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Can You Crack This Square’s Hidden Code?

Welcome to The Riddler. Every week, I offer up problems related to the things we hold dear around here: math, logic and probability. There are two types: Riddler Express for those of you who want something bite-sized and Riddler Classic for those of you in the slow-puzzle movement. Submit a correct answer for either,1 and you may get a shoutout in next week’s column. If you need a hint, or if you have a favorite puzzle collecting dust in your attic, find me on Twitter.

Riddler Express

From Tyler Barron, there’s more to this numbered square than meets the eye:

6 4 2 7
4 2 0 5
8 4 X 6
8 6 4 8

There’s a logic underlying which numbers go where in that square. Crack the square’s code, and solve for X.

Submit your answer

Riddler Classic

From Dave Moran, a tricky take on a classic cash money problem:

You’re on a game show, and you’re asked to sit down at a table covered with sealed envelopes. You are told that each envelope contains a check for an amount of money, each amount different from all the others, but you are given no other information about the distribution of amounts. (As far as you know, the biggest check on the table could be $1.06 or it could be $98,765,432,100.00.) You may pick an envelope, open it and read the amount of the check. You can then either keep that check, ending the game, or toss it away permanently and open another envelope. You can then keep that second check or toss it away and open a third envelope. And then you can keep the third check or throw it away and pick a fourth envelope. But that’s it — if you open a fourth envelope, you have to keep that check, no matter how paltry it is.

What strategy should you follow to maximize your chances of getting a nice payday?

Submit your answer

Solution to last week’s Riddler Express

Congratulations to 👏 Neema Salimi 👏 of Atlanta, winner of the previous Express puzzle!

It’s NBA Finals time! How many games would we expect to be needed to complete a best-of-seven series if each team has a 50 percent chance of winning each individual game?

On average, the series will last about 5.8 games.

We know for sure that the series is going to take 4, 5, 6 or 7 games to complete. If we figure out how likely each of those is, we can arrive at our overall average. There are a few possible tallies that a team can have accrued when the series ends. For a team to sweep, going 4-0, it clearly has to win four games in a row, which occurs with probability \(0.5^4 = 0.0625\). For a team to win 4-1, a team needs to win four times, but not before it loses once. That occurs with probability \(0.5^4 \cdot 0.5^1 \cdot 4 = 0.125\). (That “4” comes from the fact that there are four different games in the series the team could lose and still wind up the 4-1 record.) Here’s the complete list of possible outcomes, with their probabilities:

4-0: \((0.5^4) = 0.0625\)

4-1: \((0.5^4 \cdot 0.5^1 \cdot 4) = 0.125\)

4-2: \((0.5^4 \cdot 0.5^2 \cdot 10) = 0.15625\)

4-3: \((0.5^4 \cdot 0.5^3 \cdot 20) = 0.15625\)

3-4: \((0.5^4 \cdot 0.5^3 \cdot 20) = 0.15625\)

2-4: \((0.5^4 \cdot 0.5^2 \cdot 10) = 0.15625\)

1-4: \((0.5^4 \cdot 0.5^1 \cdot 4) = 0.125\)

0-4: \((0.5^4) = 0.0625\)

If we tally up the probabilities from the list above, we see that four-game series (4-0 and 0-4) happen with probability 0.125; five-game series with probability 0.25; and six- and seven-game series with probability 0.3125. A weighted average gives us \(0.125\cdot 4 + 0.25\cdot 5 + 0.3125\cdot 6 + 0.3125\cdot 7 = 5.8125\) games.

We can redo the exercise above, supposing that one team is a favorite in each individual game. If one team is favored to win 60 percent of its games, the series is expected to last about 5.7 games. If a team is favored to win 70 percent of its games, the series is expected to last about 5.4 games. It remains to be seen how many games the series lasts if one of the teams is the 2016-17 Golden State Warriors.

Solution to last week’s Riddler Classic

Congratulations to 👏 Daniel Eriksson 👏 of Stockholm, Sweden, winner of the previous Classic puzzle!

You’d like to buy a fancy painting to decorate your new mansion, and it just so happens your local auction house is selling one. However, there is another bidder: your arch nemesis. Each of you two puts a different dollar valuation on the painting, based on your haute tastes and assessment, drawn uniformly randomly between $0 and $100,000,000. You know precisely how much you value the painting but not how much the other bidder does. You submit a sealed envelope containing your bid to the auctioneer, as does your rival. Whoever submits the higher bid wins, and must pay whatever that bid was. Suppose your specific valuation of the fancy painting is $X. How much should you bid?

You should bid $X/2, or precisely half of however much you value the painting.

There’s a whole branch of game theory called auction theory. In that field, this problem is known as a first-price sealed-bid auction. You can find rigorous mathematical approaches to this bidding solution, often relying on differential equations, in online economics book chapters and lecture notes, for example.

So, why $X/2? Let’s start with a couple of observations. First, you never want to bid more than your valuation of the painting. If you lose the auction, you’re right where you started, and if you win the auction, you’ll have overpaid and lost money and be sad — you’d have been happier just staying home. Second, there’s also no point in bidding exactly your valuation. If you lose the auction, you’re right where you started, and if you win the auction, you’ve broken even and so are right where you started — again, you may as well have stayed home. The only general strategy that makes any sense is to bid somewhat below your valuation, hoping to win and therefore to profit.

Your expected payoff from employing a given bidding strategy is 1) the probability that you win the auction multiplied by 2) your “profit” if you win — the difference between how much you value the painting and the winning bid that you must pay the auctioneer. Bidding $X/2 is essentially a compromise between 1) and 2). If you bid exactly how much you valued the artwork, you’d have a good chance of winning but no payoff if you did. If you bid $0, you’d have no chance at winning but a great payoff if you did. The Nash equilibrium of this auction is both bidders bidding precisely half their true valuation, compromising between these two fields.

As the rules of the auction change, so does the optimal bidding strategy. In a second-price sealed-bid auction, for example, where the winner is the one with the highest bid but only has to pay the second-highest bid, it is optimal to bid exactly one’s valuation.

If more bidders show up to vie for the painting, you’ll have to make higher, more competitive bids. Specifically, following the same general logic above, if there are N bidders, and you value the painting at $X, you should bid (N-1)/N * $X.

Going once … going twice … sold! With math!

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  1. Important small print: For you to be eligible, I need to receive your correct answer before 11:59 p.m. EDT on Sunday. Have a great weekend!

Oliver Roeder was a senior writer for FiveThirtyEight. He holds a Ph.D. in economics from the University of Texas at Austin, where he studied game theory and political competition.