VDice - Dice Rolling for Proboards v5

[Former Member]

theshinigamieyes.proboards.com/thread/531/pta-ravens-game?page=14 That's the link. We were using different dice in the earlier pages, so page 14 is the only one with vdice. I'm using SDE 1.1.3.

[Virgil Sovereign]

@littlemissraven: I'm assuming 'Gifted' is the one whose rolls are the problem? There appears to be a repeat of 14 and 1, which is unlikely to be a coincidence.

Are you able to determine which browser she's using? If it's IE 8 or earlier, I can tell you the problem right away (IE 8 isn't supported by the plugin). Otherwise you'll have to ask her to roll again.

[Former Member]

She is using Google Chrome, and it isn't just hers, but before I realized it might be a problem, I deleted the second page of rolls. We have various people using various browsers who have rolled the same thing again and again, and if you roll the same side of dice in the same roll, it will give you a new number, but the order will be the same each time no matter the browser, and I have the latest versions of both chrome and firefox. One of my friends has Safari.

[Virgil Sovereign]

@littlemissraven: VDice is designed so that if you roll dice in a post and then edit the post, deleting the dice and putting in new ones will not change the values rolled. This is intentional to avoid cheating.

What should not happen is posters posting multiple times in different posts and getting the same values or a highly limited range of values. For example, if you post once and get the outcome 7 15 for two dice, then post again and get 7 15, and again and get 7 15, etc., there's a problem with a plugin.

Since the dice in the thread you've linked to seem sufficiently random to me, I'll need you to ask one of the posters whose dice are constantly coming up with the same value to post several times (with dice) in that thread. This will help me to determine why the values aren't being computed randomly.

If your problem is that you want players to be able to modify their posts and change their dice values, I can give you the ability to do this. But it somewhat defeats the purpose of a random roll in the sense that players can just re-roll their dice as many times as they want to get desirable outcomes.

[Former Member]

That isn't the problem I'm experiencing. In different posts, dice rolls with the same range will end up the same every time. There were additional posts that I deleted that showed that other people were experiencing the same problem as Gifted, including myself.

[Virgil Sovereign]

Understood. Just insert a few more rolls in that thread and I can diagnose the problem.

[Wormopolis]

never thought rolling a couple of cubes would become so complicated...

[Former Member]

The problem seems to have fixed itself. Sorry for all the trouble I've caused .-.

[Virgil Sovereign]

The problem seems to have fixed itself. Sorry for all the trouble I've caused .-.

Not a problem.

If it recurs at some point in the future, just let me know.

never thought rolling a couple of cubes would become so complicated.

Here's a dice brainteaser for you, Wormo:

As you know, the probability distribution for the sum of rolling two six-sided dice is

2: p = 1/36
3: p = 2/36
4: p = 3/36
5: p = 4/36
6: p = 5/36
7: p = 6/36
8: p = 5/36
9: p = 4/36
10: p = 3/36
11: p = 2/36
12: p = 1/36

The challenge is the following:

Given each die face may only have a non-negative integer value, and eliminating the restriction that all die faces must have unique values, come up with a set of two non-identical six-sided dice whose sum produces exactly the same distribution as above when rolled together.

You must assume that the two dice are uncorrelated, and that the probability of any face coming up is 1/6, the same as for regular dice.

There is a solution.

[Wormopolis]

puzzles! my vice!

one die is all odd, the other is all even

[1 3 5 7 9 11] [2 4 6 8 10 12]

s p
3 1/36
5 2/36
7 3/36
9 4/36
11 5/36
13 6/36
15 5/36
17 4/36
19 3/36
21 2/36
23 1/36

[Virgil Sovereign]

Aug 18, 2013 23:06:19 GMT -8 Wormopolis said:

puzzles! my vice!

one die is all odd, the other is all even

[1 3 5 7 9 11] [2 4 6 8 10 12]

s p
3 1/36
5 2/36
7 3/36
9 4/36
11 5/36
13 6/36
15 5/36
17 4/36
19 3/36
21 2/36
23 1/36

Ah. You get exactly half credit.

All of the probabilities are correct, but for full credit the outcomes have to match up too. That is, an outcome of 2 has to have probability 1/36, an outcome of 3 has to have probability 2/36, etc. Identical to the original distribution.

Although even getting a pyramidal distribution for another set of outcomes deserves a 'Like'.

[Wormopolis]

oh.. I didnt get the part that the sums all have to match too. Im guessing 0 is allowed, otherwise the only way to get 2 is 1 and 1

[Virgil Sovereign]

Aug 19, 2013 9:20:41 GMT -8 Wormopolis said:

oh.. I didnt get the part that the sums all have to match too. Im guessing 0 is allowed, otherwise the only way to get 2 is 1 and 1

Correct. And face values don't have to be unique. Hence a die with faces [0 1 1 3 5 7] would be permissible.

A trivial solution would simply be to add one to all faces on one die and subtract one from all faces on the second die. Hence [0 1 2 3 4 5] [2 3 4 5 6 7] would be a trivial solution. But there is a non-trivial solution too.

Hint: Multiplying two polynomials convolves their coefficients. Likewise, summing two random variables convolves their distributions. Ergo, try to recast the problem as a simple polynomial factoring problem.

[Wormopolis]

whereas I can factor polynomials, I cant make the connection to this. the best I got was the trivial solution you already posted above.

[Virgil Sovereign]

Suppose we represent a 6-sided die as a polynomial in x:

P(x) = x^a + x^b + x^c + x^d + x^e + x^f

where a, b, c, ... represent the not-necessarily-unique non-negative face values of the die. We will call this a 'die polynomial'.

For example, a standard six-sided die would be

P(x) = x¹ + x² + x³ + x⁴ + x⁵ + x⁶

Then the distribution of the sum of rolling any two dice, represented by P1(x) and P2(x), would be given by 1/36 P1(x)·P2(x), taking the order of each term as a possible sum and its associated coefficient as its probability.

For example, rolling two standard dice, we see 1/36 P(x)·P(x) gives us

1/36 x² + 2/36 x³ + 3/36 x⁴ + ... + 2/36 x¹¹ + 1/36 x¹²

Call the above polynomial P_sum(x). To solve the problem, you simply need to find two non-identical die polynomials, P_A(x) and P_B(x), whose product generates this same distribution polynomial. That is,

1/36 P_A(x)·P_B(x) = P_sum(x)

This'll put you on the right track.