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First off, yes it takes a lot of geekness to make a post like this, so yes I'm a geek...deal with it.

Anyways, in the case of the ep I watched, going into final round it was:

Leader: $15,600
2nd Place: $10,200
3rd Place: $4,200

It so happens the 3rd place person is negligible as expected but not because they can't reach the leader's starting amount with a right answer if wagering everything, but instead because they can't reach the 2nd place person's starting amount with a right answer and wagering everything. So, moving on...

What's standard strategy in this situation, constantly reaffirmed with nearly every single leader in Jeopardy doing it? Of course, leader bids $4,801 so that a right answer gets them to $20,401, $1 over twice the 2nd place person's starting amount so they can't possibly lose. The 2nd place person instinctively wagers everything even though in reality their fate rests on the leader getting the question right or wrong (though arguably not all 2nd place players do it, but probably 99% of the time they do). The thing is, the leader can wager MORE than that amount with a right answer securing a win. They don't because of fear of losing with a wrong answer. However, in some cases including the one I watched, they can wager more money with absolutely no difference in outcome with the assumption that the 2nd place player wagers everything.

In the case of this episode I watched, how about this: Wager $5,399 (the difference between leader and 2nd place - $1). Look at the scenarios again assuming 2nd place wagers everything:

Scenario 1: Leader answers correctly
If leader wagered $4,801, they win by $1 if 2nd place answered correctly or by a lot more if they answered incrrectly
If leader wagered $5,399, they win by $599 if 2nd place answered correctly or by a lot more if they answered incrrectly (and have $598 more to show for it)

Scenario 2: Leader answers incorrectly, 2nd place answers correctly:
Regardless of leader's wager, they lose because $20,400 is more than their starting amount.

Scenario 3: Leader answers incorrectly, 2nd place answers incorrectly:
As long as leader wagered no more than $5,399, they couldn't lose regardless of 2nd place's wager.

In fact, the only scenario that CAN change with 2nd place's wager is Scenario 2, and again the chances of them wagering less than everything is extremely slim anyways. In this case, the only way it makes a difference is if they wager between $1 and $599...what are the odds of that happening?

Having said all that, there are 2 other situations of $$$ going into the final round to consider such that the 2nd place person is within reach of the leader taking into account what the 3rd player has:

Situation 1:
Leader: X - let's say $15,000
2nd place: Y (something less than X whereby 2Y > X) - let's say $9,000
3rd place: Z (something less than Y whereby 2Z > Y) - let's say $5,000

In this case, while priority is still getting above 2Y with a right answer (making (2Y - X) + 1 the minimum wager the leader should do), the leader wants to make sure they stay above 2Z with a wrong answer as well if possible since the 3rd place player may wager everything too, so the wager should be (X - 2Z) - 1 or (2Y - X) + 1, whichever is larger. Example: Leader has $15,000, 2nd is $9,000, 3rd is $5,000. (X - 2Z) - 1 is $4,999, (2Y - X) + 1 is $3,001, so the leader should wager $4,999. (X - Y) - 1 would be $5,999, but if the leader wagers that and the 3rd place player wagers everything, wrong/wrong/right would make the 3rd place person the winner whereas $4,999 would still leave the leader ahead by $1.

Situation 2:
Leader: X - let's say $15,000
2nd place: Y (something less than X whereby 2Y > X) - let's say $9,000
3rd place: Z (something less than Y but whereby 2Z > X also) - let's say $8,000

This little change ironically means that the worry about staying over 2Z with a wrong answer is irrelevant because the leader can't, so the 3rd place person can effectively be ignored even though they can win if they answer right and the leader and 2nd place person answer wrong. So, back to the correct wager amount, it would be (X - Y) - 1 or $5,999. Most leaders will instead wager $3,001 and will effectively screw themselves out of another $2,998 if they got it right with no change in the realistic wrong answer scenarios.

So yes, all this to explain why the typical wagering strategy by leaders in Final Jeopardy is undercutting what they could be making with a higher wager that, again with the assumption that the 2nd place person wagers everything which they almost always do, makes absolutely no difference to the outcomes. Why I noticed this, felt the need to post this, or am even watching an old Jeopard on GSN while eating dinner I couldn't tell you. I guess just thought it was interesting because the traditional "bet $1 more than what the 2nd place person can get" strategy is about as set in stone as bidding $1 over the highest bid in The Price is Right if you're 4th and their bid isn't that high.

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