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J! related math question on 538

Posted: Fri Oct 13, 2017 9:18 pm
by mrkelley23
The Riddler Express asks you to calculate how much a person could win with a "perfect" game of Jeopardy.

Re: J! related math question on 538

Posted: Sat Oct 14, 2017 1:03 am
by TheConfessor
mrkelley23 wrote:The Riddler Express asks you to calculate how much a person could win with a "perfect" game of Jeopardy.
My 30 second scratching on the back of a Post It note came up with $427,200. Let's see if that's right. This seems very simple compared to some of the truly challenging problems they have presented in the past.

Not that a "perfect" game could never happen, for several reasons, but especially because there would never be three daily doubles in the top row.

EDIT:
Oh, I see they didn't post the answer and people are supposed to submit an answer. Anyway, I'm pretty sure my answer is right, so feel free to use it.

EDIT #2: My answer is way off. I'm not sure what I did wrong, but when I used a calculator tonight, I got a different answer. I still think it was a very easy question for anyone using basic logic and maybe a flow chart. And a calculator.

Re: J! related math question on 538

Posted: Sat Oct 14, 2017 6:34 am
by mrkelley23
TheConfessor wrote:
mrkelley23 wrote:The Riddler Express asks you to calculate how much a person could win with a "perfect" game of Jeopardy.
My 30 second scratching on the back of a Post It note came up with $427,200. Let's see if that's right. This seems very simple compared to some of the truly challenging problems they have presented in the past.

Not that a "perfect" game could never happen, for several reasons, but especially because there would never be three daily doubles in the top row.

EDIT:
Oh, I see they didn't post the answer and people are supposed to submit an answer. Anyway, I'm pretty sure my answer is right, so feel free to use it.
I came up with a different answer, and so, I found out later, did the folks at The Mary Sue. But I won't print it here yet, in case others want to try. I'm pretty sure yours is not correct.

Re: J! related math question on 538

Posted: Sat Oct 14, 2017 6:57 am
by bazodee
I got:

Spoiler
$35,600 after Single Jeopardy; $283,200 after Double Jeopardy; and $566,400 after Final Jeopardy.

Re: J! related math question on 538

Posted: Sat Oct 14, 2017 7:00 am
by silverscreenselect
Since a "perfect game" would mean that the other two players did not answer a question correctly, would they even play Final Jeopardy if only one player was in the plus column?

Re: J! related math question on 538

Posted: Sat Oct 14, 2017 7:37 am
by mrkelley23
silverscreenselect wrote:Since a "perfect game" would mean that the other two players did not answer a question correctly, would they even play Final Jeopardy if only one player was in the plus column?
Ooh! Good question! I did not think of that one.

Re: J! related math question on 538

Posted: Sat Oct 14, 2017 10:07 am
by triviawayne
mrkelley23 wrote:
silverscreenselect wrote:Since a "perfect game" would mean that the other two players did not answer a question correctly, would they even play Final Jeopardy if only one player was in the plus column?
Ooh! Good question! I did not think of that one.
They would not. I think it was just last season when there was only one player with money after double jeopardy and they played a really awkward final

Re: J! related math question on 538

Posted: Sat Oct 14, 2017 11:32 am
by Bob78164
mrkelley23 wrote:
silverscreenselect wrote:Since a "perfect game" would mean that the other two players did not answer a question correctly, would they even play Final Jeopardy if only one player was in the plus column?
Ooh! Good question! I did not think of that one.
They would and they did. It happened during the Ultimate Tournament of Champions. I happened to be in the audience, waiting to watch Leszek's first-round match. --Bob

Re: J! related math question on 538

Posted: Mon Oct 16, 2017 12:36 am
by TheConfessor
bazodee wrote:I got:

Spoiler
$35,600 after Single Jeopardy; $283,200 after Double Jeopardy; and $566,400 after Final Jeopardy.
I agree with your answer. I'm not sure what I did wrong before, other than try to do the arithmetic by hand.

Re: J! related math question on 538

Posted: Mon Oct 16, 2017 7:55 am
by BackInTex
That's a poor question. Are they asking for the maximum score or the perfect game? The perfect game would not result in the maximum score.

Re: J! related math question on 538

Posted: Mon Oct 16, 2017 8:20 am
by littlebeast13
BackInTex wrote:The perfect game would not result in the maximum score.

Please elaborate...

lb13

Re: J! related math question on 538

Posted: Mon Oct 16, 2017 5:30 pm
by K.P.
littlebeast13 wrote:
BackInTex wrote:The perfect game would not result in the maximum score.
Please elaborate...

lb13
Not sure if this is what he's getting at, but the maximum score can only be achieved by pulling multiple Clavins (overwagering at the risk of throwing away the win). Such a strategic faux pas has no place in any "perfect game".

Re: J! related math question on 538

Posted: Mon Oct 16, 2017 6:48 pm
by TheConfessor
K.P. wrote:
littlebeast13 wrote:
BackInTex wrote:The perfect game would not result in the maximum score.
Please elaborate...

lb13
Not sure if this is what he's getting at, but the maximum score can only be achieved by pulling multiple Clavins (overwagering at the risk of throwing away the win). Such a strategic faux pas has no place in any "perfect game".
That is obviously true, if "perfect" means "optimally played." The most important goal in Jeopardy! is to win each game and survive to the next game, so no competent player would ever choose to play a "perfect" game if that means playing for the theoretical maximum score. It all depends on how you define "perfect." You could say that no baseball pitcher has ever pitched a perfect game, if you define "perfect" as throwing only 27 pitches.

Re: J! related math question on 538

Posted: Mon Oct 16, 2017 6:56 pm
by Bob78164
mrkelley23 wrote:The Riddler Express asks you to calculate how much a person could win with a "perfect" game of Jeopardy.
To address some of the issues relating to the description of a game as "perfect," the column is more precise, asking for the maximum amount of money a person can win in a single game in (regular) Jeopardy!. "Perfect" was mrkelley23's paraphrase of the question. --Bob

Re: J! related math question on 538

Posted: Mon Oct 16, 2017 9:11 pm
by mrkelley23
Bob78164 wrote:
mrkelley23 wrote:The Riddler Express asks you to calculate how much a person could win with a "perfect" game of Jeopardy.
To address some of the issues relating to the description of a game as "perfect," the column is more precise, asking for the maximum amount of money a person can win in a single game in (regular) Jeopardy!. "Perfect" was mrkelley23's paraphrase of the question. --Bob
Bob apparently doesn't look at the headline on the article, which says, "How Much is a Perfect Game of 'Jeopardy!' Worth?" So it may be a paraphrase, but it ain't mine.

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 6:50 am
by littlebeast13
TheConfessor wrote:That is obviously true, if "perfect" means "optimally played." The most important goal in Jeopardy! is to win each game and survive to the next game, so no competent player would ever choose to play a "perfect" game if that means playing for the theoretical maximum score. It all depends on how you define "perfect." You could say that no baseball pitcher has ever pitched a perfect game, if you define "perfect" as throwing only 27 pitches.

What is so perfect about throwing 27 pitches? That's 27 more pitches than would be necessary to record a 9 inning no hit victory. Sure, there'd be 27 baserunners from the intentional walks who would all also get picked off, but since you seem to be singling out efficiency for some reason, why not go for the most efficient complete game that could possibly be "pitched?"

Nobody but Jeopardy enthusiasts and aspiring Jeopardy contestants would (or should) give a possum's patoot about the most "optimally played" game, nor would that come to the mind of many in a theoretical question of how much one would win in a "perfect" game of Jeopardy. I'm still curious if this was BiT's reasoning for the statement about perfect not equaling maximum I originally responded to...

lb13

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 7:24 am
by BackInTex
littlebeast13 wrote: Nobody but Jeopardy enthusiasts and aspiring Jeopardy contestants would (or should) give a possum's patoot about the most "optimally played" game, nor would that come to the mind of many in a theoretical question of how much one would win in a "perfect" game of Jeopardy. I'm still curious if this was BiT's reasoning for the statement about perfect not equaling maximum I originally responded to...

lb13
Yes. Perfect means without flaw. Wagering your entire bank 4 times is flawed play, in my opinion.

But perfection is like beauty, in the eye of the beholder. That is because calling something a flaw is subjective. If the intent is not to only win, but to do so only with maximum earnings, then wagering your entire bank is not necessarily a flaw; foolish maybe, but not a flaw.

Such as your baseball example. A pitcher could have a "perfect" game with 27 pitches, but that means the batters hit every pitch he threw. Is that perfect? Was it the pitcher's intent to allow the batters to hit the balls? The pitcher then is having to rely on his teammates on every pitch. Or is it more "perfect" to record 27 strikeouts on 81 pitches?

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 10:35 am
by mrkelley23
BackInTex wrote:
littlebeast13 wrote: Nobody but Jeopardy enthusiasts and aspiring Jeopardy contestants would (or should) give a possum's patoot about the most "optimally played" game, nor would that come to the mind of many in a theoretical question of how much one would win in a "perfect" game of Jeopardy. I'm still curious if this was BiT's reasoning for the statement about perfect not equaling maximum I originally responded to...

lb13
Yes. Perfect means without flaw. Wagering your entire bank 4 times is flawed play, in my opinion.

But perfection is like beauty, in the eye of the beholder. That is because calling something a flaw is subjective. If the intent is not to only win, but to do so only with maximum earnings, then wagering your entire bank is not necessarily a flaw; foolish maybe, but not a flaw.

Such as your baseball example. A pitcher could have a "perfect" game with 27 pitches, but that means the batters hit every pitch he threw. Is that perfect? Was it the pitcher's intent to allow the batters to hit the balls? The pitcher then is having to rely on his teammates on every pitch. Or is it more "perfect" to record 27 strikeouts on 81 pitches?
Unlike our current president, BiT and I use quotation marks with intention.

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 10:47 am
by frogman042
BackInTex wrote:
littlebeast13 wrote: Nobody but Jeopardy enthusiasts and aspiring Jeopardy contestants would (or should) give a possum's patoot about the most "optimally played" game, nor would that come to the mind of many in a theoretical question of how much one would win in a "perfect" game of Jeopardy. I'm still curious if this was BiT's reasoning for the statement about perfect not equaling maximum I originally responded to...

lb13
Yes. Perfect means without flaw. Wagering your entire bank 4 times is flawed play, in my opinion.

But perfection is like beauty, in the eye of the beholder. That is because calling something a flaw is subjective. If the intent is not to only win, but to do so only with maximum earnings, then wagering your entire bank is not necessarily a flaw; foolish maybe, but not a flaw.

Such as your baseball example. A pitcher could have a "perfect" game with 27 pitches, but that means the batters hit every pitch he threw. Is that perfect? Was it the pitcher's intent to allow the batters to hit the balls? The pitcher then is having to rely on his teammates on every pitch. Or is it more "perfect" to record 27 strikeouts on 81 pitches?
Imagine the case where the other 2 players consistently gave the wrong responses to every question first - so when the 3 DD's came up as the last clues - the opponents were all in way negative territory - you might argue that the last 2 DDs they should wager all but a $1 each time and the same for FJ to make sure that they don't end up either not playing FJ or not 'winning' - so maybe the final answer should be $7 lower, but since (in this case) the opponents were way negative, there is no reason not to bet all (or all but $1) without putting yourself at the risk of not winning. They don't even have to have missed every clue, just some and both opponents would be negative and then you would be 'safe' to make it a true DD or a true DD - $1 without risking a loss.

Now, I've never seen a DD on the top row - has anyone? I do recall it being in the 2nd row, not often but occasionally - does anyone know how the DD clues are picked - randomly (truly random or some weighted random) or do they have some rule regarding never in the top row, never more than 1 in the 2nd row, etc. If such rules exists, then the score calculated for this problem would be wrong - we would need to know the DD placement rules in order to have a more accurate answer.

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 10:53 am
by bazodee
frogman042 wrote:
BackInTex wrote:
littlebeast13 wrote: Nobody but Jeopardy enthusiasts and aspiring Jeopardy contestants would (or should) give a possum's patoot about the most "optimally played" game, nor would that come to the mind of many in a theoretical question of how much one would win in a "perfect" game of Jeopardy. I'm still curious if this was BiT's reasoning for the statement about perfect not equaling maximum I originally responded to...

lb13
Yes. Perfect means without flaw. Wagering your entire bank 4 times is flawed play, in my opinion.

But perfection is like beauty, in the eye of the beholder. That is because calling something a flaw is subjective. If the intent is not to only win, but to do so only with maximum earnings, then wagering your entire bank is not necessarily a flaw; foolish maybe, but not a flaw.

Such as your baseball example. A pitcher could have a "perfect" game with 27 pitches, but that means the batters hit every pitch he threw. Is that perfect? Was it the pitcher's intent to allow the batters to hit the balls? The pitcher then is having to rely on his teammates on every pitch. Or is it more "perfect" to record 27 strikeouts on 81 pitches?
Imagine the case where the other 2 players consistently gave the wrong responses to every question first - so when the 3 DD's came up as the last clues - the opponents were all in way negative territory - you might argue that the last 2 DDs they should wager all but a $1 each time and the same for FJ to make sure that they don't end up either not playing FJ or not 'winning' - so maybe the final answer should be $7 lower, but since (in this case) the opponents were way negative, there is no reason not to bet all (or all but $1) without putting yourself at the risk of not winning. They don't even have to have missed every clue, just some and both opponents would be negative and then you would be 'safe' to make it a true DD or a true DD - $1 without risking a loss.

Now, I've never seen a DD on the top row - has anyone? I do recall it being in the 2nd row, not often but occasionally - does anyone know how the DD clues are picked - randomly (truly random or some weighted random) or do they have some rule regarding never in the top row, never more than 1 in the 2nd row, etc. If such rules exists, then the score calculated for this problem would be wrong - we would need to know the DD placement rules in order to have a more accurate answer.
Yes, it is occasionally in the first row, once or twice a season.

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 12:04 pm
by Bob78164
frogman042 wrote:
BackInTex wrote:
littlebeast13 wrote: Nobody but Jeopardy enthusiasts and aspiring Jeopardy contestants would (or should) give a possum's patoot about the most "optimally played" game, nor would that come to the mind of many in a theoretical question of how much one would win in a "perfect" game of Jeopardy. I'm still curious if this was BiT's reasoning for the statement about perfect not equaling maximum I originally responded to...

lb13
Yes. Perfect means without flaw. Wagering your entire bank 4 times is flawed play, in my opinion.

But perfection is like beauty, in the eye of the beholder. That is because calling something a flaw is subjective. If the intent is not to only win, but to do so only with maximum earnings, then wagering your entire bank is not necessarily a flaw; foolish maybe, but not a flaw.

Such as your baseball example. A pitcher could have a "perfect" game with 27 pitches, but that means the batters hit every pitch he threw. Is that perfect? Was it the pitcher's intent to allow the batters to hit the balls? The pitcher then is having to rely on his teammates on every pitch. Or is it more "perfect" to record 27 strikeouts on 81 pitches?
Imagine the case where the other 2 players consistently gave the wrong responses to every question first - so when the 3 DD's came up as the last clues - the opponents were all in way negative territory - you might argue that the last 2 DDs they should wager all but a $1 each time and the same for FJ to make sure that they don't end up either not playing FJ or not 'winning' - so maybe the final answer should be $7 lower, but since (in this case) the opponents were way negative, there is no reason not to bet all (or all but $1) without putting yourself at the risk of not winning. They don't even have to have missed every clue, just some and both opponents would be negative and then you would be 'safe' to make it a true DD or a true DD - $1 without risking a loss.

Now, I've never seen a DD on the top row - has anyone? I do recall it being in the 2nd row, not often but occasionally - does anyone know how the DD clues are picked - randomly (truly random or some weighted random) or do they have some rule regarding never in the top row, never more than 1 in the 2nd row, etc. If such rules exists, then the score calculated for this problem would be wrong - we would need to know the DD placement rules in order to have a more accurate answer.
There are two rules relevant to your analysis. First $0 is never a winning score. Second, the minimum bet for a Daily Double is $5. So if this were an issue, you'd need to leave yourself at least $6 after the first Daily Double of Double Jeopardy! to ensure that you aren't risking a $0 score after the second, and of course you'd save a dollar after Final.

But the question as stated in the column simply asks for the maximum possible score, so none of these considerations are relevant. --Bob

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 12:09 pm
by BackInTex
frogman042 wrote: Imagine the case where the other 2 players consistently gave the wrong responses to every question first
The time would likely expire long before the all clues could be played.

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 12:29 pm
by TheConfessor
frogman042 wrote: Imagine the case where the other 2 players consistently gave the wrong responses to every question first - so when the 3 DD's came up as the last clues - the opponents were all in way negative territory - you might argue that the last 2 DDs they should wager all but a $1 each time and the same for FJ to make sure that they don't end up either not playing FJ or not 'winning' - so maybe the final answer should be $7 lower, but since (in this case) the opponents were way negative, there is no reason not to bet all (or all but $1) without putting yourself at the risk of not winning. They don't even have to have missed every clue, just some and both opponents would be negative and then you would be 'safe' to make it a true DD or a true DD - $1 without risking a loss.

Now, I've never seen a DD on the top row - has anyone? I do recall it being in the 2nd row, not often but occasionally - does anyone know how the DD clues are picked - randomly (truly random or some weighted random) or do they have some rule regarding never in the top row, never more than 1 in the 2nd row, etc. If such rules exists, then the score calculated for this problem would be wrong - we would need to know the DD placement rules in order to have a more accurate answer.
Occasionally there are categories where it would be safe to bit it all on a DD, instead of all but a dollar. Like if the category is "ANIMAL, VEGETABLE, OR MINERAL" and the clue is "POTATO" and the other two players have already guessed animal and mineral. You'd still want to keep a dollar safe on your FJ wager to preserve the guaranteed win.
EDIT: As BiT pointed out, this makes no sense. The other two players would not have a chance to be wrong on the leader's DD clue. Move along, nothing to see here.

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 12:46 pm
by TheConfessor
littlebeast13 wrote:
TheConfessor wrote:That is obviously true, if "perfect" means "optimally played." The most important goal in Jeopardy! is to win each game and survive to the next game, so no competent player would ever choose to play a "perfect" game if that means playing for the theoretical maximum score. It all depends on how you define "perfect." You could say that no baseball pitcher has ever pitched a perfect game, if you define "perfect" as throwing only 27 pitches.

What is so perfect about throwing 27 pitches? That's 27 more pitches than would be necessary to record a 9 inning no hit victory. Sure, there'd be 27 baserunners from the intentional walks who would all also get picked off, but since you seem to be singling out efficiency for some reason, why not go for the most efficient complete game that could possibly be "pitched?"

Nobody but Jeopardy enthusiasts and aspiring Jeopardy contestants would (or should) give a possum's patoot about the most "optimally played" game, nor would that come to the mind of many in a theoretical question of how much one would win in a "perfect" game of Jeopardy. I'm still curious if this was BiT's reasoning for the statement about perfect not equaling maximum I originally responded to...

lb13
I chose to posit "27 pitches" as a reasonable definition of a "perfect game" in baseball mostly to give you the opportunity to bring up situations where an out can be recorded without a bitch being thrown. I'm not an expert on such obscure rules. I can't recall the last time I watched an entire baseball game, so I'm out of the loop. Are you saying that the pitcher no longer has to throw four pitches to get an intentional walk? Can the pitcher just signal that he wants the batter to go to first base at any time, without throwing a pitch, or regardless of the current ball-strike count? I knew this had been proposed, but it offended baseball purists. Is that currently the rule in MLB, and when did that happen?

While we're at it, what would be a "perfect game" in Who Wants To Be A Millionaire? Winning a million without using a lifeline? Are there other plausible answers? Has that ever been done? If I ever get a chance to play again (which I won't), I would never answer the last question until I had used all of my lifelines.

Re: J! related math question on 538

Posted: Tue Oct 17, 2017 12:57 pm
by Bob78164
TheConfessor wrote:I chose to posit "27 pitches" as a reasonable definition of a "perfect game" in baseball mostly to give you the opportunity to bring up situations where an out can be recorded without a bitch being thrown. I'm not an expert on such obscure rules. I can't recall the last time I watched an entire baseball game, so I'm out of the loop. Are you saying that the pitcher no longer has to throw four pitches to get an intentional walk? Can the pitcher just signal that he wants the batter to go to first base at any time, without throwing a pitch, or regardless of the current ball-strike count? I knew this had been proposed, but it offended baseball purists. Is that currently the rule in MLB, and when did that happen?
The manager gives the signal, not the pitcher. The rule took effect for the 2017 season. I still don't like it. --Bob