Friday Puzzler -- More Odd Dice

Re: Friday Puzzler -- More Odd Dice

Larry Barrett

I tried a simpler version to see if there are any insights.

Imagine two dice, each with sides numbred 1,1,1,2,2,2.

If you roll these and add the result you can get 2, 3, 4 wih probabilities 9/36, 18/36, 9/36.

Changing the configuration of one or both dice leads to the same outcome but with different probabilities. Say one dice is set to 1,1,1,1,2,2 and the other is 1,1,1,2,2,2. The result is the same 2, 3, 4, but the probabilities are 12/36, 18/36, 6/36.

If a 3 is added to one dice in place of a 2, the other dice must be all 1s to get the same sum.

For example, set one dice to 1,1,1,1,1,1 and the other to 1,2,2,2,2,3. The result is 2, 3, 4 but with probabilities 6/36, 24/36, 6/36.

So these simple examples do not seem to provide any insight into how to configure the more complicated dice that Alex proposes.

A start

Alex Y

I don't think I am giving too much away here, just confirming what you probably already have figured out. (Sometimes it helps not to have to keep going back to confirm what you think you know.)

Each die has to have exactly one "1". Proof: The only two positive integers you can add together to get 2 are 1+1, and if one of the dice had more than one "1", there would be more than one roll to get a total of 2.

Hint -- the threes

Alex Y

are not formed the same way with the alternate dice.

Let there be one red and one white die in each set.

The only way to roll a two, with either set of dice, is R1-W1

With standard dice, the ways to roll a three are R1-W2 and R2-W1.

Not so with the alternative dice.

Re: Hint -- the threes

Larry Barrett

So in the alternate set one die must have no 2s, the other must have two 2s.

You can get a 3 two ways, R1 W2 and R1 W2.

On the right track

Alex Y

To make it easier to see your results as you move to higher numbers, you might want to modify your notation to indicate multiple instances of a number on a die, such as R1W2 and R1W2' as the tosses that result in a total of 3.

Re: On the right track

Larry Barrett

So far, dice with this configuration - 1,2,2,3,4,5 and 1,3,3,5,6,7 comes very close. The resulting sum frequencies are same as normal dice for sum = 2,3,4,9, 10, 11, 12 and only off by one for 5,6,7, and 8.

Still working on this. I don't see a simple way to proceed without a spreadsheet.

On my spreadsheet I compute the 36 possible sums in a 6x6 array, then use the COUNTIF(array,n) function to count the occurences of n in the array.

Re: On the right track

Alex Y

Sounds like your solution seeking spreadsheet and mine are nearly identical. Your first die has only one wrong face.

Time for more hints?

Alex Y

Larry's latest reported attempt is VERY close. He has only one face wrong in his first die.

The second die should have only one face that differs from a standard die.

Larry's attempt went wrong at the ways to roll a sum of four.