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I bumped into a probability problem, proposed by an app. Here it is!

"In front of you is an infinitely-lived machine that proposes amounts of money, which you can either accept or reject. If you accept, the machine hands over the proposed amount, but shuts down and will never give you anything else. If you reject, it'll show you a new proposal next period. Each period's proposal is an independent draw from a uniform distribution on [0, 100]. The time between periods is long — several months, say — and you are impatient: a dollar next period is worth only 0.9 to you today; similarly, a dollar two periods from now is worth 0.9*0.9 = 0.81 today, et cetera. If your strategy were to always accept, you'd expect to make 50 dollars, i.e. the mean of the first draw. If instead you decided to accept any first proposal above 50, and — in case you reject the first — any second proposal whatsoever, your expected discounted payoff would be 60 dollars. But you can do better! If you follow the strategy that maximizes your expected discounted payoff, what is the threshold above which you should accept the machine's first proposal?"

Now, I solved others similar problems, but this time the payoff gets reducing by 10% every time I reject the amount and go on with the game. If you could enlighten me, I'd appreciate it.

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The best strategy now will also be the best strategy at any future turn, assuming you reach that point. And there is not point in accepting a particular low amount now but rejecting a particular higher amount now, as you would be better off swapping those decisions. So the sensible strategy is to set yourself a threshold and accept higher offers than the threshold

If your strategy is to accept any offer over a threshold $t$ then you will accept the next offer with probability $\dfrac{100-t}{100}$ and given you do accept, it will have an expected value at that time of $\dfrac{100+t}{2}.$ If you do not accept then you expect to get the net present value of playing the game in the following time period, which we might call $\dfrac{9}{10}V$

This leads to the expected net present value of the game now being $V= \dfrac{100^2-t^2}{200} + \dfrac{9t}{1000}V$, i.e. $V= \dfrac{50000-5t^2}{1000-9t}$

This is maximised when $t\approx 62.6789$ in which case $V\approx 69.6432$

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    $\begingroup$ Yes. You just beat me to it! $t\approx 62.6789$ is $\frac{100}{9}(10-\sqrt 9)$. $\endgroup$
    – TonyK
    Commented Oct 20, 2016 at 16:24
  • $\begingroup$ That's it! I was trying to find $t$, I was only close to find answer by alternative ways, but you got it. Your reasoning is enlightening. Thanks a lot, Henry. You too, TonyK, thanks. Good job, everybody! $\endgroup$
    – user380786
    Commented Oct 21, 2016 at 10:57
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    $\begingroup$ I meant $\frac{100}{9}(10-\sqrt {19})$. $\endgroup$
    – TonyK
    Commented Oct 21, 2016 at 21:15
  • $\begingroup$ @Henry I understand how you get the first term $\frac{100^2-t^2}{200}$ but I don't understand why the second term is not $\frac{9}{1000}t*50$ since rejecting the first offer leads to accepting 50 dollars on average multiplied by $0.9$ due to value loss? $\endgroup$
    – benji
    Commented Dec 9, 2016 at 22:49
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    $\begingroup$ @C.Helling Your screenshot suggests that you typed in the expected value of the game $V$ rather than the value for the threshold $t$ $\endgroup$
    – Henry
    Commented Mar 22, 2017 at 0:20

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