[bonanova]

Alex and his friends Ian and Davey were enjoying

some cold ones at Morty's last night when Alex pulled

a small pouch out of his pocket and began to tell his

story.

So this fellow I met claimed he mastered the art of

loading dice, and he was able to separately adjust the

chances of each of the faces showing. We got to talking,

and I asked him to make a pair of dice that would make

all the possible results, 2, 3, 4, ... 10, 11 and 12,

appear with equal chances. Here they are. And

he held them up for all to see.

No way, said Davey, who was no mathematician, to

be sure, but this seemed beyond possibility.

So you're an unbeliever, replied Alex. Then watch this;

and Alex rolled the dice 5 times, getting 2, 5 12, 9 and 6.

How many times would you have to roll an honest pair

of dice to see snake eyes and box cars in 5 rolls?

Jamie heard the commotion and strolled over to the table.

Here, let me try. And Jamie rolled 3, 10, 8, 2 and 11.

Well, these are no normal dice, that's for sure, he said.

What are you going to do with them?

Haven't decided yet, Alex replied. For now, maybe just

a few little wagers with some of my, ahem, acquaintances.

Ian had been thinking for a bit. They're special dice,

to be sure, he said. But I'm quite certain they're not

what you say they are.

If you were there, who would you agree with?

[Guest]

I'll be the first to say it I have no clue what you're asking.

Am I suppose to be comparing each ones rolls ?

Maybe you should follow your sig a bit closer

[Guest]

If one of the dice was rigged so that one of the dice only ever landed on 1 or 6. If the other dice was unloaded, then every option (from 2 to 12) would have an equal chance of being rolled. However, this isn't quite correct as 7 is twice as likely to be rolled as any other individual number.

Edited December 22, 2008 by Luke Warmwater

[bonanova]

I'll be the first to say it I have no clue what you're asking.

Am I suppose to be comparing each ones rolls ?

Maybe you should follow your sig a bit closer

Read the bold type.

Alex claims his dice show all outcomes with equal chance.

Ian says that cant happen.

Who do you agree with?

[Guest]

Read the bold type.

Alex claims his dice show all outcomes with equal chance.

Ian says that cant happen.

Who do you agree with?

on whether or not the sides are numbered 1 through 6...

[Prof. Templeton]

with Ian (or d3k3).

[bonanova]

They're fair dice, cleverly weighted so the chance of each face showing has been separately adjusted.

So yes, the faces are numbered 1-6.

But your question suggests another puzzle, doesn't it?

For now, suppose Ian has said, and I can prove it.

Would you agree with Alex, who claims to have equal probability dice

[with respect to the outcomes 2, 3, 4, ..., 10, 11, 12] or with Ian, who

claims it's impossible to create such a pair of dice?

[Guest]

I would certainly doubt the possiblity that one could load dice such that snake-eyes (1+1, only one way to get that) could be equally probable with a seven (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). It could be done electronically easily enough, so I could take on faith that a mechanical solution could be devised by a very gifted person. However, since all he claims to be able to do is "separately adjust the chances of each face showing", then I would have to say that it defies logic.

One would need a 1:12 chance to get snake-eyes (1's), and a 1:12 chance to get boxcars (6's).

With even odds for both, it should be equally likely to get 1's as 6's. There are four combinations of 1's and 6's and two of those combinations add up to 7. That alone would make an even probability spread impossible. I'd have to agree with Ian and Davey.

[bonanova]

I would certainly doubt the possiblity that one could load dice such that snake-eyes (1+1, only one way to get that) could be equally probable with a seven (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). It could be done electronically easily enough, so I could take on faith that a mechanical solution could be devised by a very gifted person. However, since all he claims to be able to do is "separately adjust the chances of each face showing", then I would have to say that it defies logic.

One would need a 1:12 chance to get snake-eyes (1's), and a 1:12 chance to get boxcars (6's).

With even odds for both, it should be equally likely to get 1's as 6's. There are four combinations of 1's and 6's and two of those combinations add up to 7. That alone would make an even probability spread impossible. I'd have to agree with Ian and Davey.

Let a_i and b_i i=1,6 be the chances, respectively, of faces 1-6 appearing on die a and die b.

For snake eyes, that means a₁b₁ = 1/11.

And you'd work from there.

[Guest]

I would certainly doubt the possiblity that one could load dice such that snake-eyes (1+1, only one way to get that) could be equally probable with a seven (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). It could be done electronically easily enough, so I could take on faith that a mechanical solution could be devised by a very gifted person. However, since all he claims to be able to do is "separately adjust the chances of each face showing", then I would have to say that it defies logic.

One would need a 1:12 chance to get snake-eyes (1's), and a 1:12 chance to get boxcars (6's).

With even odds for both, it should be equally likely to get 1's as 6's. There are four combinations of 1's and 6's and two of those combinations add up to 7. That alone would make an even probability spread impossible. I'd have to agree with Ian and Davey.

Let p_x,ybe the odds of rolling y on the x^th die.

p_1,1p_2,1 = p_1,6p_2,6 = 1/11

So, the odds of rolling 6+1 can be expressed as: (p_1,1/p_1,6+p_1,6/p_1,1)/11

which is has a minimum value of 2/11 when p_1,1=p_1,6, as you point out.

[bonanova]

Let p_x,ybe the odds of rolling y on the x^th die.

p_1,1p_2,1 = p_1,6p_2,6 = 1/11

So, the odds of rolling 6+1 can be expressed as: (p_1,1/p_1,6+p_1,6/p_1,1)/11

which is has a minimum value of 2/11 when p_1,1=p_1,6, as you point out.

Since what we seek is the probability of rolling 7 to be 1/11.

That includes 2-5, 3-4, 4-3, and 5-2 also.

But showing 1-6 and 6-1 is impossibly probable does the job, nicely.

[Guest]

You'll that nowhere in the puzzle does it say that the dice are six-sided. If one of the two dice was 12-sided, with faces numbered 1-12 and the other with all faces blank. That gives you an equal chance of rolling the numbers from 1 to 12, without loading or anything. I can't think of any other configuration that results in an equal chance for all numbers 1 - 12.

There is no way to adjust or label two six-sided dice so that there is an equal chance of all numbers coming up.

[Guest]

stepping back...

Don't you have even odds on each outcome before rolling the die regardless of whether they are loaded or not? eg. If I pull out a pair of dice now and give them to Bonanova to roll, aren't the odds that he will roll snake eyes the same as the odds that he will roll boxcars?

edit: Never mind, slaps head - I saw the error there as soon as I hit post. It would only hold true for a single die, sorry for the wasted post

Edited December 22, 2008 by HokieKen

[bonanova]

You'll that nowhere in the puzzle does it say that the dice are six-sided.

If one of the two dice was 12-sided, with faces numbered 1-12 and the other with all faces blank.

That gives you an equal chance of rolling the numbers from 1 to 12, without loading or anything.

I can't think of any other configuration that results in an equal chance for all numbers 1 - 12.

There is no way to adjust or label two six-sided dice so that there is an equal chance of all numbers coming up.

True.

But there are strong hints that they are:

[1] the friend who perfected the art of loading dice, and

[2] Alex's intention of making some bets with friends

Neither would agree with a pair of dice that were not 6-sided.

It actually requires equal chance for 2-12.

That would mean an 11-sided die with equal chances for each face to show.

Can you describe such a shape?

Is there a proof of this?

[bonanova]

While the discussion continues ....

I heard this question posed, without an answer.

There are more constraints than variables.

Solve for a₁-a₆, b₁-b₆ (the face probabilities for dice a and b.):

1. p[2] = f₂(a₁-a₆, b₁-b₆) = 1/11
   
2. p[3] = f₃(a₁-a₆, b₁-b₆) = 1/11
   
3. p[4] = f₄(a₁-a₆, b₁-b₆) = 1/11
   
4. p[5] = f₅(a₁-a₆, b₁-b₆) = 1/11
   
5. p[6] = f₆(a₁-a₆, b₁-b₆) = 1/11
   
6. p[7] = f₇(a₁-a₆, b₁-b₆)= 1/11
   
7. p[8] = f₈(a₁-a₆, b₁-b₆) = 1/11
   
8. p[9] = f₉(a₁-a₆, b₁-b₆) = 1/11
   
9. p[10] = f₁₀(a₁-a₆, b₁-b₆) = 1/11
   
10. p[11] = f₁₁(a₁-a₆, b₁-b₆) = 1/11
    
11. p[12] = f₁₂(a₁-a₆, b₁-b₆) = 1/11
    
12. Sum(a₁-a₆) = 1
    
13. Sum(b₁-b₆) = 1
    

There are 13 equations in 12 unknowns.

(Even) if a solution for 12 of them exists, the 13th equation will be a killer.

Throw in the constraint a₁-a₆, b₁-b₆ >=0 for all i, as well.

Gotta go with Ian on this one.

[Guest]

But your question suggests another puzzle, doesn't it?

It isn't possible that way either, come to think of it. 11 doesn't go very nicely into 36...

[bonanova]

It isn't possible that way either, come to think of it. 11 doesn't go very nicely into 36...

Tho it's tempting - choose face numbers and probabilities.

No sums other than 2-12 only requires min[a]+min = 2 and max[a]+max = 12, allowing 0's 7's 8's, etc.

[Prof. Templeton]

Since you have to get 2 and 12 out of both dice, 1 and 6 have to show up on both dice. But now 7 is twice as likely to show as well. The only way to avoid that is if what shows on the second die is conditional on what shows on the first. No can do. I think the closest you could come would be to have 1-6 on the first die and only 1's or 6's come up on the second. Now you can roll 2,3,4,5,6,8,9,10,11,12 with equal probability but again 7 shows up twice.

[Prof. Templeton]

If we throw a zero in there we can get 1-12 with equal probability.

[Guest]

See, to me the clue is that he takes the dice out of a small pouch - just like the ones gamers use for their 20 sided, 10 sided, etc. dice. You're right that would have to be one 11-sided die and one blank die. I think that the "loaded" aspect is a red herring, and that he can still win bets if he leaves the dice -in the pouch- while making the bet.

This kind of puzzle often turns on an unspoken assumption - in this case, that the dice have six-sides.

Of course I could be wrong....

True.

But there are strong hints that they are:

[1] the friend who perfected the art of loading dice, and

[2] Alex's intention of making some bets with friends

Neither would agree with a pair of dice that were not 6-sided.

It actually requires equal chance for 2-12.

That would mean an 11-sided die with equal chances for each face to show.

Can you describe such a shape?

Is there a proof of this?

[Guest]

It isn't possible that way either, come to think of it. 11 doesn't go very nicely into 36...

I'm not following where the 36 comes in.

[Guest]

It actually requires equal chance for 2-12.

That would mean an 11-sided die with equal chances for each face to show.

Can you describe such a shape?

It's called a "regular undecagon"

Edited December 22, 2008 by Ordover

[bonanova]

It actually requires equal chance for 2-12.

That would mean an 11-sided die with equal chances for each face to show.

Can you describe such a shape?

It's called a "regular undecagon"

Yep.

[36 comes in as combinations of cubic die faces.]

[Guest]

Hey, question - is there someplace one can go for the actual answer to these? Or do I just have to listen to every episode of Car Talk?

[Guest]

Yep.

[36 comes in as combinations of cubic die faces.]

Ah, well, yes, two six-sided dice can show 36 different combinations.

More evidence for my notion of two 11 sided dice, one label 2-12, the other blank (although the blank die does not have to more than four sides) is that read closely, Alex is -telling a story- about people who rolled various things on his "loaded" dice, we're not presented with an -objective- version of events, nor are we "shown" the dice in the course of the query.

So I present my 11-sided die theory, and am eagerly awaiting others.

[Guest]

stepping back...

Don't you have even odds on each outcome before rolling the die regardless of whether they are loaded or not? eg. If I pull out a pair of dice now and give them to Bonanova to roll, aren't the odds that he will roll snake eyes the same as the odds that he will roll boxcars?

edit: Never mind, slaps head - I saw the error there as soon as I hit post. It would only hold true for a single die, sorry for the wasted post

Acutally - forget about adding the two six-sided dice together for a moment. If you don't add the numbers up, then -every combination- has the exact same probability, just like on one die. Consider:

First die /Second Die

1/1

1/2

1/3

1/4

1/5

1/6

2/1

2/2

2/3

.

.

.

.

.

6/6

are already equally probable. Perhaps that is the answer, as the question is very carefully phrased....if 3/3 and 5/1 and 4/2 are considered distinct events, not three forms of "6" then they are already equally probable as all other two-dice combos.