cool dice/probability problem

[gwnn]

Using "dice" in the singular is unfortunately not wrong, but if you look closely at your links you'll find that your MW link lists "die" as a synonym for "dice (singular)" and Oxford provides a direct link to the noun "die" where that word is defined among other meanings as singular of "dice". So it's hardly a matter of "other sources".

 

Nero Wolfe might have equally well objected to "dice" as singular as he did to "contact" as a verb. B-)

Sorry, I was unclear about what I meant by the "other sources" part. I meant that there are sources which give both forms as possible/correct and some only give a singular "die." Still, if at least a few (reputable) sources give a form as correct, then it is not "simply wrong", which is all I set out to disprove. Of course, which one one uses is a matter of simple preference. I find "one die" grating and "one dice" slightly less grating but still awkward. So I would usually try to avoid talking about dice altogether, at least when there is a reasonable chance that I will need to talk about only one of them at some point of the conversation (and I can't get around the issue by using a construction such as the one in this sentence). Needless to say, starting a thread about rolling dice is one of the less wise things I can do in this regard.

[barmar]

"throw a dice" sounds more wrong to my ears than "throw a die".

[kenberg]

"throw a dice" sounds more wrong to my ears than "throw a die".

 

The one where the correct usage still sounds wrong to me is "the data show". I usually convert it to "the set of data shows" or "the data points show".

 

If we look in

http://www.merriam-w...tionary/agendum

we find that agenda is the plural of agendum, but I am quite sure I have never heard anyone say "His agenda are".

So my agendum is to insist that anyone who corrects me when I say "the data shows" must forever after say "the agenda for this meeting are" or "the agendum for this meeting is".

 

Or we could all just agree that English is a concoction of several languages and does not really conform well to logic.

[gwnn]

Is anyone interested in the dice problem too? I'm not complaining about any hijacking since I am solely culpable for it. I just wanted to remind people that there's a reasonably cool (if useless) problem on the top of this page.

[kenberg]

Is anyone interested in the dice problem too? I'm not complaining about any hijacking since I am solely culpable for it. I just wanted to remind people that there's a reasonably cool (if useless) problem on the top of this page.

 

Yes, I fin it interesting. A couple of things:

 

a. Using three dice, you can in fact model two distinguishable dice. if the three dice come up either {1,1,2} or {2,2,1} then call this (r,g) =(1,2) while of the three dice come up {1,3,3} or (1,1,3} ca;; this (r,g)=(2,1). I haven't worked out any pattern, but clearly this can be done.

 

b. The idea of a "simple" matching is of interest, but hard to pin down. At firat I thought this could mean that we must use an algebraic expression and work modulo 6, but in fact I imagine any mathcing could be put in such a form, albeit maybe with many terms.

 

c. Generalize!

 

 

d. Spouse is getting impatient for dinner.

[gwnn]

You don't need to distinguish between (1,2) and (2,1), FWIW. The other solution I saw (the only other one in the video comments that I know works) has one of the rules as:

 

(a,b,b),(a,a,b),(6,a,b) -> (a,b), if a<>b and a,b<>6.

 

And then go on from there. Certainly it makes sense to have (1,2,2) correspond to (1,2), but it's a bit asymmetric.

 

I was thinking of defining a metric of similarity, for example the overlap between the three-dice result and the corresponding two-dice result. Then, maybe I could optimize this using a genetic algorithm? Maximum overlap is definitely not necessarily "easiest to explain", I know.

[kenberg]

You don't need to distinguish between (1,2) and (2,1), FWIW. The other solution I saw (the only other one in the video comments that I know works) has one of the rules as:

 

(a,b,b),(a,a,b),(6,a,b) -> (a,b), if a<>b and a,b<>6.

 

And then go on from there. Certainly it makes sense to have (1,2,2) correspond to (1,2), but it's a bit asymmetric.

 

I was thinking of defining a metric of similarity, for example the overlap between the three-dice result and the corresponding two-dice result. Then, maybe I could optimize this using a genetic algorithm? Maximum overlap is definitely not necessarily "easiest to explain", I know.

You do not need to distinguish between (1,2) and (2,1), but I find it interesting that you can. That is, three indistinguishable dice can be used to model the throw of two distinguishable dice. Obviously two indistinguishable dice can not be used to model two distinguishable dice. I have not thought about whether four indistinguishable dice can be used to model three distinguishable dice. I doubt it.

 

We all just take this where it leads us.

[nige1]

Thank you Gwnn for telling us about this intriguing problem and your ingenious solution.

 

It might be easier to tackle the general case i.e. using the throw of m identical dice to simulate the throw of n dice (distinguishable or not).

 

e.g. to simulate the throw of 1 die from m dice, you can use your mod 6 trick.

[hrothgar]

This is all very interesting, however, from a practical perspective, I would simple decide that I will roll the dice and then chose the left-most die as #1 and the second left-most die as #2...

 

If there is a "tie", I'll chose the bottom die before the top...

[Fluffy]

Csaba I suppose you mean the second problem. I felt that the only viable solution for me was to write down the 56 posibilities and make groups that got the same probabilities as 2 dice. In other words: brute force. Similar to the famous problem that finished with "the oldest plays piano", I see nothing mathematical or entertaining on using brute force.

 

I think your solution was good, and I don't expect the ideal solution too look much less complex

 

 

 

[gwnn]

This is all very interesting, however, from a practical perspective, I would simple decide that I will roll the dice and then chose the left-most die as #1 and the second left-most die as #2...

 

If there is a "tie", I'll chose the bottom die before the top...

I think applying the first solution twice would be my solution. Of course, a sledgehammer might be even simpler. I definitely don't claim that my solution is practicable for late-night drinking/strip games (or both).

Csaba I suppose you mean the second problem. I felt that the only viable solution for me was to write down the 56 posibilities and make groups that got the same probabilities as 2 dice. In other words: brute force. Similar to the famous problem that finished with "the oldest plays piano", I see nothing mathematical or entertaining on using brute force.

 

I think your solution was good, and I don't expect the ideal solution too look much less complex

Of course my solution was also based on brute force. I started from realizing that ssB->ss can cover all "ss" tosses and went from there and filled up the space with as simple mappings as I could. This problem strongly reminded me of bridge system design so I thought others here might enjoy it. idk

[Vampyr]

"throw a dice" sounds more wrong to my ears than "throw a die".

 

Yes, and perhaps it also depends on how often you roll dice. I play backgammon, and it is often important to distinguish between one die and two (or more) dice.

[blackshoe]

The one where the correct usage still sounds wrong to me is "the data show". I usually convert it to "the set of data shows" or "the data points show".

 

If we look in

http://www.merriam-w...tionary/agendum

we find that agenda is the plural of agendum, but I am quite sure I have never heard anyone say "His agenda are".

So my agendum is to insist that anyone who corrects me when I say "the data shows" must forever after say "the agenda for this meeting are" or "the agendum for this meeting is".

 

Or we could all just agree that English is a concoction of several languages and does not really conform well to logic.

Sorry, Ken, but I don't think "agenda" as a singular noun signifying a list of items (individually each an "agendum") is any more wrong that "die" is wrong as a singular for "dice" or "data" used in the singular is wrong.

 

"I rolled a die" and "the data are" both sound right to me. So, I confess, does "the agenda (meaning the list of items to be discussed in a meeting) is". That said, I have occasionally (mis)used "the data says". I don't recall ever saying "I rolled a dice" or "the agendum is" - the latter would refer to a single item on the agenda (list), but most people, I think, call them "agenda items", as do I. B-)

[kenberg]

Sorry, Ken, but I don't think "agenda" as a singular noun signifying a list of items (individually each an "agendum") is any more wrong that "die" is wrong as a singular for "dice" or "data" used in the singular is wrong.

 

"I rolled a die" and "the data are" both sound right to me. So, I confess, does "the agenda (meaning the list of items to be discussed in a meeting) is". That said, I have occasionally (mis)used "the data says". I don't recall ever saying "I rolled a dice" or "the agendum is" - the latter would refer to a single item on the agenda (list), but most people, I think, call them "agenda items", as do I. B-)

 

I am fine with people using agenda as a singular, as long as they don't go ballistic if I inadvertently use data as singular. I try to remember it is plural, and I usually do, but I sometimes slip. The two cases seem equivalent. So I think people should get in a dither over both "the data shows" and "the agenda is" or they should get in a dither over neither. I prefer letting it be. I suppose it is forum and fora also, but I am not living in Ancient Rome.

 

And Caesar can say"The dice is cast" if he wants to, he is Caesar, but it sounds weird.

[Zelandakh]

Csaba, just in case you are still interested in solutions, here is mine. It is unfortunately not any simpler than the others.

 

Take the 3 numbers and order them from smallest to largest as abc.

 

For a triple, a=b=c -> 6:6

For a double, a=b or b=c -> a:c

For 3 different numbers a<>b and b<>c,

- if a=1 -> b:c

- if a=2 and b=3 -> 1:c-3 (or a+b+c-8)

- if a=2 and b>3 -> 1:b+c-5 (or a+b+c-7)

- if a>2 -> double a+b+c-10

 

It would be nice to have a simple formula for this last part, especially if it could keep the property that, roughly, a better 3-dice throw gives a better 2-dice result.

 

Intuitively, the general case of any number of dice ought to have a solution, since the number of possibilities is simply a factor greater and the single outcomes always sum to a whole factor. But proving that might be quite difficult - it would be a lot easier to disprove if someone can come up with a counter example.

[mgoetze]

Intuitively, the general case of any number of dice ought to have a solution, since the number of possibilities is simply a factor greater and the single outcomes always sum to a whole factor. But proving that might be quite difficult - it would be a lot easier to disprove if someone can come up with a counter example.

Well, I don't know how general you mean the general case to be, but I think it is quite clear that you cannot accurately simulate throwing two 2-sided dice by throwing three 2-sided dice.

[barmar]

Well, I don't know how general you mean the general case to be, but I think it is quite clear that you cannot accurately simulate throwing two 2-sided dice by throwing three 2-sided dice.

2-sided dice? I think those are coins. :)

[nige1]

Well, I don't know how general you mean the general case to be, but I think it is quite clear that you cannot accurately simulate throwing two 2-sided dice by throwing three 2-sided dice.

but you can with 4 * 2-sided "dice"

HHHT -> HH
HTTT -> TT
HHHH + TTTT + HHTT -> HT

[Zelandakh]

The coin case is a good one because you can solve it using Pascal's triangle. You simply need to find a transformation between the x^th row and 2^(x-y) times the y^th, where x is the number of dice thrown and y the number required. x=3 and y=2 gives 1:3:3:1 and 2:4:2, which as Michael points out does not work, whereas, as Nigel points out, x=4, y=2 is 1:4:6:4:1 -> 4:8:4, which can be mapped by adding the 2 triples to the HHTT result (4:6+1+1:4). The same logic shows that x=4, y=3 also fails as a mapping from 1:4:6:4:1 to 2:6:6:2 is also not available. This time increasing x to 5 (1:5:10:10:5:1), 6 (1:6:15:20:15:6:1) or 7(1:7:21:35:35:21:7:1) does not give a solution. I have not checked further but it does not look good for the general case. ;)

[WellSpyder]

I play backgammon, and it is often important to distinguish between one die and two (or more) dice.

Perhaps we'll all end up just calling then cubes.....

[Cyberyeti]

2-sided dice? I think those are coins. :)

 

Roleplayers will know that you do occasionally have to roll 2 sided dice. You normally roll a 6 sided dice and go odd/even or high/low.

[Zelandakh]

Roleplayers will know that you do occasionally have to roll 2 sided dice. You normally roll a 6 sided dice and go odd/even or high/low.

My standard random roll was a d20 and a d10, which covered most possibiilites - ability checks, percentile rolls, wandering monsters, etc - without tipping off the players what was going on. The same principle applies though, 1-10 -> 1, 11-20 -> 2 and ignore the d10. That is of course also the answer to the earlier question about simulating a die of a lower number of sides. B-)

[nige1]

Perusing the

, Alan Ingleby's answer seems neat

When m = 3 and n = 2:
t = m1 + m2 + m3.
n1 = 1 + (t mod 6).
n2 = 1 + (t mod 3) + 3 * (t mod 2).

[gwnn]

That can't work. You can't map 3 dice to 2 based on the sum alone. This has been established.

[nige1]

That can't work. You can't map 3 dice to 2 based on the sum alone. This has been established.

Duh :(