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The absolute max you could win during a single game of Jeopardy!

BirdOPrey5

Staff member
Administrator
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Been wanting to calculate this for a while...

Round 1


One Daily Double somewhere on the board

Round 2

Two Daily Doubles somewhere on the board

Since this assumes ideal conditions we'll stipulate you were able to get the producers to put the daily double's in columns and values of your choosing because they owe you a favor. A big favor.

So round one I add one column, 200+400+600+800+1000 = $3000
Multiplied by 5 = $15,000
There is a 6th column with a daily double. To make the most money the daily double must be the last answer you choose AND also must be the lowest value available that round, so in round one the daily double must be the $200 answer. We wisely will have worked backwards from $1000... so 1000 + 8000 + 600 + 400 = $2800
$15,000 + $2,800 = $17,800
So we then bet it all for the daily double, win, and wind up with $35,600 at the end of round 1.

Round 2 starts with $35,600.

This time each column (400+800+1200+1600+2000) equals $6000

So $6000 x 4 complete columns plus our Round 1 winnings = $59,600

OK, I believe it is a rule that the two daily double's in round 2 can't be in the same category but that is just as well because it would be a mistake to make the two daily double's he last $400 and $200 clue in the last column. Instead they should both be behind $200 clues in different columns.

So we have $59,600 with two full columns left, each with a daily double. You might be tempted to complete a column, but then you'd have left some potential winnings.

So we choose (and win) $2000 in both columns, then $1,600 in both columns, $1,200 in both, $800 in both, and $400 in both...That all adds to an additional $12,000. $12,000 + $59,600 = $71,600. * See abqtj's post below
That all adds an additional $11,200 so $11,200 + $59,600 = $70,800.


So we have $71,600 $70,800 with two $400 clues left on the board, but behind each is a Daily Double.

First Daily Double takes as from $71,600 to $143,200! $70,800 to $141,600!

Now we play the last clue, we risk it all, and amazingly, we win, $286,400 $283,200. But it's not over...



So now neither of the other contestants can go on to final Jeopardy, at best they have zero, or they may even be in the negative if they answered incorrectly before we could answer. (The next fun calculation is to figure out the lowest possible score you could have, that is probably even more interesting, but another time for that.)

So it's just you in Final Jeopardy, you've won but you can't help it. You need to bet it all! You jot down your answer.

A heavily perspiring Alex Trebek, blown away by what he has seen today checks your answer... It is correct. He takes a gulp before checking how much you bet.

Sure enough the screen reads $286,400! $283,200!

You're final maximum possible winnings in a single game of Jeopardy under absolutely ideal conditions is a whopping...

$572,800
$537,600
$566,400.

Did you get a different number? Did I make a mistake? Could be... Let me know.
 

abqtj

I'm a damn delight!
Staff member
Administrator
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You part about the final two columns (before daily doubles) = $12,000 is wrong.

it's $5,600 per column, or $11,200
 

Ardvark

Well-known member
Glad that is crossed.off your bucket list..

Calculate the odds that a juer would end up in bed with both Olsen twins when they turned 18 and then every year after that.
 

BlackDak

Banned
VIP
You have a 1 and 870 chance of choosing the two daily doubles as your last two boxes in the second round. And a 1 and 21600 of hitting both the daily double in the first round and the two in the second as the last chooses in each round.
 

BrandonM7

MaMway Platinum Member
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Moderator
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Ultra-Premium
Glad that is crossed.off your bucket list..

Calculate the odds that a juer would end up in bed with both Olsen twins when they turned 18 and then every year after that.
Divide by zero error
 

BirdOPrey5

Staff member
Administrator
VIP
You have a 1 and 870 chance of choosing the two daily doubles as your last two boxes in the second round. And a 1 and 21600 of hitting both the daily double in the first round and the two in the second as the last chooses in each round.
Assuming their placement is random, aside from the fact in double jeopardy the two can't be in the same category, what are the chances all three will be behind the lowest value boxes for each round?
 
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