Boy, stay put!

Probability Level 4

Each minute, a little boy Ben is supposed to be next to his mom at the same vertex of a busy, triangle-shaped marketplace. This minute, he somehow ends up at B with Mom at M . But that's no problem because he is trained well enough to stay put until Mom shows up (who doesn't know at which vertex Ben is).

In fact, Mom can walk one edge of the equilateral triangle per minute, so it takes her only an average of 1.5 minutes to find Ben: 1 minute if she goes M $\to$ B , and 2 minutes if she takes the path M $\to$ T $\to$ B , hence 1.5 minutes on average.

Ben grows up and can walk just as fast as Mom, so he wouldn't do it any more. If he gets lost, now he would also look for Mom at the same time. He insists this because they could even find each other as fast as in 0.5 minutes, half way between their respective vertices, if they are lucky.

If Ben really gets lost and they each head for an adjacent vertex randomly picked every minute, how long will it take for them to get together again, on average?

1.00 minutes 1.25 minutes 1.50 minutes 1.75 minutes 2.00 minutes They'll never find each other

1 solution

Jimin Khim Staff
Sep 27, 2017

Let's go over all possible cases:

Case 1: Mom going counterclockwise, Ben going clockwise $\big($ with probability $\frac12 \times \frac12 = \frac14\big)$
$\hspace{1cm}$ This is the luckiest case where they can see each other in just $\color{#D61F06}0.5\text{ minutes}$ half way between M and B .

Case 2: Mom going clockwise, Ben going counterclockwise $\big($ with probability $\frac12 \times \frac12 = \frac14\big)$
$\hspace{1cm}$ This is also good because they can find each other in $\color{#D61F06}1\text{ minute}$ at T .

Case 3: Both going in the same direction, clockwise or counterclockwise $\big($ with probability $2\times \frac12 \times \frac12 = \frac12\big)$
$\hspace{1cm}$ This is the worst case scenario where they've only wasted $\color{#D61F06}1\text{ minute}$ and have to start all over again, as they are exactly in the same situation as 1 minute ago, now one being at T and the other at M or B .

Now, let $x$ be the expected amount of time (in minutes) that it takes for them to find each other. Then $\begin{aligned} x&=\left(\frac14\times 0.5\right) + \left(\frac14 \times 1\right) + \left(\frac12 \times (1+x)\right) \qquad (1)\\ &=\frac18+\frac14+\frac12+\frac{x}2\\\\ \frac{x}2&=\frac78\implies x=\frac74=1.75. \end{aligned}$ You may wonder what the $x$ at the end of equation (1) might be. It's the same $x$ as on the left-hand side of (1) because, with probability $\frac12,$ they have to go through exactly the same thing again (as if nothing happened) after wasting 1 minute, hence the term $\frac12\times (1+x).$

Therefore, our answer is 1.75 minutes, which is greater than 1.5 minutes. So, the conventional wisdom prevails: "Boy, stay put!"

When do I ever call you boy

Tino Tadena - 3 years, 8 months ago

Boy, stay put!

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