# 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?

### 7 Responses

1. Anonymous says:

49 secs.

• Sunil Kumar says:

How so?

• Mark says:

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.

2. Anonymous says:

I got 48

3. Anonymous says:

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.

4. Anonymous says:

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

• Thinker says:

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