Did you solve it? Are you smarter than a Navy admiral? | Mathematics
Earlier today, I asked you these three riddles. Here they are again with solutions.
1. Battleships
You are a Navy admiral, charged with an important mission. You have two choices.
a) Send a single ship whose chance of success is P percent.
b) Send two ships, each of which has a chance of success of P/2 percent. At least one ship must succeed for the mission to be a success.
What is the best option?
Solution. a) Send only one ship.
Intuition might tell you that it is better to send both ships, since two chances are better than one. However, it is easy to see that this is not the case if P = 100. In this case, a) guarantees success, but b) only gives 75 percent success. (Since the risk of failure of both ships is 50 percent x 50 percent = 25 percent.)
It turns out that for all values of P it is better to send just one ship. To leave p be the probability of success. (i.e. P/100). If two ships each have a probability p/2, the chance that both fail is (1–p/2)2so the chance that at least one succeeds is 1 – (1–p/2)2 = p – (p2)/4, which is always less than pthe chance that the only boat succeeds.
2. The two oracles
Before you are two oracles, Randie and Rando, who will answer yes or no to all your questions.
Randie randomly answers yes or no to all questions.
Rando randomly decides whether to tell the truth or lie for each question, then answers the question accordingly.
Is there a way to differentiate them? If so, what is it?
Solution. Yes, you can do it!
The trick is to realize that it is possible to ask Rando questions that you know he will always answer “YES”.
For example: “Are you answering this question honestly?” The liar and the one who tells the truth give the same answer.
So ask this question until you get a no. Once you do, you know it’s Randie. If you don’t, you can be sure it’s Rando.
3. Bad calculations
Johnny’s homework is to calculate 5548-5489. The answer is 59. He thought the 548 canceled out, leaving 59.
He tested the technique again. He typed in a subtraction of the form XXYZ – XYZW where X, Y, Z and W were separate numbers, and discovered that it was indeed XW!
How many digits in the new calculation are the same as in the old one? (i.e. X = 5, Y = 4, Z = 8 or W = 9)
Solution. Z and W are equal to 8 and 9
The calculation can be broken down as follows:
1100X +10Y + Z – 1000X – 100Y – 10Z – W = 10X + W
which reduces to
90X – 90Y = 9Z + 2W
We deduce that W must be divisible by 9, so it is either 0 or 9. In addition, 9Z + 2W must be divisible by 10.
If W = 0, then Z = 0, which is a contradiction since they call into question states from which they are distinct.
So W = 9. Then 9Z + 18 is divisible by 10, which means Z = 8. We are left with
90X – 90 Y = 90. So
X = Y + 1, which has several solutions. But W = 9 and Z = 8.
Today’s riddles are taken from an excellent new collection of puzzles, Mathematical puzzles and curiosities, by Ivo David, Tanya Khovanova and Yogev Shpilman. I edited the wording of the puzzles and solutions to fit the column.
I’ve been putting up a puzzle here every other Monday since 2015. I’m always looking for great puzzles. If you want to suggest one, send me an email.
