3 Kings

Cidu Bill on Jan 6th 2013

3kings.gif

Filed in Bill Bickel, Bravity, CIDU, Guy & Rodd & Dan, comic strips, comics, humor | 35 responses so far

35 Responses to “3 Kings”

  1. Kamino Neko Jan 6th 2013 at 11:06 am 1

    They’re three wise men, so they’re equally well matched.

    (Of course, that ignores the whole ringing in part…)

    (Also, I’m pretty sure that that number is impossible for a one-day total, even if one person gets all the questions.)

    (And I really, really like parentheses.)

  2. Elyrest Jan 6th 2013 at 11:52 am 2

    Like Kamino Neko I’m pretty sure that that number is impossible for a one-day total, but is there some significance to the number 1123581?

  3. That's Me Jan 6th 2013 at 12:01 pm 3

    It’s the first seven digits of the Fibonacci series. (The 1 on the end would be the first half of 13.) What that has to do with wise men/Kings is beyond me, though.

  4. Chakolate Jan 6th 2013 at 12:19 pm 4

    Nice catch, That’s Me. I didn’t see that at all.

  5. Elyrest Jan 6th 2013 at 12:24 pm 5

    That’s Me - Thank you! I just knew that someone here would be able to answer that question. I’m familiar with the Fibonacci series, but it’s from high school. Fibonacci is a wonderful word to say. Doesn’t help me understand the comic, but I feel better.

  6. Elyrest Jan 6th 2013 at 12:33 pm 6

    Bill - Did you post this comic today because it’s the Feast of the Epiphany - in other words Three Kings Day?

  7. DemetriosX Jan 6th 2013 at 01:40 pm 7

    @1 Kamino Neko
    That’s definitely an impossible one-day total. Not counting the Daily Doubles, there’s only $ 27,000 combined on the two boards. I think the theoretical maximum is around $ 250,000 to $ 300,00. Not only that, they all made some really freaky bets to come up with the numbers in the comic.

  8. Treesong Jan 6th 2013 at 02:38 pm 8

    Since Alex says ‘again’, presumably this is not a one-day total.
    The maximum possible one-day total is $566,400 but the maximum total for a three-way tie would be much lower (and very hard to calculate), so they’ve probably been at this for months.

  9. Cidu Bill Jan 6th 2013 at 02:48 pm 9

    Or twelve nights, anyway.

  10. Elyrest Jan 6th 2013 at 02:52 pm 10

    “Or twelve nights, anyway.”

    Perfecto!

  11. Kamino Neko Jan 6th 2013 at 03:05 pm 11

    Since Alex says ‘again’, presumably this is not a one-day total.

    That’s actually what leads to the conclusion that it’s a one-day total.

    That’s what he’d say right at the end of the game, just after Final Jeopardy! The cumulative totals are only shown at the beginning of the show and after the winner (or winners, in the case of ties) have already been announced.

  12. Cidu Bill Jan 6th 2013 at 03:16 pm 12

    Yes I did, Elyrest.

  13. Cidu Bill Jan 6th 2013 at 03:18 pm 13

    Or, Kamino, it could mean they ended up tied one day so they all had to come back the next day, and they continue to be cumulatively tied at the end of each day.

  14. Kamino Neko Jan 6th 2013 at 03:21 pm 14

    Also, the use of the Fibonacci sequence just makes it weird that they went so high.

    112,358 would be a completely reasonable total for a 3 way tie (although they’d have had to bet strangely on Daily Doubles and/or Final Jeopardy).

    1,123,581 is an unreasonably high number (isn’t that around what Ken Jennings made in his epic run?), AND it cuts off the sequence in the middle of a number in the sequence.

    11,235,813 is an even more insane total, but it would at least keep the sequence going properly.

  15. Jeff S. Jan 6th 2013 at 03:27 pm 15

    $????1 is a common total after Final Jeopardy, as the contestants try to wager one more dollar than their opponents, assuming they get the question correct. And of course, $???9 is a common amount when they wager something ending in $1 and then get the question wrong.

    BTW, the maximum amount you can win in one day is $566,400.

  16. mark d Jan 6th 2013 at 04:13 pm 16

    Three cheers for the three kings: YUCK! YUCK! YUCK!

  17. Some Old Guy Jan 6th 2013 at 04:53 pm 17

    So, Jeff S., what’s the maximum amount three players could tie at?

  18. Boise Ed Jan 6th 2013 at 05:57 pm 18

    Elyrest (#6), that sounds like a good day to be playing poker.
    That’s Me (#3), thanks for the Fibonacci note. For any three consecutive numbers {x, y, z} in the series, x + y = z. It has some actual usefulness, but I’ve long forgotten what that is. Anyway, these three wise guy must be pretty amazing if they can each get this odd score in a game that is scored in quanta of 100.

  19. Boise Ed Jan 6th 2013 at 05:59 pm 19

    Oh. I just read Jeff S. Never mind what I said about the quanta of 100.

  20. fj Jan 6th 2013 at 07:57 pm 20

    What we now call the Fibonacci series was known to Indian mathematicians as far back as 200 BCE, so it’s not impossible that the magi (who were likely astronomers/astrologers and well-versed in math) would be familiar with the series (particularly if one of them, as says tradition, was from India). But, as others have said, it doesn’t make any sense as their 3-way tie score.

    I’m surprised the cartoonist omitted the contestants’ names: Caspar, Melchior, and Balthazar.

  21. Some Old Guy Jan 6th 2013 at 11:22 pm 21

    I have actually spent way too much time today trying to answer my own question about a three way tie. I’m close to an answer (I think) on a two-way tie, but three gets into math (as opposed to arithmetic) and just reaffirms my decision (many years ago) to major in psychology instead.

  22. Proginoskes Jan 7th 2013 at 04:02 am 22

    @ Some Old Guy: That’s actually an interesting question, especially since there’s some asymmetry built in. (There’s one daily double in the first half, and two in the second half.)

    The “symmetric” solution would be:
    * Player 1 gets all the clues correct.
    * Player 2 gets all the clues in 3 categories correct.
    * Player 3 gets all the clues in the other 3 categories correct.
    In every case, the daily double would be in the top row and be the last-chosen question.
    * Final jeopardy, everyone wagers what they have and win

    However, this doesn’t work; Player 1 gets $71,200, and players 2 and 3 only $70,400. (They have to “burn” $200 in Double Jeopardy because the top-row clues are worth $200 more than the first round.)

    However, if in round 1, the daily double is in row 2, and it’s the last one chosen, then all 3 players end up with $70,400 each. I doubt that this is optimum, although it’s probably close. (Sorry, Kamino Neko, #14.)

  23. The Bad Seed Jan 7th 2013 at 07:04 am 23

    There are 2 Daily Doubles per each of the 2 rounds, and although they would all have to be in the top row, they’d only have to be the last 2 questions answered by the guy(s) who picked them (the other guys could still answer questions afterward). Maybe I’ll be bored-enough at work today to work on this problem. ;)

  24. The Bad Seed Jan 7th 2013 at 07:06 am 24

    Crap, I should have checked first, but I was so sure… Yes, only 1 Daily Double per round. I should go back to bed.

  25. Ooten Aboot Jan 7th 2013 at 08:06 am 25

    Bad Seed, I believe Jeopardy has one daily double in the Jeopardy round and two in the Double Jeopardy round. Therefore, the wise guys could have hit one each.

  26. Morris Keesan Jan 7th 2013 at 08:43 am 26

    Unless things have changed very recently, there are two Daily Doubles in the Double Jeopardy round.

  27. Proginoskes Jan 7th 2013 at 11:33 pm 27

    I solved it … I think … by building an integer program (shown below; it’ll be a long post), and it doesn’t do any better than the $70,400 total above. (The tough part is finding a model which covers as many cases as possible.)

    I assumed that when each player hits their daily double, that they have at least $1000 (first round) or $2000 (second round), so that their wager is truly at most what they have at that point.

    I also assumed that each player gets exactly 1 daily double. (If one player gets more than one daily double, then some player doesn’t get any, and it looks impossible for them to get the same total as the player who gets at least two, without that number being small.)

    I set up the variables as follows; every x is a nonnegative integer in an integer program.

    x1 * 200 = amount player 1 gets before 1st daily double
    x2 = amount of player 1’s wager on daily double
    x3 * 200 = total amount of clues that player 1 gets after the daily double
    x4 * 200 = amount that player 2 gets during the regular round.
    x5 * 200 = amount that player 3 gets during the regular round.
    x6 * 400 = amount that player 1 gets during double jeopardy
    x7 * 400 = amount player 2 gets in double jeopardy before daily double
    x8 = amount of player 2’s wager on the daily double
    x9 * 400 = amount player 2 gets after the daily double
    x10 * 400 = amount player 3 gets in double jeopardy before daily double
    x11 = amount of player 3’s wager on the daily double
    x12 * 400 = amount player 2 gets after the daily double
    x13, x14, x15 = amount of each player’s wager in Final Jeopardy

    The conditions imposed by the rules during regular jeopardy are:
    x2 <= 200 * x1
    x1 + x3 + x4 + x5 <= (1+2+3+4+5)*6 - 1

    During double jeopardy:
    x6 + x7 + x9 + x10 + x12 <= (1+2+3+4+5)*6 - 2
    x8 <= 200 x4 + 400 x7
    x11 <= 200 x5 + 400 x10

    And during Final Jeopardy:
    x13 <= 200 x1 + x2 + 200 x3 + 400 x6
    x14 <= 200 x4 + 400 x7 + x8 + 400 x9
    x15 <= 200 x5 + 400 x10 + x11 + 400 x12

    In order for the scores to be equal, you have two equations:
    x13 + 200 x1 + x2 + 200 x3 + 400 x6
    = x14 + 200 x4 + 400 x7 + x8 + 400 x9
    = x15 + 200 x5 + 400 x10 + x11 + 400 x12

    and you are trying to maximize
    x13 + 200 x1 + x2 + 200 x3 + 400 x6

    Now, I fired up Maple and used LPSolve (with the assume=nonnegint option) and got the “symmetric” solution I mentioned several posts up.

  28. tdcjames Jan 8th 2013 at 10:04 am 28

    Impossible figure or not, if there’s a tie for the lead, all the tied contestants move on to the next day, so the wise men could be colluding to extend their runs indefinitely.

  29. fj Jan 8th 2013 at 11:52 am 29

    @Proginoskes,

    I played around with this a bit, too, and likewise determined that the $70,400 “symmetric” solution has to be the highest possible single day 3-way tie. Any attempt to even the scores by allowing players 2 and 3 to score in round and/or player 1 to score in double jeopardy results in lower scores than $70,400. Having player 1 miss a top row question lowers his score too much.

    tdcjames has a good point, though.

  30. The Ploughman Jan 8th 2013 at 06:00 pm 30

    The actual idea was probably “the wisemen have ridiculously high scores because they’re, you know, wise, so therefore they get a lot of points.” But that’s still pretty unstable premise - even assuming they are all equally wise, a Jeopardy board might play to one area of strengths. What if one category was “Myrrh and Its Many Uses”?

  31. James Pollock Jan 9th 2013 at 01:45 am 31

    A few problems arise.
    First, the Bible doesn’t actually state how many wise men there were. The common assumption is that there were three because they brought three gifts, but this does not have Scriptural authority.
    Second, they wouldn’t all have the same total, even if they finished tied each day for several days. One of them would have had to have won the day before the other two came on the show, ousting the previous champion. (When they limited a player’s run to five days, a five-day winner would “retire” and come back for the tournament of champions, and three new players would start the next day; now a champion stays on until he or she loses. So one of the players would have had to oust the previous champion, then the other two could begin.)

  32. Kamino Neko Jan 9th 2013 at 02:31 am 32

    That last point is not actually true.

    If all of the players end with zero, or in the red, nobody will come back.

    It’s a rare thing, but not unprecedented. (Maybe only happened once? Can’t remember off the top of my head. But it’s something they’ve had to rule on.)

  33. Proginoskes Jan 9th 2013 at 03:16 am 33

    @ Kamino Neko: It has happened a few times. Wikipedia says; “This first happened on the second episode of the current run, on September 11, 1984, and most recently on June 12, 1998.”

    There was a 3-way tie for actual money once; it happened on March 16, 2007.

    (See http://en.wikipedia.org/wiki/Jeopardy#Returning_champions )

  34. Kilby Jan 9th 2013 at 03:57 am 34

    All of these complicated calculations trying to arrive at a three-way tie ignore the fact that the joke about omniscient Jeopardy contestants was done long ago (and in a more logical fashion) in “The Far Side“, with God as one of the participants. To paraphrase the caption: “…it looks like our current champion hasn’t scored at all yet.

  35. carrot Jan 12th 2013 at 04:50 pm 35

    i kept thinking the game show should have been price is right, and all of them guessed the correct price. anyway, the idea of them colluding to maximize the winning sounds lots better to me.

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