We're giving **£100** 🤑 to whoever submits the best (i.e., shortest, cleverest, etc.) solution to the puzzle below, kindly donated to us by Professor Peter Winkler, author of Mathematical Mind-Benders and Mathematical Puzzles: A Connoisseur's Collection:

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 $p$?

Scan your solutions to [email protected] before 21st September and we'll announce the winner 🍾 shortly afterwards. If you have any questions, post them below.

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