Friday puzzle -- A dicey one, part 1

Re: Friday puzzle -- A dicey one, part 1

I'll have to think on this sum.

Re: Friday puzzle -- A dicey one, part 1

And I'll eagerly await the product of your consideration.

Re: Friday puzzle -- A dicey one, part 1

I know this is probably incorrect decipherin' but

Dice1 + Dice2 =

1 1 2

1 2 3

1 3 4

1 4 5

1 5 6

2 2 4

2 3 5

2 4 6

3 3 6

so there are 9 chances with addition

Dice1 * Dice2 =

1 1 1

1 2 2

1 3 3

1 4 4

1 5 5

1 6 6

2 2 4

2 3 6

and there are 8 chances with multiplyin'

My money is on summin' up.

On the right track

But off in the calculation.

You are right that since the values on the target die are equally likely, you can just add the occurrences that add or multiply out to ANY number 1-6.

And you are right that the sum is more likely. But you have slightly overstated the bias.

Let's see if anyone can fine-tune it, then I will present part 2.

Re: On the right track

2+2=4 and 2*2=4

Re: Friday puzzle -- A dicey one, part 1

The possible values of the product are:

1 2 3 4 5 6 8 9 10 12 15 16 18 20 24 25 30 36

This is a total of 18.

Six of them are possible outcomes of the single dice roll,

hence the probability of the desired event is 6/18 = 1/3.

The possible values of the sum are:

2 3 4 5 6 7 8 9 10 11 12

This is a total of 11.

Five of them are possible outcomes of the single dice roll, provided that this roll will produce 2 or 3 or 4 or 5 or 6.

The probability of the desired event is (5/11)*(5/6) = 25/66.

Since 25/66 > 1/3, my money is on the sum.

Re: On the right track

Yes, that is the only case where the first die is both the sum and the product.

right winner, but wrong odds

Jim's approach is closer; just needs a little tweaking.

Hint

Follow Jim's line of reasoning, but consider the possibilities if there is one red and one blue die.

Solution

Jim was very close in enumerating the possibilities, but he missed out on some possibilities. Consider doing this with three different colored dice, a green one to set the "target", and one red and one blue one to provide the potential terms or factors. Now if the green one comes up with 2, then, as Jim said, there is only one way to get the addition--throwing snake-eyes. And while Jim's solution pointed out that 1 and 2 gave you the factors, he missed that there were two ways to get a 1 and a 2 -- 1 with 2; and 1 with 2

So to add that additional information to Jim's table, we get:

Target, Factors, Count

1, 1 1, 1

2, 1 2, 2

3, 1 3, 2

4, 1 4, 2

4, 2 2, 1

5, 1 5, 2

6, 1 6, 2

6, 2 3, 2

Total 14 possibilities for multiplication

Target, Terms, Count

1, ,

2, 1 1, 1

3, 1 2, 2

4, 1 3, 2

4, 2 2, 1

5, 1 4, 2

5, 2 3, 2

6, 1 5, 2

6, 2 4, 2

6, 3 3, 1

Total 15 possibilities for addition

So everyone who answered was right to pick addition, but the advantage was smaller than one might think.

BTW, there are 216 results for throwing the three dice, so the probability of either is relatively small. In fact, because of the small probability and closeness, you have to get a fairly large sample to get this result in a Monte Carlo test. I tossed the dice 1000 times (okay, my spreadsheet did), and multiplication was slightly favored. I had to go to a sample of 4000 tosses to get results reliably matching the probabilities. (I also don't know how good the random number generator in my spreadsheet is.)