Can you solve this math problem based on the Jeopardy television show?

They need to have 10,001 and bet it all.

What if both players answered incorrectly and the one in second place did not bet anything?

@LostInParadise Aha – there’s the rub! So the first player has to have 20,001 and not bet anything.

That is the case I already mentioned where one player has over twice as much as anyone else. By eliminating the case where the player in first answers incorrectly and the other player answers correctly we can lower the number.

Let me start you off. If they both answer correctly, the player in first needs to end up with at least $20,001 and if they both answer incorrectly, the player in first has to end up with at least $10,001.

We can ignore the case where the first player answers correctly and the other player answers incorrectly. The player in first will always win in that case. Of course to be in first they need to have at least $10,001.

Imagine 10,001 and 20,001 on a number line. What can we say about the number in the middle of these two numbers?

If he has 16,000 and bets 5,000 or 15,999, he should be golden. Right?

Actually, 15,001 with a bet of 5,000 should do it.

Well done!

Here is a way of looking at it. 15,001 is in the middle of 10,001 and 20,001. Having 15,001 and betting 5,000 will cause a move from 15,001 to either 10,001 or 20,001, depending on whether or not the question is answered correctly.

^^ Yes. That’s how I solved it – finally!

I should point out that this does not actually prove that $15,001 is the least amount that will guarantee victory. I know I promised no mathematics beyond elementary school, but there is a very simple algebraic proof of this:

Let m = money the person in first place has in final jeopardy and let b = the amount they bet in final jeopardy. In order to have victory when either both players answer correctly or both players answer incorrectly, we have:
m+b > 20,000
m-b > 10,000

Adding these inequalities gives:
(m+b)+(m-b) > 20,000+10,000
The b values cancel, leaving 2m>30,00, m > 15,000, which is what we wished to prove

The one in front would need $20,001 dollars to be assured of winning. Betting $0 would allow him to win.

Yes, but I added the condition that for this problem we should elimihate the case where the person in the lead gives the wrong answer and the other person gives the right answer. Under those circumstances, $15,001 suffices.

@LostInParadise You said the one in second has $10,000. You eliminated basically either person giving the wrong answer. If the one in second has $10,000 and bets it all, they stand to have $20,000. So $15,001 is not enough for the lead person to have. If you are assuming that they both are going to answer correctly the lead person could have $10,001 and bet it all and come out on top. But to ensure the person in the lead could not lose, he/she would have to have $20,001 and bet nothing.

Give the one in first place $15,001. If they bet $5000 and answer correctly, they end up with $15,001 + $5000 = $20001. The one in second place bets all of their $10000 gving them $10,000 + $10,000 = $20,000

if the one in first place answers wrong then they end up with $15,001 – $5,000 = $10,001. The one in second place has to answer wrong, becuase we have eliminated the case where the one in first answers wrong and the one in second place answers correctly. The best that the one in second place can do is to bet $0, giving them $10,000

@LostInParadise I guess @seawulf didn’t really read the thread or he would have seen the solution.