X has a car. Its value is unknown yet, but between 0 and 1000, uniformly distributed
You offer a price to buy the car.
If price < value, you can’t buy.
If price >= value, you can buy, give the price of money to X.
(for example, if value is 200, and you offer 300, you can buy, but you must give 300 to X)
If your buy is successful, Y will pay you the money of 1.5 * value to buy this car himself.
(for example, if value is 200, you offer 300, you can buy, give 300 to X, and then Y pays you 1.5*200 = 300, and he takes the car)
I think first of all, I need to define the optimal case, which is at the end I earn more money than the price I offered.
Assume value is 200.
If I offer too low price, nothing will happen, not make much sense
If I offer 300, buy the car and then sell the car to Y for 300, in the end, I still have 300, unchanged and makes not much sense.
If I offer 250, then I will earn 50 in the end, this makes sense.
If I offer 400, then I actually loose 100, this is even worse.
What should I do?
Update
Here is my thinking.
Let m be the price I am going to offer and v be the value of the car.
What I want is after all things (including I don't buy the car successfully), the money in my hand is still at least m.
There are three cases:
1. I can't buy the car, i.e., v > m
Because 0 <= v <= 1000, the probability of v > m is (1000-m)/1000.
2. I buy the car and I get less than m in the end, i.e., m > 1.5v
The probability is m/1500
3. I buy the car and I get more than m in the end, i.e., 1.5v >= m >= v
The probability is 1 - (1000-m)/1000 - m/1500 =m/3000.
I wish to have case 1 and case 3, so (1000-m)/1000 + m/3000 > m/1500. solve this I get 0 <= m <= 791.
As long as I offer a price between 0 and 791, I have bigger possibility to get extra money. (image if I play this for 10000 times)
