In 1918, German flying ace Manfred von Richthofen chased an inexperienced Canadian pilot out of a dogfight and up the Somme valley. It would be the last chase of his life. In this week’s episode of the Futility Closet podcast we’ll describe the last moments of the Red Baron and the enduring controversy over who ended his career.
We’ll also consider some unwanted name changes and puzzle over an embarrassing Oscar speech.
Steve, Tony, and Bruce have a plate of 1,000 cookies to share. They decide to share them in the following way: beginning with Steve, each of them in turn takes as many cookies as he likes (they must take an integer amount, greater than or equal to 1), and then passes the plate clockwise (with Tony sitting to Steve’s left, and Bruce sitting to Tony’s left). Nobody wants to feel like he hogged too many cookies, so they all want to avoid being the player at the end who has taken the most cookies. Additionally, nobody wants to feel cheated by finishing with the fewest cookies. Finally, given that the previous two conditions are definitely met, or definitely cannot be met, each player would like to maximize the number of cookies he eats. The players’ objectives can be summarized as follows:
Have one player who has eaten more cookies than you, and one player who has eaten fewer cookies than you.
Eat as many cookies as possible.
Objective #1 takes infinite priority over Objective #2. Assuming that all players are perfectly rational, that they are all aware of each other’s rationality and objectives, and that they cannot communicate with each other in any way, how many cookies should Steve take to ensure he meets both objectives and how many cookies will Tony and Bruce take if Steve takes the winning amount?
First, suppose Steve takes 335 cookies. We will argue that Tony can then win by taking 334 cookies. If Tony does take 334, Bruce will be left with a plate of 331 cookies to take from. No matter how Bruce plays from here, he cannot meet objective #1, because he will need at least 335 cookies to do so, and has only 331 to take from. Therefore, Bruce will maximize the number of cookies he eats, and take the remaining 331. Tony finishes with 334, the middle amount, so he meets objective #1, justifying his decision to take 334 cookies. (Note that if Tony had taken more than 334, Bruce would still have taken all the remaining cookies, but Tony would not have met objective #1, so taking 334 is in fact Tony’s best play.)
Therefore, Steve will fail objective #1 if he takes 335 cookies. Furthermore, we can repeat the argument above if Steve takes more than 335 cookies, by always having Tony take 1 fewer cookie than Steve (or all remaining cookies, if this amount is less than the number Steve took).
Now suppose instead that Steve takes 334 cookies. We will show that Tony cannot take any amount of cookies resulting in him (Tony) meeting objective #1. First, note that if Tony takes 333 or more cookies, Bruce will be unable to meet objective #1 (he will need at least 334, but there are at most 333 remaining), and so will take all remaining cookies, resulting in a “loss” for Tony.
Now suppose instead that Tony takes x cookies, where x ≤ 332. Then there will be 666 – x cookies remaining. Suppose Bruce takes 333 cookies. This leaves 333 – x cookies on the plate. Even if Steve takes only 1 cookie each turn from now on, Bruce can ensure that he will have fewer cookies than Steve by also taking 1. Suppose Steve takes s cookies. Then on Tony’s next turn, he has 333 – x – s cookies to take from. Even if Tony takes all remaining cookies, total over his two turns, he will have taken 333 – x – s + x = 333 – s, which is at most 332 (since s ≥ 1), so Tony will definitely finish with fewer cookies than Bruce. Thus, if Tony takes 332 or fewer cookies, Bruce can win by taking 333 cookies, since this guarantees he will finish with less than Steve and more than Tony.
Therefore, if Steve takes 334 cookies, Tony cannot meet objective #1, and so will take as many cookies as possible, leaving none for Bruce. This leaves Steve with the middle amount of cookies, so taking 334 cookies is in fact a winning strategy for Steve. As shown above, Steve cannot do any better than this.
The upper edge of the setting sun is sometimes seen to take on a green tinge, an effect of atmospheric refraction. Normally this is apparent only briefly, but for Richard Byrd’s Antarctic expedition of 1928-1930 it lasted more than half an hour:
Here the sun descends so slowly that it seems to roll along the horizon and as it will be only two days until it is above the horizon all the time for the rest of the summer it clings interminably before, with seeming reluctance, dropping from sight. As its downward movement is so prolonged the last rays shimmer above the barrier edge as it moves eastward, appearing and reappearing from behind the irregularities of the barrier surface. It trembles and pulsates, producing a vibration light of great beauty.
The night the green flash was seen some one ran into the administration building and called, ‘Come out and see the green sun.’
There was a rush for the surface and as eyes turned southward, they saw a tiny but brilliant green spot where the last ray of the upper limb of the sun hung on the skyline. It lasted an appreciable length of time, several seconds at least, and no sooner disappeared than it flashed forth again. Altogether it remained on the horizon with short interruptions for thirty-five minutes.
When it disappeared momentarily it seemed to have been shut off by a tiny spurt, an inequality in the skyline caused by the barrier surface.
“Even by moving the head up a few inches it would disappear and reappear again and after it had finally disappeared from view it could be recaptured by climbing up the first few steps of the [antenna] post.”
The Guinness record for the most fraudulent election ever reported belongs to the Liberian general election of 1927, in which President Charles D.B. King was re-elected over challenger Thomas J. Faulkner:
Charles D.B. King
Thomas J. Faulkner
As there were fewer than 15,000 registered voters, this represents a turnout of 1,680 percent — robust indeed.
I just ran across this anecdote by Jason Rosenhouse in Notices of the American Mathematical Society. In a middle-school algebra class Rosenhouse’s brother was given this problem:
There are some horses and chickens in a barn, fifty animals in all. Horses have four legs while chickens have two. If there are 130 legs in the barn, then how many horses and how many chickens are there?
The normal solution is straightforward, but Rosenhouse’s brother found an alternative that’s even easier: “You just tell the horses to stand on their hind legs. Now there are fifty animals each with two legs on the ground, accounting for one hundred legs. That means there are thirty legs in the air. Since every horse has two legs in the air, we find that there are fifteen horses, and therefore thirty-five chickens.”
(Jason Rosenhouse, “Book Review: Bicycle or Unicycle?: A Collection of Intriguing Mathematical Puzzles,” Notices of the American Mathematical Society, 67:9 [October 2020], 1382-1385.)