# The Drunkard Riddle

**Puzzle of the day 2038 – **The drunkard problem

A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s.

**How long the drunkard takes to fall in a pit 13 m away from the start?**

49 secs.

How so?

He takes 8 total steps to go 2 steps forward. He does this 6 times. The next step he takes is into the pit. 49 total steps 49 total seconds.

I got 48

Each step takes 1 second, and he takes 5 steps forward = 5 seconds. Then he takes 3 steps back, which takes 3 seconds. So in 8 seconds (5+3), he’s moved forward 2 meters (each step = 1 m, 5 m forward, 3 backwards leaves him 2 m from where he started). So, 8 seconds, 2 meters; 16 seconds, 4 meters; 24 seconds, 6 meters; 32 seconds, 8 meters.

Now, this is where it gets tricky. If we assume that once he reaches the edge of the hole, he immediately falls in, then at the end of the next 5 steps forward (8 meters + 5 meters = 13 meters), he falls in and we’re done. Which would give us a time of 32+5 seconds, for 37 seconds total.

If however, he has to step past the edge of the hole at 13 meters to fall in, then we add 1 round of 5 forward, 3 back, in which he gains 2 meters (40 seconds, 10 meters), followed by the four steps forward to put him in the hole. So, 44 seconds and he ends at 14 meters/inside the hole.

Of course, this is all assuming that he walks in a straight line towards the hole the entire time.

It doesnt ask how long it takes him to get to the pit, it asks how long it takes to fall in it…

Thats correct but the next step after reaching will make him fall in the pit