How much money is needed to play this simple number game?

[member 428835]

I pick a number $n$ from 1 to 100 (integers only). If you guess correctly you win $n$ dollars, else you get zero. How much would you pay to play?

Evidently the solution is to pick $k$ with probability $1/k$, so expected payout is $(\sum_{i=1}^{100} 1/j )^{-1} = 0.2$. But where's the $k$ info coming from? I really don't understand the solution.

 

[phinds]

Where did you get that solution?

 

[member 428835]

Where did you get that solution?

From the book that posed the question. It's an interview book for stats and quant stuff. Lot's of random problems

 

[mathman]

Missing information: cost to play, what criterion you use to pick a number.

 

[Office_Shredder]

The correct solution should make the person picking the number indifferent between picking any number. If, for example, the other player would make the most money by picking 7, then you could select 7 slightly less often, and they would still make the most money by picking it, but they would make less money.

So if $p(n)$ is the probability you pick $n$, you need to have things like $7p(7)=15p(15)$ be true for the other player to think that 7 and 15 are equally good solutions. In particular, $np(n)=p(1)$, so $p(n)= \alpha/n$ for some constant $\alpha$. The constant is definitely not 1, as if it was, the probabilities would add up to more than 1.

 

[WWGD]

Isn't the expected return the sum of n-1/n)

Missing information: cost to play, what criterion you use to pick a number.

The question is what cost are you willing to incur.

 

[WWGD]

If I understood correctly, you win any amount with probability 1/100. So your expected gain is (1+2+...+100)(1/100)=[100(100+1)]×(1/2)[1/100]=50.5.

So any value 50.5 or lower would work.

 

[member 428835]

If I understood correctly, you win any amount with probability 1/100. So your expected gain is (1+2+...+100)(1/100)=[100(100+1)]×(1/2)[1/100]=50.5.

So any value 50.5 or lower would work.

This is true if you get the amount you guess, so say a 100-sided di where you keep the roll. But that is not the game being played here, where you must guess what the other person chose.

The correct solution should make the person picking the number indifferent between picking any number. If, for example, the other player would make the most money by picking 7, then you could select 7 slightly less often, and they would still make the most money by picking it, but they would make less money.

Okay this makes sense. Let me clarify: person A selects a number 1-100, and person B is offered to play the game where they guess which number. If they guess correct they win that number. In order to calculate person B's expected value, we first must determine how player A selects their number, which brings us to your post.

So if $p(n)$ is the probability you pick $n$, you need to have things like $7p(7)=15p(15)$ be true for the other player to think that 7 and 15 are equally good solutions. In particular, $np(n)=p(1)$, so $p(n)= \alpha/n$ for some constant $\alpha$. The constant is definitely not 1, as if it was, the probabilities would add up to more than 1.

So are we saying $p(n) = \alpha/n$ is the probability player A selects number $n$. Then player B's expected value is $\sum_{n=1}^{100}np(n)^2$ where $np(n)$ is the expected value of player A's selection, and we have probability $p(n)$ selecting it. But this implies EV = $\alpha^2\sum_{n=1}^{100}1/n \neq (\sum_{n=1}^{100}1/n)^{-1}$ where $\alpha = 1/k$ from post 1. What am I missing?

 

[sysprog]

To me the problem statement looks incomplete.

I pick a number $n$ from 1 to 100 (integers only). If you guess correctly you win $n$ dollars, else you get zero. How much would you pay to play?

Evidently the solution is to pick $k$ with probability $1/k$, so expected payout is $(\sum_{i=1}^{100} 1/j )^{-1} = 0.2$. But where's the $k$ info coming from? I really don't understand the solution.

Please pose the complete question as stated where you found it..

 

[member 428835]

To me the problem statement looks incomplete.

Please pose the complete question as stated where you found it..

LOL I literally did. Sorry you don't like it. I don't either, but think Office_Shredder gets it.

 

[WWGD]

This is true if you get the amount you guess, so say a 100-sided di where you keep the roll. But that is not the game being played here, where you must guess what the other person chose.Okay this makes sense. Let me clarify: person A selects a number 1-100, and person B is offered to play the game where they guess which number. If they guess correct they win that number. In order to calculate person B's expected value, we first must determine how player A selects their number, which brings us to your post.

So are we saying $p(n) = \alpha/n$ is the probability player A selects number $n$. Then player B's expected value is $\sum_{n=1}^{100}np(n)^2$ where $np(n)$ is the expected value of player A's selection, and we have probability $p(n)$ selecting it. But this implies EV = $\alpha^2\sum_{n=1}^{100}1/n \neq (\sum_{n=1}^{100}1/n)^{-1}$ where $\alpha = 1/k$ from post 1. What am I missing?

My bad, I read it carelessly.

 

[Hornbein]

The idea seems to be that the game is set up to minimize payout. So the probabilities are 1/k times some constant so they sum to one. It makes no difference which k the contestant chooses as far as expectation goes. The constant is equal to the expectation, so that's the max a player should pay.

 

[willem2]

So are we saying $p(n) = \alpha/n$ is the probability player A selects number $n$. Then player B's expected value is $\sum_{n=1}^{100}np(n)^2$ where $np(n)$ is the expected value of player A's selection, and we have probability $p(n)$ selecting it. But this implies EV = $\alpha^2\sum_{n=1}^{100}1/n \neq (\sum_{n=1}^{100}1/n)^{-1}$ where $\alpha = 1/k$ from post 1. What am I missing?

Note that there is only one value of α that will make the sum of the sum of the p_n add up to one. (p_n = α / n).
Since the first player chose the p_n to make the expected value for the second player equal for all n, the second player can chose any n, in particular n = 1, and the expected value of player 2 is p₁ = α.

 

[Office_Shredder]

So are we saying $p(n) = \alpha/n$ is the probability player A selects number $n$. Then player B's expected value is $\sum_{n=1}^{100}np(n)^2$ where $np(n)$ is the expected value of player A's selection, and we have probability $p(n)$ selecting it.

This is mostly unnecessary. $p(n)$ is the probability of player A picking $n$. B can do whatever they want to do. In particular, they don't have to pick a random number. If they pick $k$ (always) then their expectancy is $kp(k)=k\alpha/k=\alpha$. This is the same number no matter which number they pick, which was the whole point.

So now you need to compute $\alpha$ which makes all the probabilities sum to one.

You can also do what you did. All of your math is correct, though overcomplicating B's choice. It turns out your final expression of $\alpha^2 \sum \frac{1}{n}$ is also correct. The last equation you wrote down has an unknown, how can you conclude it's wrong if you don't know what $\alpha$ is? Solve for $\alpha$ and try again :)

 

[member 428835]

This is mostly unnecessary. $p(n)$ is the probability of player A picking $n$. B can do whatever they want to do. In particular, they don't have to pick a random number. If they pick $k$ (always) then their expectancy is $kp(k)=k\alpha/k=\alpha$. This is the same number no matter which number they pick, which was the whole point.

So now you need to compute $\alpha$ which makes all the probabilities sum to one.

You can also do what you did. All of your math is correct, though overcomplicating B's choice. It turns out your final expression of $\alpha^2 \sum \frac{1}{n}$ is also correct. The last equation you wrote down has an unknown, how can you conclude it's wrong if you don't know what $\alpha$ is? Solve for $\alpha$ and try again :)

Got it! So $\sum p(n) = 1 \implies \sum \alpha/n = 1 \implies \alpha \approx 0.192776$. Plugging this into the EV $\alpha^2 \sum 1/n = 0.192776$ which agrees with the book solution. Wow, thanks for this, I learned a ton here.