# Highest Card Draw Problem [closed]

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?

• Please don't unwind my edits. You can't use dollar signs the way you are trying to; it confuses the formatting system.
– lulu
Apr 9 at 19:41
• 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.
– lulu
Apr 9 at 19:43
• Thanks sorry about that. Apr 9 at 19:44
• As to the question itself, what have you tried? What's the expected maximum of the two values you don't see?
– lulu
Apr 9 at 19:46
• Wait, are there 3 people or 4 including you? You say two things in the problem that seem to conflict. Apr 9 at 19:53

$$Game 1 =\begin{cases}20 & \text{if win}\\-10&\text{if lose}\end{cases}$$

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

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 $$r, I walk out with $$-\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

• It would be useful to know why this answer was downvoted. Apr 10 at 8:43

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.

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.

• Isn't the default value 0?
– E-A
Apr 9 at 20:06
• You have to pay $3$ for the knowledge. Apr 9 at 20:12
• 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. Apr 9 at 20:21
• Very interesting outcome took me a while to absorb.. so from a mathematical standpoint it is better off to make an ignorant bet Apr 9 at 20:47
• Apparently not, I made a mistake somewhere. Although, if the price of information were high enough, it would become not worthwhile, I assume. Apr 9 at 22:17

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 $$\frac{1}{\sqrt{3}}$$.

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.