Puzzle Solution: Kant and his Clock

Here is a potential solution to this problem.

Let "t" represent the difference between Kant's clock and the Schmidt's clock (the correct time) in minutes. Let "x" represent the amount of time it takes for Kant to walk from his house to Schmidt's.

For illustration, let us assume a concrete case: t = 60 mins, and x = 30 mins. Suppose Kant leaves his house at 1pm according to his clock, and stays at Schmidt's place for 2 hours.

Event	Kant's Clock	Schmidt's Clock
Leaves Home	1:00	2:00
Arrives at Schmidt's	1:30	2:30
Leaves Schmidt's	3:30	4:30
Arrives Home	4:00	5:00

He can't be at two places at the same time (the problem would be trivially solved if he could). The bold red times in the table above indicate the clocks that Kant has access to at any particular point.

Since Kant knows his algebra he sets up a simple system of equations. The first equation can be set by observing in sloppy notation that 1:00 + t + x = 2:30. Since "t" and "x" are measured in minutes this is better represented as t + x = 90.

Similarly, 4:30 - t + x = 4:00, can be recast as x-t = -30.

Adding the two equations gives t = 60 and x = 30.

This can be easily generalized.

Dicey puzzle: Solution

The full puzzle statement may be found here.

In short "What are the odds that n=2 v/s n=3 dice are rolled, given that the sum is 7?"

Solution:

For n=2, there are 6 ways of rolling a 7 (1+6, 2+5, 3+4, 4+3, 5+2, and 6+1), out of a total of 6^2=36 total outcomes.

Therefore p(sum = 7 | n = 2) = 6/36 = 36/216.

For n = 3, there are 15 ways of rolling a 7 (1+1+5, 1+2+4, 1+3+3, 1+4+2,1+5+5, 2+1+4, 2+2+3, 2+3+2, 2+4+1, 3+1+3, 3+2+2, 3+3+1, 4+1+2, 4+2+1, 5+1+1), out of a total of 6^3 = 216 total outcomes.

Therefore p(sum = 7 | n = 3) = 15/216.

Thus, the odds of n = 2 v/s n = 3 are 36/15. That is it is about 2.5 times more likely that n = 2.

What happened? There were more ways of getting a 7 with n = 3?