Welcome to The Riddler. Every week, I offer up a problem related to the things we hold dear around here — math, logic and probability. These problems, puzzles and riddles come from lots of top-notch puzzle folks around the world, including you, the readers. You’ll find this week’s puzzle below.

Mull it over on your commute, dissect it on your lunch break, and argue about it with your friends and lovers. When you’re ready, submit your answer using the form at the bottom. I’ll reveal the solution next week, and a correct submission (chosen at random) will earn a shoutout in this column. Important small print: To be eligible for the shoutout, I need to receive your correct answer before 11:59 p.m. EST on Sunday — have a great weekend!

Before we get to the new puzzle, let’s return to last week’s. Congratulations to 👏 Emily Schuch 👏 of Brooklyn, our big winner. You can find a solution to the previous Riddler at the bottom of this post.

Now, here’s this week’s Riddler, which comes to us from Stephen Mellendorf, a trading algorithm developer from Greenwich, Connecticut.

Two players go on a hot new game show called “Higher Number Wins.” The two go into separate booths, and each presses a button, and a random number between zero and one appears on a screen. (At this point, neither knows the other’s number, but they do know the numbers are chosen from a standard uniform distribution.) They can choose to keep that first number, or to press the button again to discard the first number and get a second random number, which they must keep. Then, they come out of their booths and see the final number for each player on the wall. The lavish grand prize — a case full of gold bullion — is awarded to the player who kept the higher number. Which number is the optimal cutoff for players to discard their first number and choose another? Put another way, within which range should they choose to keep the first number, and within which range should they reject it and try their luck with a second number?

Need a hint? You can try asking me nicely. Want to submit a puzzle or problem? Email me.

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And here’s the solution to last week’s Riddler, concerning cracking the codes of four secret messages. Just a fraction of you submitted compared to the usual Riddler — codebreaking appears to have gone out of style.

The first code was a simple Caesar cipher. Each letter in the original message is replaced with a different letter, a fixed distance away in the alphabet. In this case, the letters were shifted 538 places down in the alphabet (or, equivalently, 18 places), so “I” became “A,” “h” became “z,” and so on. The original message, therefore, was “I have successfully solved the FiveThirtyEight Riddler!” All of you who submitted got this one!

The second message was encrypted using a Vigenère cipher. These ciphers are a bit trickier as they require a keyword to crack. But the keyword here, like the number of alphabet places shifted above, is “fivethirtyeight.” (How were you supposed to know that? There were hints all over the place!) A Vigenère cipher is like a bunch of different Caesar ciphers put together. The first letter in the encoded message has been shifted according to the first letter of the keyword. If the first letter of the keyword was “a,” the letter shifts zero spaces, if “b,” it shifts one, and so on. (This is made easier with a Vigenère square handy, which I assume you all have.) The first letter of our encrypted message was “x,” which should be shifted back according to the first letter of our keyword, “f,” or five alphabet spaces, so the first letter of the decoded message is “s.” The second letter, “c,” should be shifted according to the letter “i,” or eight spaces, so it becomes “u.” And so on, repeating again with “f” after the first run through the keyword. The full secret message was, “Super Tuesday is this Tuesday. Cruz, Trump, Rubio … who will win the most delegates?” (The answer was Trump.) 88 percent of submitters got this one.

The third code used what’s called a Playfair cipher. This cipher encodes pairs of letters, rather than single letters. Again, you need a keyword to solve it, and again, that keyword was “fivethirtyeight.” (Or really, “fivethryg,” since the keyword can’t repeat letters, as we’ll see in a second.) To begin, you arrange the letters of the alphabet into a 5-by-5 square, called a “tableau,” starting with the keyword. (I and J are treated as the same letter, to keep the square nice.) In this case, it’d look like this:

F I V E T H R Y G A B C D K L M N O P Q S U W X Z

To decode, work with one letter-pair at a time. If the letters are in the same row of the tableau, shift them left, wrapping around to the other side of the square if necessary. For example, “hy” becomes “ar.” If the letters form a rectangle, use the other two “corners” of that rectangle. For example, “vg” becomes “ye.” And so on. The solution: “Are you having fun yet?” 87 percent of those who submitted got this one.

This last secret message was based on the Unicode standard for emoji faces. It’s a simple replacement cipher, so the first emoji in Unicode-order is “a,” the second “b,” and so on. The code is simple — the trick was just in figuring out how to order the smiley faces to correspond to the alphabet. The final cracked code? “OK, that’s it. Now get back to work. Hey, at least it’s Friday.” 91 percent got this one.

And lo, it’s Friday again! Have a great weekend!