Black Swans

charmaine

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I'm Charmaine

I study at St Andrews and help run Black Swans.
History

History

Solution to "Who's Doing the Dishes?"

We received 57 submissions 😲 to the Who's Doing the Dishes? puzzle and, after carefully going through each of them, have judged that the best solution was given by simon. Congratulations! 🎉🍾

Here's a reminder of the puzzle:

You have begun a series of rock-paper-scissors games with your spouse, the loser to wash the dishes. Bad news: your spouse being the better player, your probability of winning any given game is $< 0.5$. Worse, you've already lost the first game.

But there's good news too: your spouse has generously agreed to let you pick, right now, the number $n$ of games that must be won to win the series. Thus, for example, if you pick $n=3$, you must win $3$ games before your spouse does (but remember, your spouse has already won a game).

What is your best choice of $n$, as a function of $p$?

And here's Prof. Winkler's solution:

Intuitively, the lower $p$ is, the smaller you want $n$ to be. This is borne out by the fact that the probability of winning $n$ games before your spouse does is of order $p^n$. For example, if $p = 0.01$, you want $n=2$ since asking for more than two miracles is clearly not in your interest. But as $p$ goes up, your optimal target will eventually change to $3$; and by the time $p$ reaches $0.49$, you want a long match (first to win $26$ games, it turns out) to minimize your spouse's advantage from having won the first game.

To solve the puzzle, we must find the threshold values of $p$, that is, the values for which you are indifferent between $n$ and $n+1$. To do this, imagine that the decision between playing for $n$ wins and for $n+1$ wins is to be made by someone else (Sam, say), who tells you to go ahead and play regardless.

Suppose you play $2n-1$ more games before Sam weighs in. It's easy to see that unless you won exactly $n$ of them, the match is over and you don't even need to hear from Sam.

If you did win $n$ of them, and Sam says the target is $n+1$, you must of course play one more game and you will win the match with probability $p$.

But if Sam says the target is $n$, the match was already over one game ago, and the winner of the superfluous final game is the loser of the match. Since your spouse won $n-1$ of $2n-1$ games, the probability that she won the final game is $\frac{n-1}{2n-1}$ and therefore that is the probability that you won the match.

It follows that for $p = \frac{n-1}{2n-1}$ you are indifferent between playing for $n$ wins and playing for $n+1$, so the thresholds are at $p = 1/3, 2/5, 3/7, 4/9$ etc. This gives the ceiling of $\frac{1-p}{1-2p}$ as the optimal number of wins to play for, if you choose $n$ when $n$ and $n+1$ are equally good; otherwise, the (equally correct) floor of $2 + \frac{1}{1-2p}$. The two answers agree except on a set of measure zero.

Yes. But receiving quite a few submissions, so bear with us 🙏.