solve using probabilistic dynamic programming: i would like to sell my computer
ID: 3150438 • Letter: S
Question
solve using probabilistic dynamic programming: i would like to sell my computer to the highest bidder. i have studied the market, and concluded that i am likely to receive three types of offers: an offer of $200 with a probability of 2/7, and an offer of $300 with probability of 4/7 and an offer of $400 with probabilty of 1/7. i will advertise my computer for up to 3 consecutive days. at the end of each of the three days, i will decide whether or not to accept the best offer made that day. what is the optimum strategy for maximizing the expected sale price for my computer? what is the expected sale price?
Explanation / Answer
Fist of all , there's no point taking a $200 offer prior to the last day (as you can't do worse by waiting to see the third offer). On the other end, of course you should take a $400 offer as soon as you get it, as you can't do better. The only debate comes if you get a $300 offer before the last day.
Case I. suppose you have made it to day 2 and get a $300 offer. Then you look at your odds. With probability 2/7 you lose $100. With probability 1/7 you win $100 otherwise you break even. That's a bad bet...much greater chance of losing money than making any, so you should take the $300.
Case II. suppose you see a $300 offer on day 1. What happens if you wait? Well you win $100 if you get a $400 offer on day 2 (probability 1/7). If you get another $300 (probability 4/7) then (by our case I analysis) you break even. If you only get a $200 offer then you try again, at which point your expectation is:
2/7×200+4/7×300+1/7×400 = $285
Putting all this together, the expected value of waiting is:
1/7×400+4/7×300+2/7×285 = $310
this is (ever so slightly!) greater than $300 so in this case you ought to wait.
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