At one point or another almost everyone who watches Jeopardy! has wondered what the highest possible final score is, mathematically speaking. This is actually pretty easy to figure out. You just need to assume that (1) a single player provides the correct response for every square on the board, (2) all three Daily Doubles are found in the row with the lowest cash value, (3) the Daily Doubles are always the last squares selected, and (4) the player always wagers all the money he or she has. Under these possible but admittedly improbable conditions, and with the show’s current dollar values, the maximum possible score is $566,400. This is nicely illustrated on this DataGenetics blog post.
What is the highest possible score for a two-way tie, and what is the highest possible score for a three-way tie? I’ve thought about this for a while and I don’t think it’s so easy to figure out. It’s obviously not $566,400 divided by two or three (or rather, if it is $566,400 divided by two or three, it’s not obvious to me how to prove that). I thought I’d turn to the mathematical geniuses here at the Dope to see if they have any insight. What assumptions and what sequence of plays and wagers is required to achieve the maximum possible two- and three-way ties, and what are those maximum amounts?
A couple of notes:
[ul]
[li]Yes, I’m aware that there is a new tie-breaking rule. I am asking about the maximum possible tie score before the tie is broken.[/li][li]I put this in GQ instead of Cafe Society because I’m really interested in the underlying mathematical question, not a discussion on the practical impossibility of this ever happening in a real game, strategies that players could use to collude to make it happen anyway, or who would win in a death match between Ken Jennings and Alex Trebek.[/li][/ul]
The easiest way to figure a two-way tie is to have each player make the same on each of the first two rounds. That’s easy for Double Jeopardy, since there’s two Daily Doubles there, but there’s only one such on regular Jeopardy. OK, so have one player win twice as much on the non-DD first round and the other make up the difference with the DD. However, since the player winning the DD doesn’t get the $200 for that answer in addition to doubling their winnings, you have to have no answer on one of the $400 questions to even things out. If I haven’t made an error (quite possible this late at night):
It’s possible there’s a way to increase this, since one question was left unanswered, but I can’t think of one right now. And I’ll let someone else come up with a three way tie score.
That’s an easy way to make a tie, but it doesn’t come anywhere close to maximizing the score. For example, you could first have a single player sweep the Jeopardy! round, where the Daily Double is located on a $200 square, and is selected last. This way the player makes 6 × ($400 + $600 + $800 + $1000) + 5 × $200 = $17,800, then bets everything on the last clue to make $35,600. Then in the Double Jeopardy! round, the second player wins all the squares except for the two Daily Doubles on the $400 spot: 6 × ($800 + $1200 + $1600 + $2000) + 4 × $400 = $35,200. The first player then uses the first Daily Double, betting $34,800 and ending up with $70,400. The second player then uses the second Daily Double, doubling their $35,200 to $70,400. In Final Jeopardy! both players wager their full amounts, ending up with a tie at $140,800.
Now, this is a lot higher than the $116,800 from your method, but I don’t know whether it’s actually the theoretical maximum.
I am thinking that there must be some formulas or algorithms that can calculate the maximum two- and three-way ties. This would be a lot better than trial-end-error thought experiments.
I’m sure this isn’t the optimum, but I think you can do better than that Player 1 answers every question in round 1 except 1 $200 question, daily double last, for $35200, and only answers DD in round 2, getting $70400
Player 2 gets $200 in round 1 and sweeps the board in round 2 except player 1 getting the first DD for $70800. Final Jeopardy they end up with $140800 each.
Instead of leaving questions unanswered, you could just have the player who would otherwise win wager less than the maximum on Final Jeopardy. That might be better in some cases, because it puts more money into the players’ hands before the doubles come up.
The player leading round one doesn’t get to start round two. She only gets to answer a daily double if she answers another question right and then picks the daily double, so there’s less to go around for the other players.
That’s a good point. So maybe Mr Shine and I are both wrong about being able to end up with $140,800. But you could still get nearly that much money—player two would just need to get one of the $400 clues in Double Jeopardy! wrong, giving player one a chance to buzz in with the correct response and then select the Daily Double. Player one then selects and flubs a different clue to pass control of the board back to player two.
OK, not having seen the show that’s a wrinkle I hadn’t considered. However I think you can still get 140800, if p1 sweeps r1 except for a $400, answers a $400 in the 2nd round then does the DD. Player 2 gets $400 in r1 and sweeps the board except p1’s $400 and DD
p1 (17400*2)+400)*2=70400
p2 (400+34800)*2=70400
FJ total: 140800. I’m going to guess this is optimal for 2 players.
Side question that I’ve always wondered about: Are the Daily Doubles tougher the higher up on the board they are found? In other words, does the dollar amount behind the DD reflect the level of difficulty of the [DEL]question[/DEL] answer?
mmm
I think Mr. Shine is right for a normal game, but there’s another wrinkle. Jeopardy will sometimes accept two responses if there’s an ambiguous question. If, improbably, the leading players both give acceptable answers to every question in both rounds (excepting the DDs, which only one person can answer), and the timing, location, and allocation of DD clues works out optimally, P1 could have up to $35,600 in the first round and P2 could have up to $17,800. In the second round, P2 could climb to as much as $106,000 using one DD. P1 would still be higher with $140,800 using one DD. (It would be up to $141,600 except that to maximize P2’s score, there has to be at least one $400 square on the board for P1 to get credit for answering correctly, take control of the board, and get the last DD. This one won’t be in P1’s account for P1’s last DD.). If they bet to tie in Final Jeopardy, they could both wind up with a maximum of $212,000.
Only where the show runners are themselves in error. That is, the first contestant who rings in gives a correct response, Alex incorrectly says that they’re wrong, and a second contestant then gives a variant correct response that Alex accepts. The show runners realize their error, and then correct the first contestant’s score at a natural break in the game (usually after a commercial). I don’t think we need to consider such pathological cases, since they arise only when the game’s rules are inadvertently broken.
As for a 3-way, it’s easy to get each of the players to get $17600 and 1 DD, each ending up with $70,400 after FJ. This seems like the most efficient use of the DDs so should not have anything higher.
Yes, I realized that some time after I posted, but I wasn’t about to get up in the middle of the night to post a better answer.
Well, there’s still the three-way tie to consider, but it isn’t quite as easy. I have an idea, but need to work it out. And I have other things more important to do.
Looks like Mr Shine had the same idea I did. The first player sweeps the first round and the other two split the second and they’ll all end up with the same score.
Is it even possible for the Daily Double to be in the first row? It’s been a while since I’ve watched, but I remember them all being in the bottom two rows.
And who starts a round is irrelevant. The starter just gets to pick a category and value, but once that’s done, anyone can ring in to answer it. Usually the first person to answer is the person who picked, since after all they’re picking the category that’s their best strength, but someone else has to beat them occasionally or there’s no game. The only way that who starts the round is relevant is if the first question is a Daily Double, which only one person gets a chance at, but we don’t want those coming early in the round anyway.
I acknowledged that it’s improbable. Since the core question was already answered, I thought it was an interesting exercise.
Mr. Shine assumed early on that the person leading the first round would answer only a Daily Double correct in the second round. As you note, there is only one way you can answer get a Daily Double right in the second round without getting any other questions right. You have to pick the Daily Double as the first question. But, the person leading the first round doesn’t get to pick first in the second round, so who goes first affects that scenario.
Mr. Shine later pointed out that a player can miss one question in the first round, get one regular question and one Daily Double in the second round, and still match the maximum total for the other player who gets every other Double Jeopardy question right. It turns out not to matter who goes first in the second round but I didn’t realize that at first.
One of the Daily Doubles in the second round has to be the last clue picked but it doesn’t much matter when the other one gets picked. The first-round leader just needs to jump in on a $400 question, get it right, and pick the next Daily Double in a $400 slot. That could happen anytime during the Double Jeopardy round without affecting the first-round leader’s total.
