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Comment on Why Ken Jennings’s ‘Jeopardy’ Streak Is Nearly Impossible to Break

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Here’s how we figured this out. We assumed that Jennings headed into Final Jeopardy with at most one other player in contention to win. Then we assumed that in order for Jennings to lose, this sequence of events has to happen: The game must not be locked up, Jennings must get the Final Jeopardy question wrong, and the other contender must get it right. We used his stats across all 75 of his games, not just his 74 wins, to better reflect his overall skills. The probability he doesn’t have the game locked up is 13.3 percent, the probability he gets Final Jeopardy wrong is 32 percent, and the probability the other contender gets it right is 49 percent. When we combine these probabilities, we see that Jennings only has about a 2.1 percent chance of losing.

This is incorrect and overstates his dominance because (1) it assumes Jennings will be ahead, something that won't always happen against Rutter/Jennings/Collins/Chu/Craig level players and (2) the lock up and opponent knowing Final Jeopardy conditions are not independent.

Pretty sloppy basic probability by 538 and make me think I'm not missing much by not subscribing (they didn't offer full RSS last I checked).

When you say that they're not independent, do you mean that the probability that the person knows the Final Jeopardy question is higher because they've managed to keep Ken Jennings from running away with it (i.e. they're better than the average person Jennings has faced)?

Just checking my understanding.

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