### 1. Probability question

- Posted by petelomax in December
- 716 views

Just being curious.

Suppose I have a bucket containing 1001 7-sided dice. When I tip them out and count them,

what is the probability of getting exactly 143 of each face?

### 2. Re: Probability question

- Posted by lgregg in January
- 523 views

Just being curious.

Suppose I have a bucket containing 1001 7-sided dice. When I tip them out and count them,

what is the probability of getting exactly 143 of each face?

If I remember probability correctly ...

If you throw one die, the probability of one face coming up is 1/7. If you throw another die, and you want the same face to come up, the joint probability is (1/7) * (1/7), or (1/7)**2. Continuing for 1001 dice, the probability should be (1/7)**1001, or approximately zero. :)

Correct me if I am wrong!

### 3. Re: Probability question

- Posted by lgregg in January
- 519 views

Sorry, I did not check before posting. The answer would be (1/7) raised to the 1001 power. It just did not come out correctly.

### 4. Re: Probability question

- Posted by Spock in January
- 514 views

Just being curious.

Suppose I have a bucket containing 1001 7-sided dice. When I tip them out and count them,

what is the probability of getting exactly 143 of each face?

If I remember probability correctly ...

If you throw one die, the probability of one face coming up is 1/7. If you throw another die, and you want the same face to come up, the joint probability is (1/7) * (1/7), or (1/7)**2. Continuing for 1001 dice, the probability should be (1/7)**1001, or approximately zero. :)

Correct me if I am wrong!

I think your result would then need to be divided by some value since each clump of 143 has 7 possibilities that would satisfy the condition.

If there are 7 entities that each have 7 states, in the total number of permutations, how many monotonic groups would there be? 7! ?

So I'm feeling that the answer lies more towards: 1 in 7 ^ 1001 / 7!

Spock

### 5. Re: Probability question

- Posted by SDPringle in January
- 505 views

Now. There are only five possible shapes made of the same regular shape for each face. So there is are no such die. You can have 20 sided, 12 sided, six sided, eight sided and four sided die.

Now, suspending that you wouldn't have 143 on one of the faces. You would have faces labelled values between 1 through 7.

So the best answer is no answer.

### 6. Re: Probability question

- Posted by Mike777b in January
- 495 views

Well, I know where I'd start. If you first consider that the probability of each result on a 7-sided die that is fair must be 1/7th you then ask how many rolls of the die must I make to prove the die is fair? A corollary formula would be that if you roll a die sufficient times (X) the expectation is that each result will come up X/7 times. Clearly, as X approaches infinity the probability that each result comes up X/7 times must be 1 (otherwise the die is not fair).

So, the question is whether the more logical result is zero (1001 is not nearly a sufficient number of rolls) or 1 (1001 is enough). My gut tells me that it is much closer to zero than to 1.

Another part of the solution is how many total permutations are there? With a seven sided die rolled 1001 times the total universe consists of 7^1001. To say this is a large number is an understatemnt.

So, the question is how many of the 7^1001 permutations satisfy the criteria?

Where would you go from there?

### 7. Re: Probability question

- Posted by ChrisB (moderator) in January
- 496 views

Hi

Logically, only one.

However I fired up my improbability drive and got a highly confused whale.

Cheers

Chris

### 8. Re: Probability question

- Posted by petelomax in January
- 466 views

I was privately emailed

(1001! / (143! ** 7)) / (7 ** 1001) 3.6330038905013278985302381744914316707641687749129 × 10^-9

which led me to compose

include bigatom.e bigatom p = ba_div(ba_div(ba_factorial(1001),ba_power(ba_factorial(143),7)),ba_power(7,1001)) ?ba_sprint(p) ?ba_sprint(ba_div(1,p))

output:

"0.0000000036330038905013278985302381" "275254315.7508172479064601750053166"

Which seems a bit too likely to me... (until I remembered I had asked a different question to the one lgregg answered...)

If I remember probability correctly ...

If you throw one die, the probability of one face coming up is 1/7. If you throw another die, and you want the same face to come up, the joint probability is (1/7) * (1/7), or (1/7)**2. Continuing for 1001 dice, the probability should be (1/7)**1001, or approximately zero. :)

Correct me if I am wrong!

I almost agree: If you throw one die, the probability of one face coming up is 100%(!). The second matches 1/7, rather than (1/7)*(1/7), leading to (1/7)**1000:

?ba_sprintf("%10.10eB",ba_power(1/7,1000)) ?ba_sprintf("%10.10eB",ba_power(7,1000))

giving:

" 7.9792116643e-846" " 1.2532566400e+845"

which almost feels far too unlikely... (until I think of 1e1000, then it feels about right)

Unfortunately, though, it isn't the question that I asked (143 of each, not 1001 all the same).

There are only five possible shapes made of the same regular shape for each face. So there is are no such die.

You're not wrong, but... https://www.thediceshoponline.com/dice/7511/Impact-Opaque-Black-White-D7-Dice or https://www.mathartfun.com/d357.html (two types), and while I'm not convinced myself, you could try making one of these: https://www.polyhedra.net/en/model.php?name-en=tetrahemihexahedron...

### 9. Re: Probability question

- Posted by SDPringle in January
- 473 views

Seriously I would just reformulate this problem and say "given a fair random number generator that gives you between 1 and seven inclusive" rather than a die. Solving this solution is a branch of mathematics called **combinatorics**. I studied this in University. The problem is too big to be solved by iterating all of the possible combinations to roll these random number generators and counting these configurations in Euphoria or any other language. You will need to study that topic of Combinatorics. Once done you will have an expression a ninth grader can do. Then you can use Euphoria.

### 10. Re: Probability question

- Posted by ChrisB (moderator) in January
- 471 views

Hi Pete, what happened to your "at least seven"?

Anyway, the original question asked for 143 of each face, which would mean using up all the dice in an exact amount, of which there is only one way to do this - 143 of each face. Racking my brains to to see how there could could be 7 possibilities of 143 of each face, unless of course you are individually numbering each die, to get unique sets of 143.

My whale has just been passed by a bowl of petunias.

Cheers, Chris

### 11. Re: Probability question

- Posted by petelomax in January
- 451 views

Hi Pete, what happened to your "at least seven"?

I posted it before realising I had been sidetracked into thinking about a different question to the one I originally asked a month ago. Sorry.

### 12. Re: Probability question

- Posted by ChrisB (moderator) in January
- 440 views

That's ok, no need to apologise.

Anyway, I went swimming, and had some thoughts. I have 3 3 sided dice. What is the probability of getting each dice showing a unique number, ie 1 2 3.

There are 27 combinations (3^3) there are 6 chances of getting 123 (123, 132, 213, 231, 312, 321)

or 6/27, 2/9

Now this implies that there is more than one chance of getting 7 groups of 143 numbers with the 7 sided dice. So not only one then.

Still don't know what the probability is though,

### 13. Re: Probability question

- Posted by SDPringle in January
- 427 views

That's Combinatorics Chris.

The number possible combinations with three three-sided random generators is 3^3 = 27. Now the sets we are interested in are where you pick three values out of three. The first value one of the three, the second value is one of the two that are left, and the last has only one possible value after you check that. So that's 3 * 2 * 1 (=6).

Now what if you had six coins and you wanted to have of three of them heads and three of them tails. How many ways can that happen with that restriction? How many ways can you flip six coins generally speaking? The former question is harder of course.

Pete, read this site: http://www.coolmath.com/algebra/20-combinatorics/01-counting-principals-01