16 replies to this topic

[TLEberle]

PYLdude, on 20 July 2012 - 08:12 PM, said:

I still say Roger Craig dropped a relatively easy Final in his loss, but that's just my opinion.

Trust me, we all know your opinion.

[PYLdude]

TLEberle, on 20 July 2012 - 10:19 PM, said:

PYLdude, on 20 July 2012 - 08:12 PM, said:

I still say Roger Craig dropped a relatively easy Final in his loss, but that's just my opinion.

Trust me, we all know your opinion.

ALL RIGHT, it's over.

[Kevin Prather]

PYLdude, on 20 July 2012 - 11:44 PM, said:

TLEberle, on 20 July 2012 - 10:19 PM, said:

PYLdude, on 20 July 2012 - 08:12 PM, said:

I still say Roger Craig dropped a relatively easy Final in his loss, but that's just my opinion.

Trust me, we all know your opinion.

ALL RIGHT, it's over.

And the next ball is...."Taking your ball and going home".

[Mr. Armadillo]

Why shouldn't he?  No reason for anyone to get wound up here.  Let it go.

Back to the OP:  If we assume a straight 1/3 chance to win any game, given that someone wins one game, there is a 1-in-81 chance they'll win five.

Given that someone has won five games, there's a 1-in-729 chance they'll win eleven.

Therefore, since there have been fewer than 729 five-time champions, I'm not all that surprised that we haven't seen it.

(Yes, I know that a five-time champion is more than 1/3 to win any given game.  However, if we assume that there are roughly 10 five-game winners per season, then there would have been 80 of them since Ken's run, roughly.  To get a 1-in-80 chance that a five-game winner will win six more, the probability of a five-game winner to beat two random opponents would have be roughly .4817.)

[Jeremy Nelson]

Mr. Armadillo, on 23 July 2012 - 01:56 PM, said:

(Yes, I know that a five-time champion is more than 1/3 to win any given game.  However, if we assume that there are roughly 10 five-game winners per season, then there would have been 80 of them since Ken's run, roughly.  To get a 1-in-80 chance that a five-game winner will win six more, the probability of a five-game winner to beat two random opponents would have be roughly .4817.)

And the odds are even worse than your estimates- I counted 29 winners of 5+ games in the past five seasons, so even then, your estimate is almost cut in half. making the probability substantially lower.

[Matt Ottinger]

Jeremy Nelson, on 23 July 2012 - 03:01 PM, said:

Mr. Armadillo, on 23 July 2012 - 01:56 PM, said:

(Yes, I know that a five-time champion is more than 1/3 to win any given game.  However, if we assume that there are roughly 10 five-game winners per season, then there would have been 80 of them since Ken's run, roughly.  To get a 1-in-80 chance that a five-game winner will win six more, the probability of a five-game winner to beat two random opponents would have be roughly .4817.)

And the odds are even worse than your estimates- I counted 29 winners of 5+ games in the past five seasons, so even then, your estimate is almost cut in half. making the probability substantially lower.

I have a real problem with a blind statistical analysis of something that is so performance-based.  Certain people do better in this game than other people do, and that is not something you can calculate.  Even starting with the idea that three players have an equal chance to win gets you off on the wrong foot as far as I'm concerned.

[TLEberle]

Go to the Jeopardy Archive. Pick a random game from the last year or two. Try and correctly question the Final Jeopardy clue. Now do that again for the next nine FJs. That should give you a rough idea of how hard it is to retain the title belt so many times.