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July 8, 2015

Wednesday, July 8, 2015

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With the amount of work our players have given me lately, I’m tempted to ask for a raise. (It doesn’t help that my liquor cabinet is looking terribly empty these days.)The Final Wager – July 8, 2015

Let’s start with that terrible Penultimate Wager by one of our challengers.

Sawyer

Morgan

Ian

Bourland

Jenny

Thorngate

11,000 10,400 11,800
PENULTIMATE WAGER
NAME-DROPPING SONG LYRICS for $800

Given our players’ proficiency with the rest of this category and the clue’s second-level status, I’d be tempted to go for $10,201 and end things right now. Even if Jenny is wrong, she can still win in Final.

The other option is no more than $199. This places Sawyer in Stratton’s Dilemma: if he covers Ian’s wager in Final, he’ll lose to Jenny if he’s wrong.

$3,000 does nothing if she’s right and puts her in third if she’s wrong. But she did get it right, and we have these scores heading into Final:

Sawyer

Ian

Jenny

11,000 10,400 11,800

Tonight’s Final Jeopardy! category is:

THE THOMAS JEFFERSON ADMINISTRATION

First-order calculations

Second doubles up

If Sawyer doubles his score, he’ll have 22,000.

To cover this all-in wager, Jenny will need to wager 7,200.

Sawyer
Ian
Jenny
11,000 10,400 14,800
11,000 7,200
22,000 22,000
min max min max min max
7,200

An incorrect response with that wager will leave Jenny with 7,600.

To stay above her total, Sawyer can wager up to 3,400.

Ian can wager up to 2,800.

Sawyer
Ian
Jenny
11,000 10,400 14,800
3,400 2,800 7,200
7,600 7,600 7,600
min max min max min max
3,400 2,800 7,200

Third doubles up

A successful doubling will put Ian at 20,800.

To cover this, Sawyer should wager at least 9,800. In this case, though, he will lose on an incorrect response, so he might as well wager everything. (Stratton’s Dilemma!)

Sawyer
Ian
Jenny
11,000 10,400 14,800
9,800 10,400 6,000
20,800 20,800 20,800
min max min max min max
9,800* 2,800 7,200
3,400

Here’s where things stand after the first order:

Sawyer
Ian
Jenny
11,000 10,400 14,800
FIRST-ORDER SECOND-ORDER COVER ZERO
min max min max min max
9,800* 2,800 7,200
3,400

Second-order calculations

Second’s max vs. first

If Sawyer makes the rational maximum wager of 3,400, Jenny will need to wager 400 to stay above him.

Sawyer
Ian
Jenny
11,000 10,400 14,800
3,400 400
14,400 14,400
min max min max min max
9,800* 2,800 7,200
3,400 400

In that case, Sawyer would have to wager 4,200 and respond correctly, and again, an all-in makes sense.

Ian might consider risking at least 4,800 – everything.

Third’s max vs. first

Sawyer might go for 2,200.

Sawyer
Ian
Jenny
11,000 10,400 14,800
2,200 2,800 1,600
13,200 13,200 13,200
min max min max min max
9,800* 2,800 7,200
2,200 3,400 400
4,800*

That downside means Ian should cap his wager at 1,600.

Sawyer
Ian
Jenny
11,000 10,400 14,800
FIRST-ORDER SECOND-ORDER COVER ZERO
min max min max min max
9,800* 1,600 7,200
2,200 3,400 2,800 400
4,800*

Zero wagers

Score differences:
Jenny and Sawyer: 3,800
Jenny and Ian: 4,400
Sawyer and Ian: 600

Sawyer
Ian
Jenny
11,000 10,400 14,800
FIRST-ORDER SECOND-ORDER COVER ZERO
min max min max min max
9,800* 1,200 1,600 7,200 7,600
2,200 3,400 2,800 400
4,800*

Then we add a dollar to the minimum wagers, and subtract a dollar from the maximum wagers.

What actually happened

Sawyer Ian Jenny
11,000 10,400 14,800
10,900 10,400 7,201
21,900 20,800 22,001
min max min max min max
11,000 1,201 1,599 7,201 7,599
2,201 3,399 2,799 399
10,400

It looks like the writers are recycling Final Jeopardy! clues from the Teen Tournament that won’t be happening this season.

Tomorrow, Jenny will be our seventh consecutive different champion. Will we get another Penultimate Wager tomorrow? (I hope not.)

The Final Jeopardy! clue

THE THOMAS JEFFERSON ADMINISTRATION
OF THIS AGREEMENT, THOMAS JEFFERSON SAID HE “STRETCHED THE CONSTITUTION UNTIL IT CRACKED”
Correct response: Show
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5 Comments
  1. Aaron Fisher permalink

    I’ve noticed that in these situations you usually advocate for 2nd going big and 3rd place going small.

    I think that’s worth taking another look at. The second biggest piece of the pie (after first place has made a lock-out wager) is usually the triple-stumper.

    Second place has effectively has “dibs” on going small if they want it, since both second and third wagering small is a great for second and disastrous for third.

    • Kelly permalink

      Remember that in this game that Ian is capable of covering Sawyer’s maximum “unsafe” wager without making himself vulnerable to losing to Jenny’s MSBIW. In a Stratton’s Dilemma game 2nd has the strongest incentive to go “big” in either a scenario like this one (where 3rd can safely cover 2nd’s maximum “unsafe” wager without losing to the leader’s standard bet if wrong) or when 3rd cannot win without being right anyway (and thus has reason to bet it all). It’s when 3rd has enough to beat the leader on a TS, but not enough to do both that and cover an “unsafe” or even a zero wager out of 2nd that the incentives for 2nd to go “small” go up. Nonetheless, I wouldn’t fault a player in 2nd to go “small” if he/she is uncomfortable with the category in any SD case.

      • Aaron Fisher permalink

        Sawyer’s 0-599 + 1,000-1,799 + 2,201-3,400 range covers this effectively. Ian can never get both the large and the small range, and going for the small range incurs a significant risk of getting neither. (To go small, Ian needs to predict which subrange Sawyer is on, and is in awful shape or even locked out if he predicts wrong and makes a wager within 600 of Sawyer’s).

        As long as we’re on the assumption that the small wager is more valuable, Sawyer should contest it. He will usually get it, but even if he doesn’t, he’ll effectively have made the large wager.

        The major risk is Jenny going small, but not enough leaders make that play.

  2. Kelly permalink

    I found a third possibility for Jenny’s penultimate wager – between 4,701 and 4,866. That amount gives her a crush lead if right, and keeps her just a tiny bit over 2/3 of Ian’s total if wrong (still giving a bit of mind games with that scenario, but betting zero to win on a TS with proper wagering remaining a possibility). (If I did that I’d go for 4,800 to keep the numbers simple with the final calculations.)

    • Aaron Fisher permalink

      I think that’s the best wager with these scores for DDs that are very close to 50:50s. (Although I agree with Keith that this DD was likely to be an easy one, increasing the appeal of the all-in wager.)

      The all-in wager leaves you (approximately, looking at previous FJs, and wagering trends) at 100% or 0%. The crush wager leaves you at 70% or 29%. The null wager leaves you at 48% or 48%.

      This is a bit of an odd scenario, where the right wager could be any of those depending on how difficult the DD is.

What do you think?