2! * ... * ni!)
```

So we have

```
n!/(n1! * n2! * ... * n6!) = (420!)/(70!)^6
```

So that’s the total number of ways to get 70 exactly of 1 to 6.

To find the probability, divide by total number of ways to throw the dice, which is 6^420 since there are 6 options on each of 420 steps.

I get 5.99605*10^-7

In conclusion, not likely to happen

Edit: looks like I came a bit late to this thread

But it also looks like other people got the same thing. I guess this is correct then.

It had to be random or it wouldn’t have been a probability question.

Here’s a crack at it before reading responses

Answer

tl;dr final answer:

420! / (70!^6 * 6^420)

or about

6.04e-7

A little rusty on my abstract algebra but here goes.

The odds of 70 ones followed by 70 twos followed by … followed by 70 sixes is:

(1/6)^420

There are 420! permutations of this, i.e. the order of the symmetric group of order 420. However, some of these permutations don’t matter. The symmetric group is generated by individual swaps, one pair at a time. Swapping two ones isn’t actually another permutation (ones are indistinguishable), and so on with twos, threes, … These irrelevant swaps generate six distinct symmetric groups of order 70 within the larger S_420. Their direct sum is isomorphic to all elements in S_420 generated by the irrelevant swaps. Call this subgroup T<S_420. Then the right cosets of T in S_420 are the equivalence classes of indistinguishable permutations, and we need to count these. This is the index of T in S_420, which is the quotient of their orders: 420! / (70!)^6

Then multiply this by the probability for the ordered case, final result:

420! / (70!^6 * 6^420)

Stirling’s approximation on the factorials cancels lots of stuff and we get:

sqrt( 6 / (140*pi) ^ (5) ) ~ 0.000000604 (6.04e-7)

For an n sided die rolled n*k times, this becomes:

(nk)! / (k!^n * n^(nk)) ~ sqrt( n / (2*pi*k)^(n-1) )

Edit: euhlmann method for me