What are the highest, lowest, and most different scores contestants can achieve during a single game of...

Additional material

[ Editor: discussion of the entry above, rec.puzzles and mail to the
archive, June-August 1994 and June 1995.  The thread started with Jeff
Spivak <jsp@dlogics.com> asking: ]

>What is the absolute maximum dollar amount that a person could
>possibly win on one day of Jeopardy? I'm assuming that this person
>answered every question in both rounds correctly including
>the daily doubles and final Jeopardy.

Incorrect assumption; Final Jeopardy! won't happen if only one player
has a positive dollar amount.  So let's plan ahead and let one of the
other contestants answer a $100 question (question a $100 answer? :))
in the first round, just so we can get that extra doubling.

I'll make the possibly incorrect assumption that the daily doubles
don't appear above the third row; I know I've never seen them in the
first row, but I'm not sure about the second.

Jeopardy! round:

Our lucky winner answers all the questions in 5 of the 6 categories,
getting (100+200+300+400+500) * 5 = 1500 * 5 = 7500 points (they're not
actually dollars until you win).

Another chap gets 100 points, our winner gets another 200+400+500
(total now 8600), then gets the daily double, risks it all, has 17200.

Double Jeopardy! round:

Four categories, all answers, 3000 * 4 = 12000 (29200 total), two
categories, all answers but the $600 (for the daily doubles), another
2400 * 2 = 4800, 34000 total.  Two daily doubles, 68000 then 136000.

Final Jeopardy! round:

Risks it all against the person with 100 (yeah, right, it could
happen), wins the game with $272,000.

But wait - you asked how much this person could win in one day, not
show.  Jeopardy! tapes five shows in one day, and allows contestants to
participate in up to five games as they keep winning.  So if our lucky
winner starts playing with the first show taped, s/he can win
$1,360,000 in one day.  Not too shabby!

Note: if daily doubles do appear in the second row, our numbers are
$276,800 per show and $1,384,000 for the day.

-- Andrew K. Machan <machan@crl.com>

For the true answer, we must assume that the daily doubles will occur
in minimum valued questions so that the player will hit maximum scores
upon reaching the daily double(s).  It is also easy to see that the
daily doubles will ultimately have to be the last answers questioned.

A typical round will go like this...

Run One Category:            $1500
For 6 categories               x 6
                            ======
                             $9000
Less: $100 for the           - 100
Question where the          ======
daily double is hidden      $ 8900
                               x 2
Hit and make true daily     ======
double                     $17,800    TOTAL ROUND 1


Round 2:  Run 1 Category:     $ 3000
          For 6 Categories:      x 6
                             =======
                             $18,000
Less $400 (for 2 $200         -  400
questions where the daily    =======
doubles are hidden).         $17,600
                           +  17,800
Add Round one total...      ========
                              35,400
Double and then double        x    4
Again                      ==========

One Day Max:                 $141,600

Less Taxes :) :)

-- Rod Hutchings <rodney@cs.mun.ca>

Discussion regarding no final Jeopardy if only one positive amount ...

Has this ever happened in the TV version?  That is, only one player
coming out of Double Jeopardy with a positive dollar amount?

I also neglected to include the wager for final Jeopardy in my
calculations, and I was not aware that no final Jeopardy occurs if
only one player is eligible after double jeopardy... if we assume that
this rule does not hold, then the absolute maximum is 2 times my
previous answer of 141,600, that is $243,200.  Of course if you have to
spot a player $100 then that amount will be deducted from the round one
base total before the first round daily double.  Counting in the three
doubles in this game, and the doubling possibility of Final Jeopardy
and that reduces the Max I calculated by $1,600, so in this case the
actual total for the absolute max becomes a Staggering $241,600.  The
maximum a 5 time champ could win (prior to the tournament of champs) is
$1,208,000!!!

[ Editor: 2 x 141,600 = 283,200 (the figure in the entry above), and
283,200 - 1,600 = 281,600; five times that is 1,408,000. ]

-- Rod Hutchings <rodney@cs.mun.ca>

In article <1994Jun30.203030.6869@questrel.com>, chris@questrel.com
(Chris Cole) writes [ Editor: quoting the entry above ]:

> (2) The daily doubles (1 in the Jeopardy! round, 2 in the Double Jeopardy!
>     round) all appear under an answer in the $100 or $200 rows.

Which, as someone has already pointed out, NEVER happens.  Try not to
use bad assumptions when solving puzzles.  This is definitely a bad
assumption.

> Round 3 (Final Jeopardy!):
>                    Our (very greedy! :) hero now bets the whole thing, to
>                    see just how much s/he can actually win.  Assuming that
>                    his/her answer is right, the final amount would be
>                    $283,200.

Again, as posted earlier, this is faulty reasoning.  If both other
contestants have $0, they don't play the Final Jeopardy round.

> But the contestant can only take home $100,000; the rest is donated to
> charity.

So, in spite of all the calculations, bad assumptions, etc., the
correct answer is $100,000.  So why bother with all that preliminary
stuff?

The lesson we learn from all of this is "Just because it's in the FAQ,
doesn't mean it's correct."

-- <blask@gpx01.d39.lilly.com>

>Again, as posted earlier, this is faulty reasoning.  If both other contestants
>have $0, they don't play the Final Jeopardy round.

I believe if a contestant buzzes in with a question that is initially
judged incorrect, another may then answer and be awarded the points,
(and control).  If during the commercial break, the first response is
also ruled valid, both contestants can end up with points from that
answer.

-- Dean Inada <dmi@dmi.cerfnet.com>

As far as I know there won't be a final jeopardy round if only one
person has money.

There is a 'dirty' way around this though ... during either of the two
rounds when the 'hero' gives an answer it's different than what the
host expected so the host says they're wrong and they deduct the amount
of the question.  One of the other contestants gives the expected
answer and is then given that amount of money.

Before the round is finished the judges inform the host that the answer
given by the 'hero' is correct.  To adjust the scores they will
reimburse the 'hero' for double the amount of the question.  It's
double because they give back what they took away for the wrong answer
then they add the amount for the right answer.

You now have two contestants so there will be a final jeopardy.

-- John Latala <john.Latala@Waterloo.NCR.COM>

Chris Cole scribes:
> What about the part about the Daily Doubles all appearing under the $100
> row on Single Jeopardy or the $200 row on Double Jeopardy.  Is that
> impossible or just very unlikely?

If I remember correctly if you get a daily double you can wager up to a
certain amount, even if you don't have that much.  It's my
understanding that the reason for allowing a player to do that is if
they get a daily double in the top half of a column before they have
any money accumulated it makes for more 'interesting' play.

As far as I know there's no reason that the DD's couldn't be at the
top.

-- John Latala <john.Latala@Waterloo.NCR.COM>

>>> (2) The daily doubles (1 in the Jeopardy! round, 2 in the Double Jeopardy!
>>>     round) all appear under an answer in the $100 or $200 rows.

>> Which, as someone has already pointed out, NEVER happens.

It also NEVER happens that someone gets all the responses right, or not
yet, anyway.  The question is what's theoretically possible.  As you
know.

> My correspondents are unclear on what the rules of Jeopardy! say about the
> placement of the Daily Doubles.  Some claim that they have seen them on the
> first and second row; others swear they NEVER occur there.

It's rare, but yes, I have seen them on the first row.  I've never seen
more than one on the first row in the same show.

> Has anyone ever played on the show,

There has been at least one person on Usenet who has played on the
show: Miriam Nadel.

> and if so, was the placement of the Daily Doubles
> specified in the rules?

I believe I've read or heard [note the precise citation there! :-)]
that each Daily Double is simply placed in the correct row according to
its difficulty.  Naturally, they tend to choose the DD's from among the
harder questions.

>>Again, as posted earlier, this is faulty reasoning.  If both other contestants
>>have $0, they don't play the Final Jeopardy round.

I dispute this; I think I've seen a Final Jeopardy! with one player.
Not recently, but I don't have time to watch the show much any more.
Probably what this means is that the rules have changed.  If so, the
answer should mention that, ideally.

You know, I bet if you wrote the Jeopardy! folks a letter explaining
what you needed to know about the rules and why, they'd tell you.

> Several correspondents have pointed out that it is possible for two
> contestants to get credit for the same question.  This occurs if the
> first given answer is initially scored as incorrect, but during a
> commercial the judges reverse the host.

Good point.

>>> But the contestant can only take home $100,000; the rest is donated to
>>> charity.
>>
>> So, in spite of all the calculations, bad assumptions, etc., the correct
>> answer is $100,000.  So why bother with all that preliminary stuff?
>
> I'm sure this answer would not have satisfied the questioner's intent.

If one is going to take that point of view, I believe the IRS gets some
too.  As you might have predicted I was going to say: perhaps the
question should be reworded.

-- <msb@sq.com> [ Editor: Mark Brader ], July 1994

> What are the highest, lowest, and most different scores contestants
> can achieve during a single game of Jeopardy?

There should be a "!" after Jeopardy.  I think you have all the other
instances of the name correct.

> The caveat here is that you *must* be the person sitting in the left-most
> seat ...

Seat?  There are no seats.  At least, it doesn't look as if there are.

-- <msb@sq.com> [ Editor: Mark Brader ], July 1994

[ Editor: Chris Cole asked a former contestant, Leslie E. Frates,
whether the Daily Doubles can occur anywhere on the board, in
particular in the $100 (or $200) row.  She replied: ]

Yes, the Daily Doubles can occur anywhere on the board.  In the past 10
years I have been watching this incarnation of the show, I have seen a
Daily Double occur twice in the $100 spot, and once in the $200 spot
(but I also admit to missing a few shows here and there, so the actual
number of appearances may be more).  When my husband accompanied me to
Los Angeles for the tapings of my shows, a member of the studio
audience asked Alex Trebek if a Daily Double could occur more than once
in the same category, and Alex said no, that it was "categorically"
impossible.  From my own experience, the DD's seem to occur most often
in the $400-500 and $800-1000 spots on the board, but they can (and do)
appear everywhere else.

-- Leslie E. Frates <lfrates@nis.CalState.EDU>, August 1994

**********************************************************************

[ Editor: a later version of the question, June 1995, from
smartech@bga.com: ]

>1.  What, in theory, is the largest amount of money a single contestant
>can win in a single 1/2-hour [Jeopardy!] show?
>
>2.  What is the largest amount a 2nd place contestant can accumulate, yet
>still only come in second? And the largest for 3rd, yet still finish 3rd?
>
>[...]
>In the first round, there is one "Daily Double." In the second, two.
>   When a contestant finds a hidden "Daily Double" under one of the questions,
>   s/he may wager any or all of their accumulated winnings to that point.

The last time the Jeopardy! question came around, dedicated viewers of
the show pointed out that the Daily Doubles never occur in the top row
of the board (or maybe it's the top two rows), so the solution in the
archive is not valid.

With that in mind, here's my attempted <SPOILER> ...

Assume the Daily Double (DD) never occurs in a square labeled less than
$m in the Jeopardy! (1J) round, or less than $n in Double Jeopardy!
(2J).  I believe n = 2m in this case, but let's not assume that.

(Question 1:)

Then the total value of the 1J clues *other* than the DD is
$(9000 - m), and the total value of 2J clues other than DDs is
$(18000 - 2n).  The maximum score is achieved by a contestant who wins
all the clues, always hitting the DDs after all ordinary clues are
gone, always bets everything on each DD, and bets everything in Final
Jeopardy! (assuming they still run FJ even though both other
contestants are eliminated).  The winnings in dollars are

  ((9,000 - m)*2 + 18,000 - 2n)*2^3 = 288,000 - 16m - 16n.

Assuming m = 100, n = 200 this comes out to $283,200 (the archive
answer); assuming m = 300, n = 600 this comes out to $273,600.

(Question 2:)

To maximize the second place score, assume the third contestant is
shut out as before (otherwise his or her clues could be won by one of
the other two to get at least as good a solution).  Again postpone the
DDs as late in each round as possible and have the contestants bet
everything for maximal winnings.  Moreover we get more "bang" out of
the first DD if one player sweeps, or nearly sweeps, the 1J round while
the other player makes up the difference (if necessary) by sweeping
the 2J round before the DDs.  At the end of 2J we can either have two
players each with 50% of the money so far hit DDs (technically, one
player has to hit a DD, then miss a regular clue that the other player
then wins followed by the other DD), for a combined total of 200% of
the money not counting the last two DDs, or one player with 20% of the
money so far who hits both DDs, resulting in a combined total of 160%.
(Modifying the distributions just benefits the winning player at the
expense of 2nd place, so we don't want to do that.)  Clearly it's
better for each player to hit one DD.

Suppose one player sweeps all but $k of 1J and then wins the DD.  So
the players enter 2J with (18000 - 2m - 2k, k).  There is a total of
18000 - 2n available before the last two DDs.  If we distributed
m - n + 3k/2 to the first player and the rest to the second player in
2J, they would end up with (18000 - m - n - k/2, 18000 - m - n - k/2)
before the DDs, and (36000 - 2m - 2n - k, 36000 - 2m - 2n - k) after.
Obviously we want k as small as possible consistent with
m - n + 3k/2 >= 0.  Assuming m = 300, n = 600, we can set k = 200, so
we get both contestants with $34,000 going into Final Jeopardy!  In
order to maximize second place, one player bets (and wins) everything,
while the other holds back $1 but wins, ending up with $67,999.

I haven't tried to analyze the maximum third place score so carefully,
but assume one player sweeps 1J, including the DD, except for $k in
ordinary clues, but the other two players each get a DD near the end of
2J (after all their other winning responses).  Assuming the first
player gets $r in 2J, he or she goes into FJ with
$(18,000 - 2m - 2k + r), and the other two each have
$(18,000 - 2n - 2r + k).  We want these all as close as possible, so
try to set 2m + 2k - r = 2n + 2r - k, equivalently k - r = 2(n - m)/3.
Assuming m = 300, n = 600, we can set k = 200, r = 0.  So each
contestant enters FJ with $17,000.  Two contestants bet and win
everything, and so are tied for first; the other contestant holds back
$1 but responds correctly, ending up in third place with $33,999.

-- David A. Karr <karr@cs.cornell.edu>, June 1995

In article <D9wv75.Hp0@mv.mv.com> pwm@mv.mv.com (Pete Mitchell) writes:
>  judges decide that player 1's answer was acceptable:  player 1 receives
>  credit for the answer (adjusting his/her score accordingly), but player
>  2 does NOT lose the money that he/she was awarded for (also) answering
>  the question correctly!  [...]
>  ps:  Yes, I realize that this is probably not in the spirit of the original
>  question, [...]

Why not?  The original question asks us to find a situation that is
extremely unlikely but logically possible.  I think this fits the bill.

The responses that don't fit the spirit of the question are those that
point out the technical difference between the final score and what
the first-place player actually is allowed to take home.

In fact it considerably simplifies the solution if you assume that all
three players profit from every clue other than the DDs.  I believe
this could happen, in principle at least, if a player rings in but is
judged incorrect, then another player rings in and is judged
incorrect, then the third player rings in and is judged correct, after
which the judges decide that actually the first two responses were
correct as well.  (I've never seen this situation come up with all
three players, only with two at a time.)

In this case, the third-place player will be one who didn't get a DD
in 2J.  The best he can do is a DD at the end of 1J.  He ends up with
at most $2*(2*(9000 - m) + 18000 - n) = $(72000 - 4m - 2n).  That's
$70,400 if (as you say) the DD's occur in the second row.

The best the second-place player can do is one DD at the end of 2J; if
she did better than that (i.e., two daily doubles), neither of the
other two could possibly catch up, and she'd take first place.  She
ends up with at most $4*(9000 - m + 18000 - n) = $(108000 - 4m - 4n).
That's $105,600 assuming DD's occur in the second row.

Note that the assumption that two players cannot both benefit from a
single DD affects the scores a lot more than the assumption that two
players do not both benefit from a single ordinary clue.

>ObPuz:  You have twelve coins, one is either light or heavy and only Monty
>Hall (who lies if it's light or tells the truth if it's heavy) knows which
>one it is.  What is the minimum number of yes or no questions that can
>positively identify the coin?

There is no minimum.  Monty answers every question by telling you what
his favorite color is.  (If the coin is light, he picks the same
incorrect color every time.)

Assuming Monty's *answers* must be "yes" or "no," the minimum is one,
you just have to be very lucky.  If you want to be able to identify the
coin under any of the possible circumstances, the minimum is five,
since there are 24 possibilities to distinguish and only two possible
outcomes per question.  Ask questions that cut the state space in half
each time, and use wording analogous to what you'd use to find the
correct path to the truth-telling village in one question.

-- David A. Karr <karr@cs.cornell.edu>, June 1995

[ Editor: for Final Jeopardy! wagering see
competition/games/jeopardy.wager. ]