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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 dollars with probability $2/7$, an offer of \$300 dollars with probability $4/7$, and an offer of \$400 dollars with probability $1/7$.

I will advertise my computer for up to three 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 this expected sale price?

hardmath
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foley
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  • This is not clear. How many offers can I expect in a day? Is the number of offers itself probabilistic? Are the various offers independent of each other? – lulu Mar 29 '16 at 18:59
  • @lulu i dont know but i am thinking the 7 stands for weekly with those probailities – foley Mar 29 '16 at 19:05
  • That doesn't make sense. If I expect multiple independent offers then the probabilities should not sum to $1$ (as you might get all three types of offers, for example). – lulu Mar 29 '16 at 19:10
  • I suspect the probabilities are for the highest dollar bid received in a single day. – hardmath Mar 29 '16 at 19:10
  • @hardmath That's at least possible, but it's a bit of a stretch to get there from the wording. – lulu Mar 29 '16 at 19:12
  • @lulu: The title mentioned (before I edited it) "dynamic programming", so that suggests (taken together with the wording "at the end of each of the three days, I will decide...") that the probabilities refer to events on each day independently. Of course one needs to choose the best offer of any day when accepting an offer, if the expected sale price is to be maximized. – hardmath Mar 29 '16 at 19:18
  • @hardmath u think u can help with these because i am kind of lost here – foley Mar 29 '16 at 19:19
  • I'll write up a solution based on @hardmath 's interpretation. – lulu Mar 29 '16 at 19:20
  • What does the "dynamic programming" approach mean to you? What if there were only one day of "advertising"? – hardmath Mar 29 '16 at 19:20
  • @hardmath dynamic programming for me is to get the best solution of any particular promlem .to maximize the sale price for this problem i would say – foley Mar 29 '16 at 19:29
  • It is actually a specific approach to finding "the best solution", which I hope your classroom instruction will help you appreciate. – hardmath Mar 29 '16 at 19:32

1 Answers1

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Using the interpretation of @hardmath, we may as well assume we only get one offer each day, with the stated probabilities.

Now, 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 $\frac 27$ you lose $\$100$. With probability $\frac 17$ 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 $\frac 17$). If you get another $\$300$ (probability $\frac 47$) 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 $$\frac 27\times 200+\frac 47\times 300+\frac 17 \times 400=\$285$$ Putting all this together, the expected value of waiting is $$\frac 17 \times 400+\frac 47\times 300 +\frac 27\times 285=\$310$$

this is (ever so slightly!) greater than $\$300$ so in this case you ought to wait.

lulu
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