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Suppose there is a game where you and two people you don’t know each other pay $\$10$ each, and then you get a random number between $0$ and $1$. Whoever has the largest random number among the three of you can get $\$30$. At this time, the game has a privilege, that is, you can see what the random number is, but you can't see the numbers of the other two people, for only $\$3$. If you are not satisfied with your numbers, you can opt out of this game and your $\$10$ can be returned to you. So, is this privilege worth buying?

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    $\begingroup$ Please don't unwind my edits. You can't use dollar signs the way you are trying to; it confuses the formatting system. $\endgroup$
    – lulu
    Apr 9 at 19:41
  • $\begingroup$ If you click on edit now, you can see the syntax I used to get your dollar signs in, Modify it from here if you like, but I suggest keeping that core syntax intact. $\endgroup$
    – lulu
    Apr 9 at 19:43
  • $\begingroup$ Thanks sorry about that. $\endgroup$ Apr 9 at 19:44
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    $\begingroup$ As to the question itself, what have you tried? What's the expected maximum of the two values you don't see? $\endgroup$
    – lulu
    Apr 9 at 19:46
  • $\begingroup$ Wait, are there 3 people or 4 including you? You say two things in the problem that seem to conflict. $\endgroup$ Apr 9 at 19:53
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$Game 1 =\begin{cases}20 & \text{if win}\\-10&\text{if lose}\end{cases}$

$Game 2 =\begin{cases}17 & \text{if win}\\-3&\text{if withdraw}\\-13&\text{if lose}\end{cases}$

The assumption is that expected earnings go up in second game. Expected earnings in first game is clearly 0. Let us consider the second game. We need a strategy as well as calculation of probabilities given the strategy.

Strategy: Exists a threshold number $r^*$, above which i'll play, else i'll withdraw. Given I play, I can win or lose, with certain probabilities. Given I don't play, which happens for all $r<r^*$, I walk out with $-\$3$. Let $r_o$ indicate the maximum of the others. The distribution of the max of 2 Uniform variables has pdf $f_X(x) = 2x$

Given this strategy, the expected earnings is \begin{align*} \mathbb E[earnings]&=-3(r^*)+\int\limits_{r=r^*}^1 \left[17\int\limits_{r_{0}=0}^r 2r_0dr_odr -13\int\limits_{r_{0}=r}^1 2r_0dr_odr\right]\\ &=-3(r^*)+\int\limits_{r=r^*}^1 \left[17r^2dr -13(1-r^2)dr\right]\\ &=-3(r^*)+\int\limits_{r=r^*}^1 \left[-13dr+30r^2dr\right]\\ &=-3(r^*)-13(1-r^*) +10(1-(r^*)^3)\\ &=-10(r^*)^3+10(r^*)-3\\ \end{align*} We can optimize for the above, by setting the derivative to 0. \begin{align*} \frac{d}{dr^*}\mathbb E[earnings]&=0\\ \frac{d}{dr^*} -10(r^*)^3+10(r^*)-3&=0\\ -30(r^*)^2+10&=0\\ r^*&=\sqrt{\frac{1}{3}}\tag{0.57735} \end{align*} What is our expected earning?

\begin{align*} \mathbb E[earnings]&=-10\left(\sqrt{\frac{1}{3}}\right)^3+10\left(\sqrt{\frac{1}{3}}\right)-3\\ &=-10(1/3)^{1.5}-10(1/3)^{0.5}-3\\ &= 0.8490 \end{align*}

Therefore it is worth taking the option.

See here for distribution of max of 2 uniform random variables:maximum of two uniform distributions

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  • $\begingroup$ It would be useful to know why this answer was downvoted. $\endgroup$ Apr 10 at 8:43
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Let us assume that after looking at the number you cancel playing and lose 3 if the number less than some threshold $t$. Otherwise you continue to play and either win 20 or lose 10. Then the expected gain is: $$G(t)=-3+\int_t^1 [20x^2-10 (1-x^2)]dx=-3+10t-10t^3. $$ This curve has a maximum at $t=\frac1 {\sqrt3}\approx0.577$ which corresponds to the expected gain of $G\approx0.849$. The gain is positive, so the option worth buying (with a correctly chosen threshold) since the expected gain is $0$ in the standard game. It is possible that there exist even better strategies to increase the gain.

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EDIT: Warning, my answer is apparently incorrect. end EDIT

In order to use the information, you must decide whether to withdraw. If you get a high value, you'll want to continue, and if you get a low value, you'll want to withdraw. So you have to decide upon a cutoff $c$.

Suppose your number is $x$. If $x < c$, you withdraw and have a value of $-3$. This occurs with probability $c$. Otherwise, $x > c$ and you continue. At this point, you either win or lose.

$P(x$ is the largest of the three$) = x^2$ and the payoff is $17$.

$P($lose$) = 1-x^2$ and the payoff is $-13$.

So the expected value of this game is $$c\cdot (-3) + (1-c)x^2\cdot 17 + (1-c)(1-x^2)(-13)$$ given knowledge of $x$.

The question is, can you choose a value of $c$ such that the expected value without knowing $x$ in advance is higher than the default value of $0$?

It seems you want to integrate over the possible values of $x$. If I integrate $$\int_{0}^1 -3c + 17(1-c)x^2-13(1-c)(1-x^2) dx$$ I get $$ -10.5 + 7.5c$$ which is strictly worse than the ignorant bet for all values of $c$.

EDIT: Apparently my answer is wrong. I thought there might be something fishy about conditional probabilities in there but couldn't nail it down. I don't actually follow the other answers, and they match each other, so I will concede to them.

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    $\begingroup$ Isn't the default value 0? $\endgroup$
    – E-A
    Apr 9 at 20:06
  • $\begingroup$ You have to pay $3$ for the knowledge. $\endgroup$ Apr 9 at 20:12
  • $\begingroup$ Ah, I see what you mean now. I had to keep going through my work changing 4 to 3 and forgot that one. Thanks for the catch. Fixed. $\endgroup$ Apr 9 at 20:21
  • $\begingroup$ Very interesting outcome took me a while to absorb.. so from a mathematical standpoint it is better off to make an ignorant bet $\endgroup$ Apr 9 at 20:47
  • $\begingroup$ Apparently not, I made a mistake somewhere. Although, if the price of information were high enough, it would become not worthwhile, I assume. $\endgroup$ Apr 9 at 22:17
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While some other answers here do a fine job of going through the computations, I thought I would add one more rigourously explaining this problem in the language of probability theory.


Consider the probability space $([0,1]^3,\mathcal{F},P)$, where $\mathcal{F}$ is the Borel $\sigma$-algebra, and $P$ is the Lebesgue measure. Let $(x,y,z)\in[0,1]^3$ denote the numbers each player receives, and let $\xi$ be a random variable equal to your winnings, given by $$\xi = 20\:\!\mathbb{1}_{\{x>y\vee z\}} - 10\:\!\mathbb{1}_{\{x\le y\vee z\}}$$

Buying the privilege gives us access to the value of $x$, which we can model by letting $\mathcal{G}=\mathcal{B}([0,1])\times[0,1]^2\subset\mathcal{F}$ be the Borel $\sigma$-algebra of the set of $x$-values $[0,1]$ times $[0,1]^2$, i.e. slices of the cube $[0,1]^3$ that are parallel to the $yz$-plane. From this, we can use the conditional expectation of $\xi$ given $\mathcal{G}$, which is given by $$\mathbb{E}(\xi|\mathcal{G}) = 20x^2 - 10(1-x^2) = 30x^2 - 10$$

This tells us the expected value of your winnings for your value of $x$. Since this is monotone in $x$, we choose to play the game only if $x>k$ for some fixed $k$ to be determined later. Therefore, your new winnings are given by the random variable $\eta$, which is given by $$\eta = \xi\mathbb{1}_{\{x>k\}} - 3$$ where the indicator function $\mathbb{1}_{\{x>k\}}$ represents the fact that you only play if $x>k$. Since $\mathbb{1}_{\{x>k\}}$ is $\mathcal{G}$-measurable, your expected winnings as a function of $x$ are given by $$\mathbb{E}(\eta|\mathcal{G}) = \mathbb{E}(\xi|\mathcal{G})\mathbb{1}_{\{x>k\}} - 3$$

In order to determine whether buying the privilege is advantageous, we compute the expectation of $\eta$ as follows:

\begin{align} \mathbb{E}\eta &= \mathbb{E}(\mathbb{E}(\eta|\mathcal{G}))\\ &= \mathbb{E}(\mathbb{E}(\xi|\mathcal{G})\mathbb{1}_{\{x>k\}} - 3)\\ &= \int_k^1(30x^2 - 10)\,\mathrm{d}x\, - 3\\ &= 10(k-k^3) - 3 \end{align}

At $k=\frac{1}{\sqrt{3}}$, this reaches a maximum of $\mathbb{E}\eta\approx0.849$. Since $\mathbb{E}\xi=0$, we conclude that it is advantageous to buy the privilege and only play the game when your number is at least $\frac{1}{\sqrt{3}}$.

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A simple proof that it is worth paying to find your number is to display a strategy where it gains. We are not asked to find the optimum strategy. If we don't pay to find our number the game is fair and our expected result is $0$. My strategy is to pay to find the number, then drop out if it is less than $\frac 12$. I then lose $3$ when the number is less than $\frac 12$, win $30$ when it is greater than $\frac 12$ and I have the highest number, and lose $13$ when my number is greater than $\frac 12$ but I lose. If I keep my number $x$ I win with probability $x^2$ because both other players have to have a lower number, which they do with probability $x$. My expected value is then $$\frac 12(-3)+\int_{\frac 12}^130x^2dx-\int_{\frac 12}^1(-13)(1-x^2)dx\\ =-\frac 32+\left.(10x^3-13x+\frac{13}3x^3)\right|_{\frac 12}^1\\ =-\frac 32+\frac {43}3\cdot \frac 78-\frac {13}2\cdot \frac 12\\ =\frac{-36+301-78}{24}\\ =\frac{187}{24}$$ As this strategy shows a gain over doing nothing, it is worth paying the $\$3$ to find our number. There may be (and another answer shows there is) a better threshold for dropping out, but that can only improve the result further. I chose $\frac 12$ as a guess that seemed reasonable and easy to calculate with.

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