This delightful problem is courtesy of Diana Davis (don’t follow the link unless you want to see the solution, though!), although I did take the liberty of formulating it slightly differently. =)

One bright and sunny day, you’re walking down the street thinking about fractals, and you are accosted by a strange-looking woman with clothes covered in printed triangles, protractors, and compasses (a la Ms. Frizzle from the Magic School Bus). “I have a mathematical proposition to make!” she says. Your initial reaction is to walk right by mumbling something about thanks but I already have a protractor… but something makes you pause to listen instead. “We will pick a random triangle,” she continues, “and if it’s obtuse, you pay me $1, but if it’s acute, I’ll pay you $2!” “Hmm…” you say, “what do you mean by a ‘random triangle’?” “Oh, we will just roll these two special dice here,” she replies. “They have infinity sides, and when you roll them, they give you a completely random number between 0 and 180 — like 28.3, 94.11106749…, and so on. We’ll use them to pick two of the angles, and of course, we can figure out the other angle since all the angles of a triangle have to add up to 180 degrees.” She rolls her special dice a few times to show you, and although you can’t quite understand how dice could have infinity sides, they do seem to give random numbers between 0 and 180 as she claims.

Should you take the bet?

## About Brent

Assistant Professor of Computer Science at Hendrix College. Functional programmer, mathematician, teacher, pianist, follower of Jesus.

No way would I take that bet! Interesting formulation you have there.

Thanks! Might I interest you in a bet involving… random quadrilaterals? =)

Ms. Fuzzymath had me going for a fleeting moment there, until she explained the “random” factor. Given that one roll has just under a 50% chance of surpassing 90 degrees, I’d have to be a pretty high-risk bettor. I suppose a right triangle means a draw, but if one of the dice rolls 0, does it count as an obtuse triangle?

Hi, Brent! It’s been a while.

Hi Steve! It has indeed been a while. Are you still in the DC area? I thought I saw you on the Metro once a few months ago but didn’t have a chance to say hi.

Your comment astutely hits upon one of a few things I have sort of swept under the rug here: you could, of course, get a RIGHT triangle, or a triangle with one of its angles equal to zero, and other such degenerate cases. However, these cases happen

with probability zero: that is, if you pick a completely random real number between 0 and 180 inclusive, the chance that you will get exactly 0, 90, or 180 is infinitely small. So it turns out that you can just ignore such cases in the analysis, since they don’t affect the solution in any appreciable way.Yes, I’m still in the area, and I take the Metro five days a week (from my new Tenleytown basement suite to my Dupont Circle office). With my English degree from Oberlin, I’ve been a copy editor first for the American Society for Microbiology and now for Health Systems Research, Inc. We can catch up on more in RL if you know a good time and place to meet.

i’m going to say no because wouldnt you have to measure the largest angle as obtuse? then you would need to get the two to add up to over 90 but not have any number over 90 in the set (although those dice are pretty smart to be able to give you two numbers between 0 and 180 yet always add up to less than or equal to 180)

there seems to be a flaw in the dice (i mean you should roll two dice that give numbers between 0 and 90, not 180, or else you might end up rolling 130 and 95 (which add up to 225, well over 180 degrees))

jon: you’re right about the dice; I should have explained it more clearly. As it turns out, you can just specify that if the dice add up to more than 180, you simply throw away that roll and roll the dice again. Alternatively, you could specify that when the dice add up to more than 180, it isn’t an obtuse triangle (since it isn’t a triangle at all!). You’ll get different answers but they’re essentially interchangeable, since you will get a dice roll of more than 180 exactly half the time on average, so you can factor that into your solution in either case.