You and your two friends are playing a game where you each draw a random real number, $r_i$ , between 0 and 1. Call this your score. You win the game if you draw the largest score of all: $\max \{r_1,r_2,r_3\}$

In two of every three games, you lose the game. However, on the occasions you do win you tend to go big.

Assume that you and your two friends play this game long enough to collect accurate statistics. Out of all the occasions in which you win the game, what is your average score?

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Assumptions
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- There is no betting component to this problem.

The answer is 0.75.

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Still not able to understand, Can someone explain in any better way?

Hari Om Swarnkar
- 7 years, 2 months ago

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Exactly what are you not able to understand?

Snehal Shekatkar
- 7 years, 2 months ago

Why doenst $\int_{0}^{1}rp(r)$ gives the expected value already? Does it is cames from a conditional probability? P(X1>r|max(X2,X3) = r )

Manoel Abranches
- 5 years, 8 months ago

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The average value is the integral $\int_0^1 \int_0^1 \int_0^1 max(x,y,z) \, dx \, dy \, dz$ (divided by the volume of the unit cube, which is 1). By symmetry this is $3$ times the integral over the region where $z$ is the max. So we get $3 \int_0^1 \int_0^z \int_0^z z \, dx \, dy \, dz = 3 \int_0^1 z^3 \, dz = \fbox{3/4}$ .

The same proof shows that the average value for $n$ friends is $\frac{n}{n+1}$ .

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Letting $r_0=0$ , we are looking for $E( \max \{ r_1,r_2,r_3\}) =E(M_0)$ where

$M_i= {\max}_j \{ r_i-r_j - \lfloor r_i-r_j \rfloor \}$

To complete the solution we need two observations:

First, $E(M_0)$ is unaffected by the value of $r_0$ . In particular, then, instead of $r_0=0$ we could assume that $r_0$ is chosen at random just like $r_1,r_2,r_3$ .

Second, $M_0+M_1+M_2+M_3=3$ .

Together, these imply that $E(M_0)=E(M_1)=E(M_2)=E(M_3)=\frac34$ .

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Can you explain $M_i = \max_j\{r_i-r_j-\lfloor r_i-r_j\rfloor\}$ and $r_0$ ?

M0+M1+M2+M3 = 3. i dont get anything from that. in fact i started feeling like im zero in this subject. Can u provide links where i could learn the basics on that and also explain it again in simplest way if possible. Thanks

Suresh Kumar
- 7 years, 2 months ago

I got the answer on the 3rd try, with no computation but pure logic, wow!

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so the mean score of each player is 0.5 because I assumed the distribution is uniform. If I am to beat the two players my score would need to range between 0.5 to 1. In the long run, the average would be in between that range, and that is 0.75.

Leo Draper
- 7 years, 2 months ago

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My sense is that my average winning score will be (N/N+1) where N is the number of players. So, while .75 is in fact correct for 3 players (3/(3+1)), it would be different for 4 or 5 or more players...

Josh Grotstein
- 7 years, 2 months ago

thats how i solved. i dono frm whr do these get those formulas & Equations.

Suresh Kumar
- 7 years, 2 months ago

If I am to beat the two players my score would need to range between 0.5 to 1

What range would your score need to be in if you were only playing against one player?

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2/3 (or .666), no?

Josh Grotstein
- 7 years, 2 months ago

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@Josh Silverman – The former: that the average winning score would be .666 (or 2/3) in the two player variant of the game.

Josh Grotstein
- 7 years, 2 months ago

Care to share?

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The average for all players must be same so lets calculate average for first player only. The probability that $r_{1}$ is greater than $r_{2}$ is $r_{1}$ because distributions are uniform. Similarly, probability that it is greater than $r_{3}$ is also $r_{1}$ . Hence the probability that it is greater than both of them is $r_{1}^{2}$ . Thus the density function in this case is:

$p(r)=r^{2}$

Hence the average value is given by:

$<r>=\frac{\int_{0}^{1}rp(r)}{\int_{0}^{1}p(r)}$

Note that in this case $\int_{0}^{1}p(r)\neq 1$

This gives us $\boxed{Average=0.75}$