## Proportional Dice

So, here’s a nice probability problem that recently made it to my Twitter friends page:

(By the way, I’m @Nebusj on Twitter. I’m happy to pick up new conversational partners even if I never quite feel right starting to chat with someone.)

Schmidt does assume normal, ordinary, six-sided dice for this. You can work out the problem for four- or eight- or twenty- or whatever-sided dice, with most likely a different answer.

But given that, the problem hasn’t quite got an answer right away. Reasonable people could disagree about what it means to say “if you roll a die four times, what is the probability you create a correct proportion?” For example, do you have to put the die result in a particular order? Or can you take the four numbers you get and arrange them any way at all? This is important. If you have the numbers 1, 4, 2, and 2, then obviously 1/4 = 2/2 is false. But rearrange them to 1/2 = 2/4 and you have something true.

We can reason this out. We can work out how many ways there are to throw a die four times, and so how many different outcomes there are. Then we count the number of outcomes that give us a valid proportion. That count divided by the number of possible outcomes is the probability of a successful outcome. It’s getting a correct count of the desired outcomes that’s tricky.

## Thumbup 3:09 pm

onWednesday, 10 February, 2016 Permalink |LikeLike

## Joseph Nebus 3:45 am

onTuesday, 16 February, 2016 Permalink |Oh, dear, they’re staring at me!

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## Thumbup 2:34 pm

onTuesday, 16 February, 2016 Permalink |UH OH! It does look like it!

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## Joseph Nebus 5:58 am

onSaturday, 20 February, 2016 Permalink |Maybe the dice need some sunglasses or something, to look cooler.

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## Thumbup 6:03 am

onSaturday, 20 February, 2016 Permalink |LikeLike

## howardat58 3:14 pm

onWednesday, 10 February, 2016 Permalink |Vegetarians clearly have different definitions.

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## Joseph Nebus 3:47 am

onTuesday, 16 February, 2016 Permalink |Well, we have to be a bit more careful than average, is all. There’s troublesome stuff where you’d never expect it.

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## Chiaroscuro 4:00 pm

onWednesday, 10 February, 2016 Permalink |So, let’s make these A/B=C/D for the dice, assuming in-order rolls. 1296 possibilities.

If A=C and B=D, it’ll always work. So that’s 36.

Additionally: If A=B and C=D, it’ll always work (1=1). So that’s 36,. minus the 6 where A=B=C=D.

Then 1/2=2/4 (and converse and inverse and both), 1/2=3/6 (same), 2/4=3/6 (same), 1/3=2/6 (same). 4, 4, 4, 4.so 16 total.

36+30+16=82, unless I’ve missed some. 82/1296, which reduces to 41/648.

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## Chiaroscuro 4:02 pm

onWednesday, 10 February, 2016 Permalink |Ooh! I missed 2/3=4/6. (and converse, and inverse, and both). So another 4, meaning 86/1296.

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## Joseph Nebus 3:51 am

onTuesday, 16 February, 2016 Permalink |There you go. For what it’s worth, I make out the same count of in-order rolls. And a little program I wrote to count them comes up with 86 as well.

Rolls where order doesn’t matter, and you can rearrange the dice to find a pattern that fits … well, obviously there’s more of them.

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## Proportions Follow-Up | A Portrait of the Math Teacher as an Aging Man 1:05 am

onSunday, 14 February, 2016 Permalink |[…] this intriguing problem. I was not the only one thrilled by the problem. Later on Wednesday I saw another post by Joseph Nebus (@nebusj) featuring the same problem! You should definitely check out his post […]

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## How February 2016 Treated My Mathematics Blog | nebusresearch 9:18 pm

onTuesday, 1 March, 2016 Permalink |[…] Proportional Dice […]

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