Statistics/Probability Quiz

[Ether]

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

[Hitchhiker 42]

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.

[smitikshah]

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.

[Ether]

Originally Posted by Hitchhiker 42

= 3.55*10^(-52)

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

[GeeTwo]

How about this:

Throwing the die 420 times, there are 6[sup]420[/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!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.

[Basel A]

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.

[smitikshah]

Originally Posted by Basel A

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.

[Ether]

Originally Posted by GeeTwo

...I get 5.99 x 10^-7...

Originally Posted by Basel A

...5.9960458e-7...

Nice work guys.

[smitikshah]

Originally Posted by Ether

Nice work guys.

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

[Ether]

Originally Posted by smitikshah

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.

[Hitchhiker 42]

Originally Posted by Ether

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.

[GeeTwo]

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.

[euhlmann]

Originally Posted by Ether

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

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.

[Ether]

Originally Posted by euhlmann

...

Nicely done!

So we have identified 3 different solutions.

[Caleb Sykes]

Originally Posted by smitikshah

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

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