From original by James Lovejoy

Cover up the end of this column before reading on!

1. A friend has one black and two white balls. First, she puts one ball into a bag, not letting you see the color. Then she puts a white ball in, so now the bag contains either one black and one white ball, or two white balls. You put your hand in and take one ball out: it’s white. What are the odds that the remaining ball in the bag is white?

(If you’re thinking, “Either the ball is black or white, so the probability is 1/2,” consider how many scenarios are possible.)

2. The scene is a pebbly beach, where an evil prince has kidnapped a young woman. He says, “In this bag, I will put one white pebble and one black pebble. You will put your hand in and draw out one pebble. If you pick the white one, you go free, but if you pick the black one, you have to marry me.” She suspects that he cheated and put two black pebbles into the bag. What does she do?

3. Four prisoners are shown two black hats and two white hats before being blindfolded. Their jailer seats three of them in line, one behind the other, with the fourth in another room (see picture). He puts either a black or a white hat on each of them, and removes the blindfolds. No communication between the prisoners is allowed; they can’t look behind them; all that is permitted is one of them saying, “I am wearing a [?] color hat.” If correct, all four will be set free. If incorrect, they will die. What strategy will free them?

ANSWERS

1. Two-thirds. Call the three balls A and B (white) and C (black). You have three starting possibilities in the bag (A and B, A and C, B and C) giving you six possible endings after one ball has been removed (B or A, C or A, C or B). Four of these six leave you with a white ball. (This is a great project to actually test, in a fifth-grade classroom, say.)

2. She puts her hand in the bag, grabs one pebble and throws it away, saying, “Whatever color is left is the opposite of what I picked.”

3. If prisoner A sees either two black or two white hats, he announces that he must be wearing a hat of the opposite color and they all go free. Otherwise, he stays silent. Prisoner B waits awhile, knowing that if A saw two hats of the same color, he’d have spoken up. If A remains silent, B realizes that his own and C’s hat must be different. If C’s hat is white, he says his is black, and vice versa. Freedom awaits. (The fourth prisoner is just a hat-wearer.)

Barry Evans (barryevans9@yahoo.com) tries not to see the world in black and white. His first 80 Field Notes are on sale at Eureka Books.

 

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3 Comments

  1. Well that’s embarrassing! How about 1/3? I neglected to toss out the times when you take out a black ball. So…

    Call the three balls A and B (white) and C (black). You have three starting possibilities in the bag, each of which will occur 1/3 of the time.
    1. A and B in the bag. 1/6 of the time you’ll leave A and 1/6 of the time you’ll leave B.
    2. A and C in the bag. 1/3 of the time you’ll leave C (after removing white ball A).
    3. B and C in the bag. 1/3 of the time, you’ll leave C (after removing white ball B).
    So 1/3 of the time you’ll end up with a white ball in the bag.

    (Thanks for the catch, John!)

  2. I believe this applies…in highschool, the riddle for the advanced math class (that I wasn’t in) was this:

    You’re on a gameshow and given the choice of what’s behind 3 closed doors, only one of which is the winner.

    You choose a door, and it’s not the winner. The host is generous and you’re given a second shot, choosing between the remaining two doors.

    What are the odds now? Most math geeks argued that the odds are still 1/3. I stand that it’s a matter of philisophy more than math, and the odds are 50/50. It all depends where you place yourself in the context of time and space…opening all kinds of doors to paradox.

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