Three condemned prisoners share a cell. A guard arrives and tells them that one has been pardoned.
“Which is it?” they ask.
“I can’t tell you that,” says the guard. “I can’t tell a prisoner his own fate.”
Prisoner A takes the guard aside. “Look,” he says. “Of the three of us, only one has been pardoned. That means that one of my cellmates is still sure to die. Give me his name. That way you’re not telling me my own fate, and you’re not identifying the pardoned man.”
The guard thinks about this and tells him, “Prisoner B is sure to die.”
Prisoner A rejoices that his own chances of survival have improved from 1/3 to 1/2. But how is this possible? The guard has given him no new information. Or has he?
Answer: This is same as the Monty Hall problem. A's chances of dying are still 1/3
Monty Hall: 3 doors, Goats behind two and treasure behind the third. Pick - and then the host will reveal one door and ask if you want to switch? Should you switch? (Answer: Yes)
What's Next
f4e, s9, se5ev - what comes next?
five, six, seven, eight!
Crossed Wires
There are 100 wires running from town A to town B. They are all mized up so we don't know which end is which. How many trips do you need to make to figure out. Note that wires can either be switched on or off - no other signals are possible.
7 trips. First turn half the wires ON in town A (think of this as first bit ON). Go to town B, and check for ON wires - label them first bit ON. Now, turn half of the ON wires OFF and half of the OFF wires ON - measure again and label. So on, you can get all the 7 bits position and label the wires accurately.
Equal Chances, or Not?
A man has two girlfriends, and wants to see them equally often. Trains leave from his station to each of their stations every one hour. So he decides that he is going to land up at the station at a random time, and take whichever train comes first. Fair? But he finds out that he is seeing one girlfriend three times as much as the second. Why?
Answer: The train to first arrives 15 minutes after the train to second
Homework Problem:
Find a better solution to the wire crossing problem above
1 round trip solution.
In city A, divide the 100 wires into 50 pairs, connect 49 of the pairs, leaving one pair disconnected. No need to mark these wires. Travel to city B. Use battery + bulb to identify the connected pairs. Label the two wires of the first pair 1a, 1b. For the second pair 2a, 2b, up through 49a, 49b. Label the disconnected pair 50a, 50b.
Connect 1b to 2a, 2b to 3a, 3b to 4a, ... 49b to 50a, leaving only 50b and 1a disconnected. Travel back to city A and identify which of the two free wires makes no circuit with any of the other pairs. This wire maps to 50b on the other side, and the other free wire maps to 50a. Label them as such.
Now take 50a and find the pair it makes a circuit with. Disconnect the pair and find which wire of the pair it connects with. This wire maps to 49b on the other side and it's pair to 49a. Now find the wire that makes a circuit with 49a. This maps to 48b and it's pair to 48a. Repeat until all wires are identified.