Last night's Final Jeopardy clue: "With 301 miles, it has the most coastline of current states that were part of the original 13 colonies." (Thanks to the Jeopardy! forum for the wording.)
This agrees with the Wikipedia list which is sourced from official US government data.
But as Mandelbrot told us, coastlines are self-similar . (Link goes to the paper How Long is the Coast of Britain as reproduced on Mandelbrot's web page, which unfortunately doesn't have pictures. I'm not sure if the original version of this paper in Science did. Wikipedia's article on the paper does.) That is, the length of a coastline depends on the size of your ruler. Furthermore, I would suspect that the fractal dimension of some states' coastlines is larger than others. Wikipedia states that "Measurements were made using large-scale nautical charts" which seems to imply that all the measurements were done at the same length scale, but if you did the measurements at a smaller scale, the states whose coastline have higher fractal dimension would move up the list.
So last night I spent twenty seconds yelling this at the TV, and ten seconds getting out the answer Maryland. Which is wrong. Also wrong: Maine, New York. (Maine used to be part of Massachusetts; the wording is a bit ambiguous.) It appears that only the south shore of Long Island counts as "coast"; the north shore, which borders Long Island Sound, doesn't.
And of course Chesapeake Bay doesn't count either.
Showing posts with label Jeopardy. Show all posts
Showing posts with label Jeopardy. Show all posts
30 March 2011
10 February 2009
Two quasi-mathematical Jeopardy! clues
Two clues from yesterday's episode of Jeopardy!. (I meant to post this yesterday, and in fact wrote the post yesterday, but didn't want to give out spoilers, so you're getting it now.)
1. "In math, it's the degree of correctness of a quantity or expression." Answer: "What is accuracy?" (This was in a category which carried the stipulation that every answer had to begin with A and end with Y.)
2. "While writing Principia Mathematica, this twentieth-century British thinker was a lecturer at Cambridge." Answer: "Who is Bertrand Russell?"
My objection to the first one is more that this isn't a mathematical concept; it seems to me that that use of "accuracy" is more common in, say, the experimental sciences.
And to the second one, I believe Russell was the answer given by a contestant and accepted, but Whitehead also might fit this description. (Whitehead left Cambridge just as the first volume of the Principia was published, in 1910.)
1. "In math, it's the degree of correctness of a quantity or expression." Answer: "What is accuracy?" (This was in a category which carried the stipulation that every answer had to begin with A and end with Y.)
2. "While writing Principia Mathematica, this twentieth-century British thinker was a lecturer at Cambridge." Answer: "Who is Bertrand Russell?"
My objection to the first one is more that this isn't a mathematical concept; it seems to me that that use of "accuracy" is more common in, say, the experimental sciences.
And to the second one, I believe Russell was the answer given by a contestant and accepted, but Whitehead also might fit this description. (Whitehead left Cambridge just as the first volume of the Principia was published, in 1910.)
05 January 2009
Shadow lengths
From a Jeopardy! clue: to get vitamin D, walk around in the sunlight for at least twenty minutes a day -- but only when the sun is low in the sky. "Low in the sky" is defined as when your shadow is longer than you. That is, when the altitude of the sun is less than forty-five degrees.
The time of Muslim afternoon prayers are also defined in terms of simple arithmetic on shadow lengths.
The time of Muslim afternoon prayers are also defined in terms of simple arithmetic on shadow lengths.
06 May 2008
Something I learned from Jeopardy!
On tonight's Jeopardy, the last clue in a category about awards referred to the Fermat Prize. The clue was something like "The Fermat prize is given in this field for work involving 'variational principles'" (I don't have the exact wording). The question, of course, is "What is mathematics?"
Now, I didn't know there was such a prize. It turns out that it's awarded for work in any of the following three fields in which Fermat's work was important: "Statements of Variational Principles", "Foundations of Probability and Analytical Geometry", or "Number theory". From the past prize winners it looks like it's been awarded in all of those categories, although when I heard the clue it sounded like the prize was only awarded for work in the calculus of variations. (Sometime in the next couple days, the exact wording should be available at the Jeopardy! archive.)
Also, Alex Trebek pronounced the "t" in Fermat. No.
And the second-to-last clue in the same category asked for the subject in which the Turing Award is given. (I'd never heard of that award, either, but the answer is obvious if -- and probably only if -- you've heard of Turing.)
Now, I didn't know there was such a prize. It turns out that it's awarded for work in any of the following three fields in which Fermat's work was important: "Statements of Variational Principles", "Foundations of Probability and Analytical Geometry", or "Number theory". From the past prize winners it looks like it's been awarded in all of those categories, although when I heard the clue it sounded like the prize was only awarded for work in the calculus of variations. (Sometime in the next couple days, the exact wording should be available at the Jeopardy! archive.)
Also, Alex Trebek pronounced the "t" in Fermat. No.
And the second-to-last clue in the same category asked for the subject in which the Turing Award is given. (I'd never heard of that award, either, but the answer is obvious if -- and probably only if -- you've heard of Turing.)
20 December 2007
A sighting of mathematics on Jeopardy!
Today's Final Jeopardy question (alas, I didn't write down the category, but it was something like "brands"):
"Each unit in this brand, introduced in 1968, is a hyperbolic paraboloid, & they fir together for perfect storage."
The answer: What are Pringles? Two of the three contestants got it right; the third answered "What is Orville popcorn?"
I wonder if the inclusion of the "hyperbolic paraboloid" makes it harder or easier. I think it made it harder for me, because I got confused and was picturing a hyperboloid of one sheet instead. Fortunately I realized that those wouldn't fit together in any reasonable way, at least if they were all the same size. I don't know the background of the contestants tonight; I suspect most people, even most people who know enough random things to appear on Jeopardy!, would probably just filter out those words, or at best replace them with "funny shape" (because they've heard of hyperbolas and parabolas). "Funny shape" is probably the right way to think of it here; the fact that Pringles are oddly shaped is a much more salient fact about them than the precise shape, unless of course you work in a Pringle factory.
(Before you ask: No, I've never tried out for Jeopardy. Yes, I probably should.)
"Each unit in this brand, introduced in 1968, is a hyperbolic paraboloid, & they fir together for perfect storage."
The answer: What are Pringles? Two of the three contestants got it right; the third answered "What is Orville popcorn?"
I wonder if the inclusion of the "hyperbolic paraboloid" makes it harder or easier. I think it made it harder for me, because I got confused and was picturing a hyperboloid of one sheet instead. Fortunately I realized that those wouldn't fit together in any reasonable way, at least if they were all the same size. I don't know the background of the contestants tonight; I suspect most people, even most people who know enough random things to appear on Jeopardy!, would probably just filter out those words, or at best replace them with "funny shape" (because they've heard of hyperbolas and parabolas). "Funny shape" is probably the right way to think of it here; the fact that Pringles are oddly shaped is a much more salient fact about them than the precise shape, unless of course you work in a Pringle factory.
(Before you ask: No, I've never tried out for Jeopardy. Yes, I probably should.)
05 October 2007
Probability on Jeopardy!
No, this isn't about the probability of winning at Jeopardy or any such thing.
On Wednesday night's Jeopardy!, there was a category "Fun With Probability". It's always funny to see math categories appear on the show, because the contestants seem to avoid them. (They shy away from science categories as well, but not to the same extent; you can fake your way through a science category by having memorized a bunch of things, but you can't do that with a math category. By the way, any puns on the word "category" I might make in this post are unintentional.)
You can see the entire game at the Jeopardy! archive. The questions were as follows:
Contestants answered "36" and "32". I think "32" was a wild guess. But "36" is actually a fairly reasonable answer here. The odds of hitting your lucky number on a single-zero wheel are, in fact, 36 to 1. A single-zero wheel has thirty-seven spots -- the zero and the numbers one through 36 -- and a bet of $1 pays $36 if your number comes up. The house edge here is just 1/37, as opposed to 2/38 on the wheels with both zero and double-zero. (Is there a form of roulette where there are no zeroes? This seems like it could exist if not played in a casino. Poker is a fair game when played socially but in casinos the house takes a percentage of each pot. Then again, I don't see people getting together socially to spin a wheel and hand each other money; poker involves infinitely more skill than roulette. I mean the word "infinitely" here literally, because roulette takes zero skill.)
One in thirty-two, of course; thirty-two is 25. (Trick question that I might give if I ever find myself teaching basic probability: the odds of getting heads on a given coin flip of a certain unfair coin are 2 to 1 against. What are the odds of getting heads five times in a row? The answer is 242 to 1 against, but I suspect a lot of people would hear "coin" and just start multiplying out the twos.)
Not really a probability question; you really just had to kind of guess your way through this one. There are three reasonable objects -- comets, meteors, and asteroids. The three contestants, in turn, guessed all three possible pairs of them; binomial coefficients in action! (There are plenty of Jeopardy! clues where there are three possible answers, and the best strategy seems to be to wait for the other two people to get the wrong answer. There are also a fairly large number where there are two possible answers; Sweden/Norway and Oxford/Cambridge seem like common pairs of this sort. At least for me.) If you know a bit of astronomy, though, you know that meteors are not that big, and hit the earth all the time in meteor showers.
A lot of Jeopardy! clues have a bunch of extraneous verbiage; the question here is really "what's the kind of allele that's not dominant?", which might be surprisingly easy for a $1600 clue. But at this point I feel obliged to mention that genetics is one of the other disciplines where probability and statistics were applied early on, or so a friend of mine tells me; I suppose this is more reputable than gambling and less boring than insurance.
(There was a video here; it was pretty clear that this was the probability of the second card being an ace, given that the first card also was an ace.) There are three aces left among the fifty-one remaining cards, so it's three in 51, which is one in 17. I was screaming at the TV -- none of the contestants got it -- but then again I do this stuff for a living, and as far as I know none of them do, so I really shouldn't be too hard on them. Incidentally, this is a "baby version" (as one of my the principle at work in card counting in blackjack; if you know the deck is rich in high cards then the dealer is more likely to bust, so you bet more.
On Wednesday night's Jeopardy!, there was a category "Fun With Probability". It's always funny to see math categories appear on the show, because the contestants seem to avoid them. (They shy away from science categories as well, but not to the same extent; you can fake your way through a science category by having memorized a bunch of things, but you can't do that with a math category. By the way, any puns on the word "category" I might make in this post are unintentional.)
You can see the entire game at the Jeopardy! archive. The questions were as follows:
$400: High rollers get to play roulette on single-zero wheels, where the chance of hitting your lucky number is 1 in this
Contestants answered "36" and "32". I think "32" was a wild guess. But "36" is actually a fairly reasonable answer here. The odds of hitting your lucky number on a single-zero wheel are, in fact, 36 to 1. A single-zero wheel has thirty-seven spots -- the zero and the numbers one through 36 -- and a bet of $1 pays $36 if your number comes up. The house edge here is just 1/37, as opposed to 2/38 on the wheels with both zero and double-zero. (Is there a form of roulette where there are no zeroes? This seems like it could exist if not played in a casino. Poker is a fair game when played socially but in casinos the house takes a percentage of each pot. Then again, I don't see people getting together socially to spin a wheel and hand each other money; poker involves infinitely more skill than roulette. I mean the word "infinitely" here literally, because roulette takes zero skill.)
$800: The chance of getting heads on any given coin flip is 1 in 2, so the chance of getting heads 5 times in a row is 1 in this
One in thirty-two, of course; thirty-two is 25. (Trick question that I might give if I ever find myself teaching basic probability: the odds of getting heads on a given coin flip of a certain unfair coin are 2 to 1 against. What are the odds of getting heads five times in a row? The answer is 242 to 1 against, but I suspect a lot of people would hear "coin" and just start multiplying out the twos.)
$1200: NASA's Spaceguard Survey watches for the "extremely small" probability of these 2 objects coming to smash Earth
Not really a probability question; you really just had to kind of guess your way through this one. There are three reasonable objects -- comets, meteors, and asteroids. The three contestants, in turn, guessed all three possible pairs of them; binomial coefficients in action! (There are plenty of Jeopardy! clues where there are three possible answers, and the best strategy seems to be to wait for the other two people to get the wrong answer. There are also a fairly large number where there are two possible answers; Sweden/Norway and Oxford/Cambridge seem like common pairs of this sort. At least for me.) If you know a bit of astronomy, though, you know that meteors are not that big, and hit the earth all the time in meteor showers.
$1600: Offspring of heterozygous parents have a 50-50 chance of getting a dominant vs. this type of allele
A lot of Jeopardy! clues have a bunch of extraneous verbiage; the question here is really "what's the kind of allele that's not dominant?", which might be surprisingly easy for a $1600 clue. But at this point I feel obliged to mention that genetics is one of the other disciplines where probability and statistics were applied early on, or so a friend of mine tells me; I suppose this is more reputable than gambling and less boring than insurance.
$2000: The probability of the first card dealt being an ace is 4 in 52, so the probability of the second card being an ace is 1 in this number
(There was a video here; it was pretty clear that this was the probability of the second card being an ace, given that the first card also was an ace.) There are three aces left among the fifty-one remaining cards, so it's three in 51, which is one in 17. I was screaming at the TV -- none of the contestants got it -- but then again I do this stuff for a living, and as far as I know none of them do, so I really shouldn't be too hard on them. Incidentally, this is a "baby version" (as one of my the principle at work in card counting in blackjack; if you know the deck is rich in high cards then the dealer is more likely to bust, so you bet more.
29 July 2007
How much is it worth to win at Jeopardy!?
Most weekday evenings at 7pm I watch the television show Jeopardy! One day I'll be on the show, if I actually get around to auditioning and then I get lucky enough to get picked. For now, I just scream "how could you not know that!" from the comfort of my living room. I bet all the contestants do that.
For those of you who aren't familiar with the game, it is a game where three players competing against each other answer trivia questions (although Jeopardy! has this silly trope where your answer has to be "in the form of a question", that's entirely irrelevant here). If they get the questions right, they gain money; if they get them wrong, they lose money. Those players who have a positive amount of money after the first sixty questions (which occur in two rounds of thirty; each round has six categories of five questions, worth varying amounts) get to participate in "Final Jeopardy!". Most players end up with a positive amount of money, since they know themselves well enough to only attempt to answer questions which they think they know the correct answer to. The players are told the category of the question; they then can wager any or all of their money that they'll get it right. Then the question is stated and they have thirty seconds to write down the answer.
Fans of the show have put together a Jeopardy! archive. There's a wagering calculator available online that takes into account common wagering strategies. But this does not take into account the following facts:
The first two here, I plan to address in a later post. It's the third one I want to talk about right now.
So first we must answer the question -- what's the probability of winning one's second game, given that you've won the first? Of winning one's third game, given that you've won the first two? It's obvious that a defending champion is probably better than the average player (because they've won at least one game), but how much better?
Fortunately, the Jeopardy! Archive can help us answer that. It would be enough to know what proportion of champions win one game, two games, three games, and so on. The archive compiles an impressive set of statistics for each season, but this is not one of them, so I have to do it myself. There's a natural cutoff point in the data -- for a long time Jeopardy forced its champions to leave after five games, but they don't do that any more, since the beginning of the 2003-04. (This led to Ken Jennings' historic 75-game run in 2004.) I originally planned to go back to the beginning of that season, but they only go as far back as the beginning of the Ken Jennings run, near the end of that season.
Since June 2, 2004, there have been:
155 one-game winners
61 two-game winners
21 three-game winners
6 four-game winner
9 five-game winners
1 six-game winner
1 eight-game winner
1 nineteen-game winner
1 74-game winner
Now, if there was no effect like the one I just mentioned, you'd expect there to be three times as many one-game winners as two-game winners, two-game winners as three-game winners, and so on. It actually seems like if there is such an effect, it's not that strong. I attribute the big peak at five games to psychology; it's probably hard to win a sixth game because you're going into some sort of "unknown" territory. Notice that only four players -- Ken Jennings, David Madden, Tom Kavanaugh, and Kevin Marshall -- have won their sixth game.
So I'll assume that there is no "memory effect" -- that if you win today, you have a one-in-three chance of winning tomorrow. This seems believable -- the categories can be very different from day to day -- but I've never seen this analysis before. (It wouldn't surprise me if other Jeopardy! hopefuls have done it, though, because they seem to be That Sort Of People.)
Thus, when wagering in Final Jeopardy, one should wager as if the prize is not just the money you're going to win -- but one and a half times that much, since you can expect to win half a time more. The average champion is a one-and-a-half game winner.
But there are two problems:
- how do you use that information? Does the amount of money one expects to win really affect proper wagering strategy?
- more importantly, you only get to play at Jeopardy! once. I think the rules say that; in any case, I've never heard of somebody who's played twice on the Alex Trebek version of the show. (There are a few cases of people who played on the Alex Trebek version and on some prior version.) So anything you say about "expected value" is meaningless! What's the point of an operation that talks about the average amount you expect to win if you don't get to play long enough for that average to take effect?
For those of you who aren't familiar with the game, it is a game where three players competing against each other answer trivia questions (although Jeopardy! has this silly trope where your answer has to be "in the form of a question", that's entirely irrelevant here). If they get the questions right, they gain money; if they get them wrong, they lose money. Those players who have a positive amount of money after the first sixty questions (which occur in two rounds of thirty; each round has six categories of five questions, worth varying amounts) get to participate in "Final Jeopardy!". Most players end up with a positive amount of money, since they know themselves well enough to only attempt to answer questions which they think they know the correct answer to. The players are told the category of the question; they then can wager any or all of their money that they'll get it right. Then the question is stated and they have thirty seconds to write down the answer.
Fans of the show have put together a Jeopardy! archive. There's a wagering calculator available online that takes into account common wagering strategies. But this does not take into account the following facts:
- only the player who wins the game gets to keep their money (probably an average of $20,000 or so); the second- and third-place players get $2,500 and $1,000 respectively. (Incidentally, the show's host, Alex Trebek, seems to refer to in-game totals as "points", which I suspect is tied to this;
- perhaps wagering strategy should depend on how likely one thinks one is to get the question right, and how likely one thinks one's opponents are
- the player who wins get to come back the next day; the others don't.
The first two here, I plan to address in a later post. It's the third one I want to talk about right now.
So first we must answer the question -- what's the probability of winning one's second game, given that you've won the first? Of winning one's third game, given that you've won the first two? It's obvious that a defending champion is probably better than the average player (because they've won at least one game), but how much better?
Fortunately, the Jeopardy! Archive can help us answer that. It would be enough to know what proportion of champions win one game, two games, three games, and so on. The archive compiles an impressive set of statistics for each season, but this is not one of them, so I have to do it myself. There's a natural cutoff point in the data -- for a long time Jeopardy forced its champions to leave after five games, but they don't do that any more, since the beginning of the 2003-04. (This led to Ken Jennings' historic 75-game run in 2004.) I originally planned to go back to the beginning of that season, but they only go as far back as the beginning of the Ken Jennings run, near the end of that season.
Since June 2, 2004, there have been:
155 one-game winners
61 two-game winners
21 three-game winners
6 four-game winner
9 five-game winners
1 six-game winner
1 eight-game winner
1 nineteen-game winner
1 74-game winner
Now, if there was no effect like the one I just mentioned, you'd expect there to be three times as many one-game winners as two-game winners, two-game winners as three-game winners, and so on. It actually seems like if there is such an effect, it's not that strong. I attribute the big peak at five games to psychology; it's probably hard to win a sixth game because you're going into some sort of "unknown" territory. Notice that only four players -- Ken Jennings, David Madden, Tom Kavanaugh, and Kevin Marshall -- have won their sixth game.
So I'll assume that there is no "memory effect" -- that if you win today, you have a one-in-three chance of winning tomorrow. This seems believable -- the categories can be very different from day to day -- but I've never seen this analysis before. (It wouldn't surprise me if other Jeopardy! hopefuls have done it, though, because they seem to be That Sort Of People.)
Thus, when wagering in Final Jeopardy, one should wager as if the prize is not just the money you're going to win -- but one and a half times that much, since you can expect to win half a time more. The average champion is a one-and-a-half game winner.
But there are two problems:
- how do you use that information? Does the amount of money one expects to win really affect proper wagering strategy?
- more importantly, you only get to play at Jeopardy! once. I think the rules say that; in any case, I've never heard of somebody who's played twice on the Alex Trebek version of the show. (There are a few cases of people who played on the Alex Trebek version and on some prior version.) So anything you say about "expected value" is meaningless! What's the point of an operation that talks about the average amount you expect to win if you don't get to play long enough for that average to take effect?
Subscribe to:
Posts (Atom)
