GeekDad Puzzle of the Week Answer: Present Game

Reading Time: 2 minutes

Cups and balls, shell game, Monty Hall, or picking presents — whatever you call it, it’s all about probability, baby. Image: Flickr/TheodoreScott

It turns out my kids are experts at determining expected value: the size, weight and number of presents under the tree have now been precisely determined and weighed against each other to determine fairness, which, it also turns out, is an extremely complex concept and one into whose waters I highly recommend parents do not wade.

In any case, here was this week’s puzzle:

Imagine a row of seven presents. Their values are $1, $2, $3, $4, $5, $6 and $40. You don’t know what’s inside and so don’t know which present matches each value. It costs $7 to pick a present at random. Is it a good bet? Now imagine you remove a present at random from the line-up. Is it now a good deal to spend $7 on the blind bet?

There are a couple correct-ish answers, depending on how you define “good bet”. First, there’s a 6-in-7 chance that you will lose money — so if you can only play once, you could make a semi-reasonable case that it’s not a good bet to play at all. But the more widely held definition of “good bet” is a situation in which you expect to win more than you lose. In our first case, the value of the presents added together is $61 and 61/7 = $8.71. So your $7 buys an expected $8.71 and it’s a good bet (really, that’s the answer).

Now the second part gets trickier. What happens when you pull a present at random? Is it still a good bet? There are two ways to think of it, and here are both:

1. By pulling a present at random, you remove an expected $8.71 from the system. So now the value remaining is $61-$8.71=$52.29 with six presents left. 52.29/6=$8.71. Dang! You still get $8.71 for your $7 bet!

2. Imagine there’s a 1/7 chance you pulled the $1 present (yay!) and a 1/7 chance you pulled the $40 present (d’oh!) and, in fact, a 1/7 chance you pulled any of the others, as well. In the best case, you get 60/6 and in the worst case you get 21/6 and all cases are equally likely, so (60/6 + 59/6 + 58/6 + 57/6 + 56/6 + 55/6 + 21/6) / 7 = 61/7 = $8.71. Still a good bet!

The winner, drawn from the many correct entrants, is frequent entrant Andy! Congrats and thanks for all your correct answers over many, many weeks! Andy is the proud winner of a ThinkGeek $50 gift certificate. Sorry, the discount code we had for ThinkGeek has expired; we’ll have a new one soon! And remember: only 126 days until Arbor Day!

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