I've been working on this problem for a while, and cannot think of an elegant solution. This is a simple example. Any pointers would be appreciated.

Numbers are evenly distributed in the interval $[0, 1]$. $5$ of these numbers can be averaged to score points. If the average of the $5$ numbers is less than $0.5$, $2$ points are scored. Otherwise, $1$ point is scored.

I am investigating the relative 'value' of these numbers. Ie, how much more useful $0.1$ is than $0.9$ (clearly an average under $0.5$ can be formed more easily with $0.1$).

I believe it follows $0.4$ is at least twice as good as $1$ for scoring points - since $0.4$ s can be averaged for $2$ points rather than $1$. I am trying to answer the general case of how much more useful one number is than another.

Any help formulating this problem more formally would be great.

Thanks very much

  • 1
    $\begingroup$ I do not see why you would want to assign a "profit value" to a number nor how you would use it in any consistent way. The question makes no sense. $\endgroup$
    – David K
    Jul 31, 2017 at 21:01
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    $\begingroup$ A minor quibble: near the start of the question you say that if the average of the five numbers is less than $0.5,$ the profit is $1.$ But later you take five numbers whose average is $0.4$ and say the profit is $2,$ contradicting the earlier rule. (Note that even if you fix this inconsistency, the question still makes no sense.) $\endgroup$
    – David K
    Jul 31, 2017 at 21:03
  • $\begingroup$ Thanks for pointing that out. I've changed the question's phrasing. Is that more clear? $\endgroup$ Jul 31, 2017 at 21:43
  • $\begingroup$ OK, the edit changes my opinion about the question. Previously it was unanswerable; now it presents what I think is a bad idea, but at least it is clear and consistent enough that I can explain why I think it's a bad idea. That makes it a reasonable question, and I have tried to answer it. $\endgroup$
    – David K
    Aug 1, 2017 at 2:04
  • $\begingroup$ I think this should be construed as "How many points is it worth paying to get a particular number as one of the five?". $\endgroup$ Aug 1, 2017 at 3:40

1 Answer 1


Let me formalize and generalize the problem. You have a sequence of $n$ random variables, $X_1, X_2, \ldots, X_n,$ such that $X_i \sim U[0,1]$ for each $i.$ Let $X = \sum_{i=1}^n X_i.$ That is, $X$ is a uniform sum distribution with parameter $n.$

Then the number of points scored is $1 + Y,$ where $$ Y = \begin{cases} 1 & \text{if } X < n/2, \\ 0 & \text{if } X \geq n/2. \end{cases} $$

That is, $Y=1$ if the mean of the $n$ uniform random variables is less than $\frac12,$ and $Y=0$ otherwise.

In order to keep the discussion relatively simple, I'll assume the values of $X_1, X_2, \ldots, X_n$ are revealed one at a time, and that the "point value" of a number $q$ at any given time is the amount you should be willing to "pay" in order to replace the next random variable with the number $q.$

You can also apply formulas based on this interpretation to a situation in which you have some kind of opportunity to choose the values of one or more of $X_1, X_2, \ldots, X_n$ and then allow the remaining variables to be assigned randomly.

Now suppose the values of $h$ of the variables are known (already randomly revealed or chosen by you), that is, we know that $X_1 = x_1, \ldots, X_h = x_h,$ and suppose you have the opportunity to set an additional $k$ variables to a certain sequence of values, $X_{h+1} = x_{h+1}, \ldots, X_{h+k} = x_{h+k}.$ The "point value" of the sequence $(x_{h+1}, \ldots, x_{h+k})$ under those conditions is the difference between the expected value of $1+Y$ given that all $h+k$ of these variables are known to have the specified values and the expected value of $1+Y$ given only the first $h$ variables. But the (conditional) expected value of $1+Y$ is just $1$ plus the (conditional) expected value of $Y$, so \begin{multline} V(x_{h+1}, \ldots, x_{h+k}) = E(Y \mid X_1 = x_1 \land \cdots\land X_{h+k} = x_{h+k}) \\ - E(Y \mid X_1 = x_1 \land \cdots\land X_h = x_h) \end{multline} In other words, the point value of the $k$ additional numbers is the amount by which we increase the conditional expectation of $Y$ when we include those $k$ numbers in the condition.

For example, with $h=0$ and $k=1$ we are assessing the point value of choosing the value of $X_1$ before any of the values are known. The point value of the numerical value $x_1$ in those circumstances is $$ V(x_1) = E(Y \mid X_1 = x_1 ) - E(Y) . $$

The conditional expectation $E(Y \mid X_1 = x_1)$ is the probability that after the remaining $n-1$ variables are chosen randomly, the total (including $x_1$) will not exceed $n/2$; that is, $$ E(Y \mid X_1 = x_1) = P\left(X_2+\cdots+X_n < \frac n2 - x_1\right).$$

The probability on the right-hand side depends on a uniform sum distribution with parameter $n-1.$ For convenience, let $F_m$ be the cumulative distribution function of a uniform sum distribution with parameter $m.$ Then $$ P\left(X_2+\cdots+X_n < \frac n2 - x_1\right) = F_{n-1}\left(\frac n2 - x_1\right). $$ Of course $$E(Y) = P\left(X_1+\cdots+X_n < \frac n2\right) = F_n\left(\frac n2\right) = 0.5.$$

The following table gives some approximate values of this probability when $n=5$ (as in the original question):

\begin{array}{lc} x_1 & E(Y \mid X_1 = x_1) \\ 0 & 0.7995\\ 0.2 & 0.6920\\ 0.4 & 0.5663\\ 0.5 & 0.5000\\ 0.6 & 0.4337\\ 0.8 & 0.3080\\ 1 & 0.2005\\ \end{array}

That is, in these circumstances the point value of $0$ is approximately $0.2995$ and the point value of $1$ is approximately $-0.2995,$ that is, we should desire to be paid at least $0.2995$ before we would be willing to choose to set $X_1 = 1.$

Going back to the general case in which $h$ variables have already been set to the values $x_1, \ldots, x_h,$ we can show that $$ V(x_{h+1}, \ldots, x_{h+k}) = F_{n-h-k}\left(\frac n2 - (x_1 + \cdots + x_{h+k})\right) - F_{n-h}\left(\frac n2 - (x_1 + \cdots + x_h)\right). $$

This is a general formula covering all cases of this kind; the fly in the ointment is that the formula for $F_m$ can be a bit annoying to work with. For the first few uniform sum distributions, the cumulative probability functions are:

\begin{align} F_1(x) &= \begin{cases} 0 & \text{if }x<0,\\ x & \text{if }0\leq x\leq 1, \\ 1 & \text{if }x > 1, \end{cases} \\ F_2(x) &= \begin{cases} 0 & \text{if }x<0,\\ \frac12x^2 & \text{if }0\leq x<1,\\ 1 - \frac12 (2 - x)^2 & \text{if }1\leq x\leq 2,\\ 1 & \text{if }x > 2, \end{cases} \\ F_3(x) &= \begin{cases} 0 & \text{if } x<0,\\ \frac16x^3 & \text{if } 0\leq x<1,\\ -\frac13 x^3 + \frac32 x^2 - \frac32 x + \frac12 & \text{if } 1\leq x<2,\\ 1 - \frac16 (3 - x)^3 & \text{if } 2\leq x\leq 3,\\ 1 & \text{if }x>3, \end{cases} \\ F_4(x) &= \begin{cases} 0 & \text{if } x<0,\\ \frac{1}{24}x^4 & \text{if } 0\leq x<1,\\ -\frac18 x^4 + \frac23 x^3 - x^2 + \frac23 x - \frac16 & \text{if } 1\leq x<2,\\ \frac18 (4 - x)^4 - \frac23 (4 - x)^3 + (4 - x)^2 - \frac23 (4 - x) + \frac76 & \text{if } 2\leq x<3,\\ 1 - \frac{1}{24} (4 - x)^4 & \text{if } 3\leq x\leq 4,\\ 1 & \text{if } x > 4. \end{cases} \end{align}

(Source: Wolfram Alpha.)

  • $\begingroup$ Not an overinterpretation. Very interesting answer. In my case, i am looking for: "How many points is it worth paying to get a particular number as one of the five?" (as another commenter said) and to then extent that question to: "How many points is it worth paying to get a particular number as one of the 4 remaining random numbers?" (once one has already been selected by the previous question). I would consider those 2 questions separate. Im looking for the function which maps real numbers $[0, 1]$ to "the number of dollars" which could be payed for them, in the context of either question. $\endgroup$ Aug 1, 2017 at 18:59
  • $\begingroup$ OK, I think I get the question now. I have updated the answer to give general formulas for calculating the "number of dollars" in each case. $\endgroup$
    – David K
    Aug 2, 2017 at 2:12
  • $\begingroup$ That answers the question perfectly, thanks. I've extended the question, hopefully in a clear way, in a separate thread (math.stackexchange.com/questions/2379477/…) can you see how the two link together? Thanks again $\endgroup$ Aug 2, 2017 at 2:22

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