What is a perfect score on Jeopardy!? All ~~great sports~~ fun activities have perfect scores: bowling is 300, batting average is 1.00, golf (I suppose) is 18. Okay, 18 in mini-golf is more likely and has actually been achieved twice in competitive history. That shouldn’t really surprise you. What should surprise you is that there’s a competitive ~~mini-golf~~ putt-putt scene. But I digress, so let’s get back to Jeopardy!.

This question was posed by FiveThirtyEight as one of their Friday Riddler questions. I quickly wrote and answer for that and thought it might be fun to post it here. Not because the code is in any way difficult, but just because it’s fun. And its Friday. And I can handle reviewing two more papers until I finish this coffee, so I’m procrastinating.

First, see if you can figure this out on your own. For those three of you millennials who don’t know the rules to Jeopardy!, they’re simple. But the trick to getting the perfect score is as much luck as it is skill.

First, you pretty much have to be playing against Sean Connery and Turd Furguson, because you have answer every question correctly and either buzz in first every time or rely on your competitors to answer incorrectly. So that’s skill. But you also have to

- find all the Daily Doubles,
- find them on the last clues of the round,
- find them all under the lowest dollar value clues,
- make them all true Daily Doubles.

# The dollar values for all the clues in each column in both rounds. j <- seq(200,1000,200) j2 <- j*2

Calculate the max score for the first round:

# Answer all clues correctly (6 columns), minus the daily double clue (the lowest value on the board, since you don't get the clue $$ for the daily double clue, even if you get it right) j.sum <- sum(j*6)-j[1] # Then, you get the daily double under the last clue and win a true Daily Double (double your $$) j.sum.dd <- j.sum*2 j.sum.dd [1] 35600

So you can take $35,600 in the first round. Now calculate the max score for the Double Jeopardy! round:

# Answer all clues correctly except for *two* of the lowest value clues, which contain the Daily Doubles. j2.sum <- sum(j2*6)-2*j2[1] # Add this to your total from the first round j2.sum.total <- j2.sum + j.sum.dd # You now have $70,800 when you hit the Daily Doubles # Make the last and second last clue true Daily Doubles (double your $$) j2.sum.dd.1 <- j2.sum.total*2 j2.sum.dd.2 <- j2.sum.dd.1*2 j2.sum.dd.2 [1] 283200

So you leave Double Jeopardy! with $283,200. Now, we have to assume that the show would actually let you play Final Jeopardy! with both your competitors at $0. I’m not sure this has ever happened, though there have been some weird scenarios. I can imagine the uproar if they didn’t let you try to double your money again, so let’s assume they would.

# You wager everything (for some reason) and get it right, doubling your $$ final.sum <- j2.sum.dd.2*2 [1] 566400

So you take home $566,400. Not bad.