Tag: Advanced


David Justice and Derek Jeter were professional baseball players. In 1997 they had the following conversation:

David: Did you know that in both 1995 and 1996 I had better batting averages than you?

Derek: No way, my batting average over the last two years was definitely higher than yours!

It turned out that both of them were right. How is it possible?

This is the so called Simpson’s paradox. The reason it occurred is that during 1996 both players had high averages and Derek Jeter had many more hits than David Justice. In 1996 both players had low averages and David Justice had many more hits than Derek Jeter. You can see their official statistics below.

Derek Jeter12/48 (.250)183/582 (.314)
195/630 (.310)
David Justice104/411 (.253)45/140 (.321)149/551 (.270)

Alice secretly picks two different integers by an unknown process and puts them in two envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss) and shows you the number in that envelope. Now you must guess whether the number in the other, closed envelope is larger or smaller than the one you have seen.

Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?

Choose any strictly decreasing function F on the set of all integers which takes values between 0 and 1. Now, if you see the number X in Bob’s envelope, guess with probability F(X) that this number is smaller. If the two numbers in the envelopes are A and B, then your probability of guessing correctly is equal to:

F(A) * 0.5 + (1 – F(B)) * 0.5 = 0.5 + 0.5 * (F(A) – F(B)) > 50%.


These are a few enigmas from the puzzle book CODEX ENIGMATUM. What is the answer to puzzle #9?

You can buy the complete book, created by Rami Hansenne, from this LINK.

  • Puzzle #2 After turning the first wheel 22 times to the right, then 19 times to the left, then 15 times to the right, and finally 11 times to the left, the final wheel will spell the word EXIT.
  • Puzzle #3 The total number of spots on the hidden sides of the die on the left is 6, which corresponds to the sixth letter in the alphabet – F. Therefore, the four dice on the right correspond to the letters K, I, N, G.
  • Puzzle #4 In the mosaic on the right, you can find a little star which contains pieces with letters H, I, D, E.
  • Puzzle #6 If you trace the signature on the paper, starting from the large C, you will pass through the letters C, O, N, T, I, N, U, O, U, S.
  • Puzzle #8 The picture on the left and the answer to puzzle #6 (“continuous”) suggest that we have to consider the images on the right which can be drawn continuously, without taking off the pencil from the paper or passing through any segment twice. These images are labeled with the letters N, O, S, E.
  • Puzzle #9 The first 2 letters from the word NOSE spell NO. The last letter from the word EXIT is T. The first letter from the word HIDE is H. The last three letters from the word KING spell ING. When you combine all of them, you get the word NOTHING.


One person went to the store and bought groceries for $13.59 total. He paid with a $100 bill, took his change, and left the store. There was something special about this transaction. What is it?

The person paid with a $100 bill. The cashier returned him a $50 bill, a $20 bill, a $10 bill, a $5 bill, a $1 bill, a quarter, a dime, a nickel, and a cent. The transaction consisted of exactly one of each (frequently used) denominations.


What number corresponds to 1985?

0 0 0 0 – 4
1 7 5 2 – 0
1 8 7 9 – 3
2 0 6 1 – 2
3 1 4 1 – 0
4 0 9 6 – 3
7 7 7 7 – 0
9 9 7 3 – 2
1 9 8 5 – ???

The numbers on the right count the total number of “holes” in the digits on the left. “1”, “2”, “3”, “4”, “5” and “7” have 0 holes in them. “0”, “6” and “9” have 1 hole in them. “8” has 2 holes in it. Therefore, the corresponding number is 3.


With 12 matches you can easily create a shape with area 9 and a shape with area 5, as shown on the picture below. Can you rearrange the 12 matchsticks, so that they encompass an area of 4?

Remark: You should have only one resulting shape and no matches should be unused.

First, create a Pythagorean triangle with sides 3, 4, 5, and area 6. Then simply flip its right angle inwards, so that the area decreases by 2.