mabfan (Michael A. Burstein) (mabfan) wrote,
mabfan (Michael A. Burstein)

My Jeopardy Prediction

[Spoilers follow behind the cut only if you think I have a remote chance of being right]

I've been following the discussion regarding tonight's Jeopardy game with great interest, and not just because I've recently joined the National Puzzlers' League. Although I haven't watched Jeopardy consistently my whole life, there are periods of time where I do return to the show for weeks or months on end to challenge myself. And to enjoy it, of course. I've also had two friends appear on the show recently, so I've been making more of a point to keep up with the program.

Anyway, a lot of people on the blogosphere are talking about tonight's show for one particular reason. The teasers for the show say that something happens tonight that has never happened before. What could that be?

It can't be that all three contestants end up with $0 and leave. From what I've read, that happened before.

Well, it could be that all three players end up with negative dollar amounts before Final Jeopardy. That would be another case where all three would have to leave the program.

But, on the other hand, one of the players is current two-game champion Scott Weiss, also known as Squonk in the NPL. (On LiveJournal, he can be found at squonk_npl.) Weiss is clearly good at trivia and puzzles, and I can't imagine a situation arising where a player as good as he ends up in negative dollar amount territory.

So if there's a good player in the game, another possibility is that he runs the board. That is, he hits his buzzer for each clue before either other player, gets each clue right, and ends up as the only one playing Final Jeopardy.

But...but...not even Ken Jennings managed to do that, and he got to play over seventy games. This seems just far too unlikely, in my humble opinion.

So here's my prediction. Since running the board is so highly unlikely, even for the current champion, my guess is that there will be a three-way tie tonight, but one in which the tie is for a positive amount of money, not for $0. In that case, I presume, all three players will get to return for the Monday show.

As for who gets to select first on Monday – well, that would normally be determined by a coin toss when there are two returning champions, but I haven't seen a three sided coin in a while. My guess is that before the show, they'll do a run-off with A vs. B on the first toss, followed by the winner of that vs. C on the second toss. And since that gives C a slight advantage, I imagine that Scott Weiss will be C. Either that or they'll use a six-sided die, where A gets it if the die falls on 1 or 2, B gets it if the die falls on 3 or 4, and C gets it if the die falls on 5 or 6.

We'll see what happens tonight. (Well, the rest of you will. In our time zone, shabbat starts just after 6:30, so gnomi and I won't get to see the program until Saturday night. Fortunately, we don't mind spoilers.)
Tags: television

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