r/Jeopardy • u/Ok_Case_6660 • 19h ago
Why most don't understand the lock/tie scenario
Example: Leader has $20k, 2nd has $10k, 3rd has $2k.
Most approach this thinking 2nd will always wager everything, which of course they should. Therefore the leader can wager any amount between $1 and $9999 and it will show the same winning chances. Seems simple right?
But it's not, because for reasons only known to math teachers across the country, 2nd place doesn't always wager everything. Here are the statistics derived from the archive of this scenario over the years.
53 lock/tie scenarios.
40 times 2nd place wagers everything.
3 times 2nd place wagers all but $1.
10 times 2nd place wagers some other random number.
So while it's hard to believe, almost 25% of the time 2nd place has failed to wager everything over the years. Almost 6% of the time 2nd place has wagered all but $1.
The above paragraph is why wagering $0 shows basically the same winning chances as wagering $1, because 25% of the time wagering $0 guarantees victory where it should not assuming 2nd place wagers rationally.
And because of the propensity of 2nd place to wager all but $1 one reason why any wager greater than $1 shows much less chances to win. The leader can wager $1 and still tie when 2nd place makes the all but $1 wager.
Furthermore, because of the other seemingly random wagers from 2nd, if the leader wanted to wager more than $1 they would have no clue as to how much was a safe wager.
Summary, if 2nd place always wagered rationally, betting $1 from the lead would be the correct play. Because 2nd does not always wager rationally, a $0 wager from the lead shows similar winning chances to $1. And this irrational behavior is also why any wager greater than $1 from the lead lessens winning chances tremendously. I've also skipped all the detailed math because nobody really cares about it. And if you do, it's out there in various places.