A Pocket Puzzle: Solution

This post contains a solution for last week’s Pocket Puzzle post. If you want to try the puzzle before seeing the solution, don’t scroll past the jigsaw pieces.

Usually, there are a lot of ways to view at a particular idea. It could be an idea about Canadian politics, photosynthesis, or calculus– you can turn the idea around and look at its different facets. Like shining light through a rotating crystal, this can give you unexpected new perspectives. And often, some of those perspectives are more illuminating than others.

I chose this puzzle because I heard about a solution with a nice perspective. It’s sort of an unexpected shortcut that gives you a new way to look at this type of problem.

As a reminder, here was the coin puzzle from last post:

Alice and Bob are bored and want to play a game. “I have an idea,” Alice chimes in. “How about you flip this coin 2009 times and I’ll flip it 2010 times and whoever gets more Heads wins?”
Bob replies, “No, that’s not fair! You’re probably going to win since you get more flips!”
“Fine!” answers Alice. “How about this? You flip the coin 2009 times and I’ll flip it 2010 times and if I get more Heads, I win. If you get more Heads, you win. And, if there’s a tie, we’ll say that you win, too.” Bob shrugs his shoulders and agrees to play.
What’s the probability that Alice wins the game? Prove your answer.

Here is one perspective on the puzzle, which I heard from a professor of mine. I think it’s pretty clever.

Alice flips her coin 2010 times. Let’s break that into two stages: we’ll look at just her first 2009 flips, and then after that, we’ll look at her last flip.

After Alice and Bob each flip 2009 times, there are three possibilities:

  1. Alice has more heads
  2. Bob has more heads
  3. Alice and Bob have the same number of heads

We don’t know what the odds of each of these three outcomes are. We do know that #1 and #2 are equally likely– they’re the same sort of outcome, except Alice and Bob are swapped.

Now let’s look at what happens when Alice flips her final coin.

Let’s say Alice flips a head. Let’s combine this with our three possibilities from the first 2009 flips.

  1. Alice had more heads, and she still does. Alice wins.
  2. Bob had more heads. Now either Bob still has more heads, or there is a tie. Bob wins.
  3. There was a tie. Now Alice has more heads. Alice wins.

Let’s say, instead, Alice gets tails.

  1. Alice had more heads, and she still does. Alice wins.
  2. Bob had more heads, and he still does. Bob wins.
  3. There was a tie, and there still is. Bob wins.

So what happened here?

We started with a fair game, where Alice and Bob each flipped 2009 coins. If Alice flips more heads during that part of the game (scenario 1), she’ll win no matter what the last flip is. If Bob flips more heads during that part of the game (scenario 2), he wins no matter what.

The only time the last flip matters is scenario 3– where Alice and Bob were tied. And as we saw, that last flip is just a tie breaker. Heads, Alice wins; tails, Bob wins.

So the game is still fair.

The unusual approach here was to break up the game up into two stages: the first 2009 flips, and the last flip. It’s a pretty creative idea. What’s cool about this perspective is that it let us check that the game is fair without computing any probabilities at all!

Anyway, I thought that was a pretty nice idea. Let me know if you have questions! And if you’d like to try another puzzle, check out The Mutilated Chessboard.

8 thoughts on “A Pocket Puzzle: Solution

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