# If in the final round of Jeopardy, there is one one other person left, with \$8000, if we ignore the case where you guess wrong and your opponent guesses right, how much do you need to assure winning?

This makes for a good high school level algebra problem. Obviously, if you had more than \$16,000, you could be always win by not wagering anything. If we eliminate the possibility of you guessing wrong and your opponent guessing right, you need a somewhat smaller amount to guarantee a win.

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\$16,001 I would say because they can double their score. Although many people with 8,000 would bet 7,999 on Final Jeopardy; I’m not sure why. Many you are penalized if you have 0 money rather than \$1. so the answer is either \$16,000 or \$16,001.

janbb (62550)

You can’t guarantee a win. It could always be a tie because your opponent has the same constraints. Apparently they get another question if that happens and the first to answer wins.

Maybe I was not clear. I want to know how much you need going into the final round. You need to cover two cases – if you and the other person guess correctly with the other person betting the full \$8000 or if the two of you guess incorrectly with the other person not wagering anything.

I think you would have to have at least \$8,001 so if they bet all and double and you do the same, you have beaten them. And then you bet \$8,000 on Final Jeopardy. Oh – but if you both get it wrong and they bet nothing, they have won. I don’t think there’s a way to guarantee a win then.

janbb (62550)

There is a way of covering both cases if you start with \$12,001 and wager \$4,000.

Here is an algebraic way of deriving the answer. Let a=the amount you have and x=the amount you wager. Don’t be bothered by having two variables. We will be getting rid of the x.

To cover the case where you both win we must have:
a + x > 16,000 or x > 16000 – a

To cover the case where you both lose we must have:
a – x > 8000 or x < a – 8000

Combining the two expressions for x :
16000 – a < x < a – 8000

We can always choose such an x provided the expression on the left is less than the expression on the right.
16000 – a < a- 8000 or 2a > 24000 or a > 12000

If the other person starts with b in place of 8000 then we get a > 3/2 b

Well, if you start with a winning amount, sure you can play it so you can’t lose. But you’re right – that would work.

janbb (62550)

Not quite sure what you are saying. \$12000 still does not provide a win if you guess wrong and the other person guesses right and wagers more than \$4000. An intuitive non-algebraic way of looking at it is to say that 12000 splits the difference between 8000 and 16000. Betting \$4000 will either take you up to 16000 or down to 8000

I’m saying you’re right.

janbb (62550)

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