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jeff6286 wrote:

SkoolRN wrote: Re #1. A staff member also walks behind the podium and writes down each person's wager on a very high-tech pad of paper on a clipboard. I assumed that was in case of a malfunction of the electronics.

If it's being done as a backup in case of an electronic malfunction, wouldn't they want to use an extremely low-tech pad of paper?

I've found the high tech paper to be more reliable than the low tech variety, but since I think this is a write once/read many situation, both are adequate.

ACW wrote:1: For FJ, how do they make sure the contestants don't change their bids?

2: Is there a screen to see the other scores when wagering for DDs or FJ?

3: For the tourneys, how is it decided who selects first?

Thanks

The title of this thread has been bothering me for two weeks now. Wouldn't "3 utterly random questions" be something like:

1. What will be the distance from Earth to Ceres, in cubits, on March 2, 2012, at 2:12:30 GMT?

2. How did Garth McKenzie (1687-1729) lose his left ring finger?

3. Where did I leave my car keys?

Rex

Rex Kramer wrote:

ACW wrote:1: For FJ, how do they make sure the contestants don't change their bids?

2: Is there a screen to see the other scores when wagering for DDs or FJ?

3: For the tourneys, how is it decided who selects first?

Thanks

The title of this thread has been bothering me for two weeks now. Wouldn't "3 utterly random questions" be something like:

1. What will be the distance from Earth to Ceres, in cubits, on March 2, 2012, at 2:12:30 GMT?

2. How did Garth McKenzie (1687-1729) lose his left ring finger?

3. Where did I leave my car keys?

Rex

1. 42.

2. He never had one.

3. See answer to question 1.

Rex Kramer wrote:1. What will be the distance from Earth to Ceres, in cubits, on March 2, 2012, at 2:12:30 GMT?

Rex

This gives me the chance to share a story. A couple weeks ago the trivia team of which I'm a member tied for third place in a local contest. There was also a tie for first place, and the moderator announced that the tie would be broken by asking one question of the two teams. The first and second place teams were placed at separate tables and asked to name the 13 original U.S. colonies. When that didn't break the tie, the moderator decided that the winner would be the first team to bring him the answer to this question: "What is the Roman numeral equivalent of 1980?"

After that tiebreaker was resolved it was our turn. The moderator again said the winning team would be the first to bring him the correct answer to this question: "Given that the moon is, on average, 238,857 miles from the Earth, what is the distance in centimeters?" I grabbed a piece of paper and began furiously scribbling down the numbers and doing the multiplication (I found out later that the other team, mostly younger people, spent two minutes or so trying to figure out how to approach the question). With another team member checking my math as I went, I came up with the answer. It seemed like forever but I was told later it was only about 2 minutes. Fortunately I was right the first time and we claimed the 3rd place prize.

Spoiler

My process was:

2.54 (centimeters in an inch) X 12 (inches in a foot) X 5280 (feet in a mile) X 238,857 miles

Afterward I thought that it might have been easier to just multiply 238,857 X 1.609 (kilometers in a mile) and then move the decimal the appropriate number of places, but it turns out that I would have been off by a few digits since both 2.54 and 1.609 are approximations.

earendel wrote:

Rex Kramer wrote:1. What will be the distance from Earth to Ceres, in cubits, on March 2, 2012, at 2:12:30 GMT?

Rex

Spoiler

My process was:

2.54 (centimeters in an inch) X 12 (inches in a foot) X 5280 (feet in a mile) X 238,857 miles

Afterward I thought that it might have been easier to just multiply 238,857 X 1.609 (kilometers in a mile) and then move the decimal the appropriate number of places, but it turns out that I would have been off by a few digits since both 2.54 and 1.609 are approximations.

Spoiler

2.54 cm in an inch is not an approximation, it's codified to be that.

http://en.wikipedia.org/wiki/Inch#cite_note-5 ... phpbb ate the original link, so here's a link to the link so you can click while you click.

earendel wrote:
After that tiebreaker was resolved it was our turn. The moderator again said the winning team would be the first to bring him the correct answer to this question: "Given that the moon is, on average, 238,857 miles from the Earth, what is the distance in centimeters?" I grabbed a piece of paper and began furiously scribbling down the numbers and doing the multiplication (I found out later that the other team, mostly younger people, spent two minutes or so trying to figure out how to approach the question). With another team member checking my math as I went, I came up with the answer. It seemed like forever but I was told later it was only about 2 minutes. Fortunately I was right the first time and we claimed the 3rd place prize.

Spoiler

My process was:

2.54 (centimeters in an inch) X 12 (inches in a foot) X 5280 (feet in a mile) X 238,857 miles

Afterward I thought that it might have been easier to just multiply 238,857 X 1.609 (kilometers in a mile) and then move the decimal the appropriate number of places, but it turns out that I would have been off by a few digits since both 2.54 and 1.609 are approximations.

You forgot to give us the answer! Unless of course you don't remember it off the top of your head. That is quite a drastically different tiebreaker than naming the 13 colonies or 1980 in Roman numerals.

Spoiler

I came up with 38,440,307,980.8 cm. I don't know if it's exactly right, but at least the first 8 or 9 digits are, as I can get the answer in scientific notation using google search (or most likely, a calculator) but I'm not sure if I know of a way to get the answer to display all 12 significant digits.

(uh oh, you awoke the science teacher.) while i suspect that the good folks at JPL actually do know the average earth-moon distance to 12 sig figs, it wasn't given in the problem, and there is no reason to believe it is 238,857.000000 miles. in fact, it's vanishingly unlikely. so doing the problem on a regular calculator (or google) will, in fact, get you the right answer. (off the top of my head, i don't see how rounding to 6 sig figs would help you do it faster by hand.)

jeff6286 wrote:

SkoolRN wrote: Re #1. A staff member also walks behind the podium and writes down each person's wager on a very high-tech pad of paper on a clipboard. I assumed that was in case of a malfunction of the electronics.

If it's being done as a backup in case of an electronic malfunction, wouldn't they want to use an extremely low-tech pad of paper?

You couldn't see my air quotes as I typed.

SkoolRN wrote:You couldn't see my air quotes as I typed.

I can do air quotes.

I can type.

I can't do both at the same time.

Brian

jeff6286 wrote:

earendel wrote:
After that tiebreaker was resolved it was our turn. The moderator again said the winning team would be the first to bring him the correct answer to this question: "Given that the moon is, on average, 238,857 miles from the Earth, what is the distance in centimeters?" I grabbed a piece of paper and began furiously scribbling down the numbers and doing the multiplication (I found out later that the other team, mostly younger people, spent two minutes or so trying to figure out how to approach the question). With another team member checking my math as I went, I came up with the answer. It seemed like forever but I was told later it was only about 2 minutes. Fortunately I was right the first time and we claimed the 3rd place prize.

Spoiler

My process was:

2.54 (centimeters in an inch) X 12 (inches in a foot) X 5280 (feet in a mile) X 238,857 miles

Afterward I thought that it might have been easier to just multiply 238,857 X 1.609 (kilometers in a mile) and then move the decimal the appropriate number of places, but it turns out that I would have been off by a few digits since both 2.54 and 1.609 are approximations.

You forgot to give us the answer! Unless of course you don't remember it off the top of your head. That is quite a drastically different tiebreaker than naming the 13 colonies or 1980 in Roman numerals.

Spoiler

I came up with 38,440,307,980.8 cm. I don't know if it's exactly right, but at least the first 8 or 9 digits are, as I can get the answer in scientific notation using google search (or most likely, a calculator) but I'm not sure if I know of a way to get the answer to display all 12 significant digits.

If you do it the old-fashioned way, with pen and paper, you get this:

Spoiler

38,440,417,980.8

I still have the piece of paper that I wrote the numbers on.

jpahk wrote:(uh oh, you awoke the science teacher.) while i suspect that the good folks at JPL actually do know the average earth-moon distance to 12 sig figs, it wasn't given in the problem, and there is no reason to believe it is 238,857.000000 miles. in fact, it's vanishingly unlikely. so doing the problem on a regular calculator (or google) will, in fact, get you the right answer. (off the top of my head, i don't see how rounding to 6 sig figs would help you do it faster by hand.)

Well, aside from the fact that calculators weren't allowed, the figure that I gave was the one that the moderator had (I don't know whether he Googled it or had done the math himself some other time). It seemed more than a bit ridiculous that this would be a tie-breaker, given the simplicity of the first-place tiebreaker.

Yeah, but all you had to know was 2.54 and how to multiply. That's not a high hurdle, either.

earendel wrote:

jpahk wrote:(uh oh, you awoke the science teacher.) while i suspect that the good folks at JPL actually do know the average earth-moon distance to 12 sig figs, it wasn't given in the problem, and there is no reason to believe it is 238,857.000000 miles. in fact, it's vanishingly unlikely. so doing the problem on a regular calculator (or google) will, in fact, get you the right answer. (off the top of my head, i don't see how rounding to 6 sig figs would help you do it faster by hand.)

Well, aside from the fact that calculators weren't allowed, the figure that I gave was the one that the moderator had (I don't know whether he Googled it or had done the math himself some other time). It seemed more than a bit ridiculous that this would be a tie-breaker, given the simplicity of the first-place tiebreaker.

My objection is that that isn't a trivia-based tiebreaker at all, but just a speed math exercise (unless one team is too dim to know how to solve it). Given that I'm prone to making careless arithmetic errors, I'd be pissed as hell if that's the tiebreaker they threw at me. I'd probably have retorted by asking to how many sig-figs they wanted it and just approximating the answer (i.e., 2.4E5 x 1.6E5 = 3.84E10)

earendel wrote:

jeff6286 wrote:

earendel wrote:
After that tiebreaker was resolved it was our turn. The moderator again said the winning team would be the first to bring him the correct answer to this question: "Given that the moon is, on average, 238,857 miles from the Earth, what is the distance in centimeters?" I grabbed a piece of paper and began furiously scribbling down the numbers and doing the multiplication (I found out later that the other team, mostly younger people, spent two minutes or so trying to figure out how to approach the question). With another team member checking my math as I went, I came up with the answer. It seemed like forever but I was told later it was only about 2 minutes. Fortunately I was right the first time and we claimed the 3rd place prize.

Spoiler

My process was:

2.54 (centimeters in an inch) X 12 (inches in a foot) X 5280 (feet in a mile) X 238,857 miles

Afterward I thought that it might have been easier to just multiply 238,857 X 1.609 (kilometers in a mile) and then move the decimal the appropriate number of places, but it turns out that I would have been off by a few digits since both 2.54 and 1.609 are approximations.

You forgot to give us the answer! Unless of course you don't remember it off the top of your head. That is quite a drastically different tiebreaker than naming the 13 colonies or 1980 in Roman numerals.

Spoiler

I came up with 38,440,307,980.8 cm. I don't know if it's exactly right, but at least the first 8 or 9 digits are, as I can get the answer in scientific notation using google search (or most likely, a calculator) but I'm not sure if I know of a way to get the answer to display all 12 significant digits.

If you do it the old-fashioned way, with pen and paper, you get this:

Spoiler

38,440,417,980.8

I still have the piece of paper that I wrote the numbers on.

Well according to my calculations, (and Google's) you're off by like a million centimeters. https://www.google.com/webhp?sourceid=c ... .osb&cad=b

So if that answer won you the tiebreaker, then I guess Woof had the right idea by going with an estimated answer and not bothering with all that messy arithmetic.

Volante wrote:

Spoiler

2.54 cm in an inch is not an approximation, it's codified to be that.

http://en.wikipedia.org/wiki/Inch#cite_note-5 ... phpbb ate the original link, so here's a link to the link so you can click while you click.

Spoiler

OMG. It's codified to be *exactly* 2.54? Why does that make me think that sometime in my lifetime they'll be codifying pi to be exactly 3.14? (Why change the measurement of an inch???)

dhkendall wrote:

Volante wrote:

Spoiler

2.54 cm in an inch is not an approximation, it's codified to be that.

http://en.wikipedia.org/wiki/Inch#cite_note-5 ... phpbb ate the original link, so here's a link to the link so you can click while you click.

Spoiler

OMG. It's codified to be *exactly* 2.54? Why does that make me think that sometime in my lifetime they'll be codifying pi to be exactly 3.14? (Why change the measurement of an inch???)

Unsure if clicked all the way through...

The (international) inch has been exactly 25.4 mm since July 1959. At this point in time the (international) yard was redefined as 0.9144 metre- until this time the ratio between the US yard and the metre was different to the ratio between the UK yard and the metre. For more information, see Engineering Metrology by K J Hume (2 ed) Macdonald London 1967. The American inch changed by 2 millionths of an inch and the UK inch by 1.7 millionths of an inch. The international inch falls mid way between the old UK and US inch.

Thankfully pi is a ratio, not an arbitrary unit where the value is decided by the majority.

Volante wrote:Thankfully pi is a ratio, not an arbitrary unit where the value is decided by the majority.

Don't be so hasty:

Did a state legislature once pass a law saying pi equals 3?

alietr wrote:

Volante wrote:Thankfully pi is a ratio, not an arbitrary unit where the value is decided by the majority.

Don't be so hasty:

Did a state legislature once pass a law saying pi equals 3?

I see your Cecil and raise you:
http://io9.com/5880792/the-eccentric-cr ... alue-of-pi

alietr wrote:Yeah, but all you had to know was 2.54 and how to multiply. That's not a high hurdle, either.

You also had to know how many feet in a mile. I think the problem for the other team was that they were mostly twenty-somethings and may not have known how to do math without a calculator. I'm told they spent two minutes figuring out how to solve the problem.

Woof wrote:

earendel wrote:

jpahk wrote:(uh oh, you awoke the science teacher.) while i suspect that the good folks at JPL actually do know the average earth-moon distance to 12 sig figs, it wasn't given in the problem, and there is no reason to believe it is 238,857.000000 miles. in fact, it's vanishingly unlikely. so doing the problem on a regular calculator (or google) will, in fact, get you the right answer. (off the top of my head, i don't see how rounding to 6 sig figs would help you do it faster by hand.)

Well, aside from the fact that calculators weren't allowed, the figure that I gave was the one that the moderator had (I don't know whether he Googled it or had done the math himself some other time). It seemed more than a bit ridiculous that this would be a tie-breaker, given the simplicity of the first-place tiebreaker.

My objection is that that isn't a trivia-based tiebreaker at all, but just a speed math exercise (unless one team is too dim to know how to solve it). Given that I'm prone to making careless arithmetic errors, I'd be pissed as hell if that's the tiebreaker they threw at me. I'd probably have retorted by asking to how many sig-figs they wanted it and just approximating the answer (i.e., 2.4E5 x 1.6E5 = 3.84E10)

I agree that this really isn't trivia, but that's what we were given. There were a lot of exclamations of surprise when the question was read.