$1000 gamble... How much would you bet?

I would bet no more money than I am prepared to lose, which would be about $1. Gambling is a fool's game.

I have no incentive to put up more than 1 cent, since I get the same $1000 regardless of the size of my bet.

Is it just you betting, or are there others making the same bet with the holder of the $1000?
What is the other person suggesting you bet?

This is an expected value problem (perhaps for homework).

The answer is probably supposed to be $500. Because if you paid $500 dollars and got $1000 back half the time you would break even (the expected value would be 0). (When I think about these problems, I consider how I would come out over 1,000,000 plays which makes it a bit easier for me to conceptualize).

However, this is a poorly worded problem which doesn't have enough information for a real answer.

First... the time playing the game has value. You wouldn't waste 30 minutes (or 10 minutes, or maybe not even 1 minute) to play a game that you expect to break even. Poker players calculate this and won't play unless they are doing well enough (i.e. better than breaking even) to justify their time.

Second... as the other response says... risk also has a cost (i.e. negative value). Some people would think that even if the probability favored you, the risk makes the deal not worthwhile. Some people would not spend $300 for a 50% chance to win $1000, even though if they did this 1,000,000 times they would almost certainly come out ahead, the chance they might lose their money makes it not worth it. An example of this is insurance. You are making a bet with a negative expected value (out of 1,000,000 people who play the game, on average they pay more money than they get back). Insurance is worth it to us (even though mathematically it is a bad deal) because it gets rid of some of the risk.

Third... in game theory there are situations where it might be logical to pay more than $500 for a 50% chance to win $1000. The canned example often given is that if you need $1.00 to get home on the bus, but only have $.75. It might be logical to risk the $.75 on a coin flip to get the $1.00 since in this situation, $.75 is as good as nothing and $1.00 has a infinitely higher value.

To answer the question. Assuming I had the money to lose (i.e. I wasn't taking food off my families table if I lost). I would risk up to $400 on this deal (at this point the deal is too good to pass off).

If this were a repeating game, i.e. I new I could play 200 times at the same odds, and if I had a bankroll (i.e. enough money so I could loose 20 times or so without stopping to even out the statistics) I would risk more money-- perhaps $480 or $490.

I would put up the smallest amount acceptable up to $499. Obviously if I could get a chance for cheaper, I would. At $500, it's not worth playing. I like this take better: There is going to be a drawing for $1000 and 500 $1 tickets have been sold. How many $1 are you willing to buy given that no more will be sold?

expectedValue(x) = 1000 * x / (500+x) -x

That is a cool function engineer.

For the viewers at home, Engineer's problem has a correct answer (if your goal is to maximize your expected winnings). There is a maximum at just above 210 tickets (assuming you can't buy fractional tickets)-- you would expect to come out about $86 on top (on average) each time you played this lottery (where as if you just bought 1 ticket, you would net just under $1 on average),

Cool, no?

Taking the problem at face value;

I'd bet 1c in both cases.
(Unless I could bet even less than that)

In case 1: if I win, I'm down 1c and up $1000. If I lose, I'm down 1c.
In case 2: if I win, I'm up $1000. If I lose, I'm down 1c.
Betting more gains me nothing if I win, it just adds to my potential losses.