I am trying to solve this probability puzzle: https://math.stackexchange.com/questions/1977360/expected-value-from-a-game-which-gives-random-numbers-from-1-to-100-with-reducin#:~:text=%22In%20front%20of%20you%20is,a%20new%20proposal%20next%20period.
The puzzle 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?"
For a start, I am trying to calculate the expected winnings if I choose the threshold to be 50 dollars. I understood the solution mentioned in the above link but I came up with another approach which does not give the correct value when the threshold is 50 dollars. As per the question, for this threshold, the expected discounted payoff will be 60 dollars. But as per the formula derived in the solution, it comes out to be 68.18 dollars.
$$ V = \frac{(50000-5t^2)}{(1000-9t)} = 68.18, \text{for } t = 50$$.
Does the question contain the wrong value?
Now here is my approach:
Let $U_1, U_2, ... , U_N$ be i.i.d. drawn from $U(0,100)$ distribution. Here N is a random variable following geometric distribution with p = 0.5 i.e. N is the first period where we observe the proposed money to be more than 50 dollars. We then need to find $E[0.9^NU_N]$ i.e. the present value of our winnings. Now conditioning of N we can write,
$$E[0.9^NU_N] = E[E(0.9^NU_N|N)] = E[0.9^NE(U_N|N)] = 75E[0.9^N]$$
As 75 will be the expected value of $U_N$ if its more than 50 and less than 100. Now,
$$E[0.9^N] = \sum_{n=1}^\infty \frac{0.9^n}{2^n} = \frac{9}{11}$$
as $P(N = n) = 1/2^n$ i.e the probability of amount less than 50 in first $n-1$ draws and amount more than 50 in the $n^{th}$ draw. So the answer I get is 61.36 dollars which is incorrect. What is wrong with my solution?
If you accept the first offer when it is over $50$ then there is a probability of $\frac12$ of accepting an expected amount of $75$.
If you otherwise accept the second offer unconditionally then there is also a probability of $\frac12$ of accepting an expected amount of $50$ next time but that is only worth $0.9\times 50=45$ now. So the overall expected gain is $\frac{75}{2}+\frac{45}{2}=60$ as in the question.
But if you otherwise always accept later offers when they exceed $50$ then expected value now is $$\frac12\times 75 + \frac14 \times 0.9^1\times75 + \frac18 \times 0.9^2\times75+\cdots \\= \frac{75}{2}\sum_{n=0}^\infty \left(\frac{0.9}{2}\right)^2 \\=\frac{75}{2}\frac{1}{1-\frac{9}{20}} =\frac{750}{11} \approx 68.1818$$ as my formula in the linked answer suggested.