# Shakespeare-themed Math Puzzles

I didn’t know I had a Shakespeare-themed-match itch to scratch.

1. Hours and hours

How many hours are there in a week? And when you’ve worked it out, can you now figure out how Shakespeare expressed that number in words? He did it using only 15 letters, true to his line “Brevity is the soul of wit.”

2. Duke of hazard

Hazard was a gambling game with dice that was one of the most popular recreations in Elizabethan England. Shakespeare refers to it indirectly several times. Usually two dice were involved, but sometimes three – which threw up this question, a particular source of debate at the time.

Three dice are thrown. Which total is the more likely (and thus a better bet)?

a. Nine

b. Ten

c. Nine and ten are equally likely.

## 4 thoughts on “Shakespeare-themed Math Puzzles”

1. Joshua Sasmor says:

I asked the nine vs ten question in my probability classes regularly. Here’s another variation: Is a nine on two dice more or less likely than a ten on three dice?

• As a high school sophomore, I played the mayor in “The Music Man,” who has a bit where he keeps trying to recite the Lincoln Gettysburg Address, but he keeps getting interrupted after he says just “Fourscore…” I knew as a prank that one of the many people who were supposed to interrupt me wouldn’t, so I spite-memorized most of that speech. (Or at least enough to make their prank fizzle.)

24 hours * 7 days in a week = 168 hours, and eightscore is 160, which is pretty close.

“Eightscore and eight” is too long for the clue.

“Eightscore eight” is 16 characters, but if you go “using only 15 letters,” hm, that’s just 9 letters (he uses some more than once). But by crossword puzzle rules, “eightscoreeight” is 15 boxes, so I guess that’s correct.

• 1 + 1 + 1 = 3

Okay, that’s going to be just as rare as 6 + 6 + 6 = 18.

Just 1 way to get 3.
Just 1 way to get 18

1 + 1 + 2 = 4
1 + 2 + 1 = 4
2 + 1 + 1 = 4

2 + 2 + 1 = 5
2 + 1 + 2 = 5
1 + 2 + 2 = 5
3 + 1 + 1 = 5
1 + 3 + 1 = 5
1 + 1 + 3 = 5

4 + 1 + 1 = 6
(3 variations)

3 + 2 + 1 = 6
(3 variations)

2 + 2 + 2 = 6
(just 1 way to do that)

7 ways to get 6

5 + 1 + 1 = 7
(3 ways)

4 + 2 + 1 = 7
(3 ways)

3 + 2 + 2 = 7
(3 ways)

9 ways to get 7.

Extrapolating from what I’ve done, I see this pattern:

1 way to get 3
3 ways to get 4
6 ways to get 5
7 ways to get 6
9 ways to get 7
X ways to get 8
Y ways to get 9
Z ways to get 10
Z ways to get 11
Y ways to get 12
X ways to get 13
9 ways to get 14
7 ways to get 15
6 ways to get 16
3 ways to get 17
1 way to get 18

I presume that X < Y < Z, though that's a presumption. It may be that X = Y and/or Y = Z, but I doubt it -- it wouldn't be called a "probability curve" if it went flat. So, Josh, without doing all the scratch work, and without looking it up, I'm going to pick answer B, a ten is more likely than a nine.

• Joshua Sasmor says:

You’re right! I did it a different way, but the results are similar. There are 36 total results for two dice, and four of them add up to nine, so the chance is 4/36 = 11.1% There are 27 ways to get a 10 on three dice, and a total of 216 results, for a chance of 27/216 = 12.5% :)