# How to generate a good guess from other guesses, if the average has already been taken?

There are votes two days from now* and someone made an excel sheet where people at work would make their guesses on the outcome. Since I'm not from this country, I had no idea what to do and decided to be super clever and make my collumn just the average of the others. Well, it turned out that I'm not the only foreigner and a college did just that before me. I tested to see what excel would do if I take the average anyway and not surprisingly the program killed both entries, which tried to compute an average $m_1$ from other entries of which one is average $m_1$ involving $m_2$ itself. So what I ended up with manipulating his entry to exclude mine and added a positive epsilon so there is a 50%-50% change I'm closer. So much for the motivation. Now this got me thinking:

Lets say $n+2$ people, the last two being called Alice and Bob, respectively make a guess $G_i$ on the outcome of some experiment. They make their guesses publicly, one after the other, and say for all $i\in \{1,2,3,\dots,n-1,n,n+1,n+2\}$, the guess $G_i$ is some rational number. Now after the first $n$ people have publicly made their guess, alice thinks she's clever and just chooses to make a guess which is just the average of all other guesses, including bob who still has to guess: $G_{n+1}:=\frac{(\sum_{i=1}^n G_i)+G_{n+2}}{n+1}$.

Bob thinks taking an average will be a good move, but now that this is done he tends to make a slightly bigger guess than what he sees the average to be. He will give a procedure to compute his $G_{n+2}$ which shall be slightly above $G_{n+1}$. The task is to compute a number which is as close to $G_{n+1}$ as possible but not equal to it. He can only use one additional number $\varepsilon$ and there is a smallest value for this $\varepsilon$ he can manually enter into the code. So again: You can use the program to compute a number involving one instance of $+\varepsilon$ with fixed positive value at some point, and for $\varepsilon=0$ you naturally want the computed function to coincide with the average (Alice guess).

Now I've had two ideas which both seem natural: Either set $G_{n+2}:=G_{n+1}+\varepsilon$, or set $G_{n+2}:=\frac{(\sum_{i=1}^n G_i)+G_{n+1}}{n+1}+\varepsilon$. I tried some random numbers for $n$, $\sum_{i=1}^n G_i$ and $\varepsilon$ and the second approach for $G_{n+2}$ seems to be quite closer to $G_{n+1}$, but moreally, I don't really know why. Does anyone see why the one approach turns out to be superiour to the other, and are there better places to put the $\varepsilon$.

*Merkel will win again.

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Using $G_{n+2}:=\frac{(\sum_{i=1}^n G_i)+G_{n+1}}{n+1}+\varepsilon$, we have that $$\displaystyle G_{n+1}=\frac{\sum_{i=1}^n G_i+\frac{\sum_{i=1}^n G_i+G_{n+1}}{n+1}+\varepsilon}{n+1}=\frac{(n+2)\sum_{i=1}^n G_i+G_{n+1}+(n+1) \varepsilon}{(n+1)^2}$$

and then, solving for $G_{n+1}$,

$$\displaystyle (n+1)^2 G_{n+1}=(n+2)\sum_{i=1}^n G_i+G_{n+1}+(n+1) \varepsilon$$

$$\displaystyle n(n+2) G_{n+1}=(n+2)\sum_{i=1}^n G_i+(n+1)\varepsilon$$

$$\displaystyle G_{n+1}=\frac{\sum_{i=1}^n G_i}{n}+\frac{n+1}{n(n+2)}\varepsilon$$
From this, substituting in $G_{n+2}=\frac{(\sum_{i=1}^n G_i)+G_{n+1}}{n+1}+\varepsilon$, we get
$$\displaystyle G_{n+2}=\frac{\sum_{i=1}^n G_i}{n}+[\frac{1}{n(n+2)}+1]\varepsilon=\frac{\sum_{i=1}^n G_i}{n}+\frac{(n+1)^2}{n(n+2)}\varepsilon$$ so that the difference between $G_{n+2}$ and $G_{n+1}$ is $$\displaystyle \frac{(n+1)^2-(n+1)}{n(n+2)} \varepsilon =\frac{n+1}{n+2} \varepsilon$$ Because $\frac{n+1}{n+2} <1$, this shows why this approach gives a $G_{n+2}$ closer to $G_{n+1}$ than that obtained using the approach $G_{n+2}=G_{n+1}+\varepsilon$. Also note that the two approaches lead to progressively more similar results as $n$ increases.