I'm trying to understand the concept of the expected value. Especially, what bothers me is the expected value for discrete random variables. I will try to formulate it by examples:

  • The expected value when throwing a six-sided die is $E(X) = 3.5$, which as I see it, is not even part of the probability space $\Omega = \{1,2,3,4,5,6\}$. Why is this not a problem at all? Is here anything wrong with my assumptions about the expected value?

  • Is there an expected value when throwing a coin? I cannot calculate a weighted mean on head and tail? Is there an expected value? Is it the maximum likely value, and what if there are multiple maximum likely values?

  • 2
    $\begingroup$ The expected value does not mean the value that you expect to see on the next roll of the die; it is what you would expect to be the average value of the outcomes on a large number of rolls. Roll the die $100$ times. You are likely to see a total of $350$ dots for an average of $350/100 = 3.5 = E[X]$. This is not guaranteed, you might get some number smaller than $350$ (in fact the minimum is $100$, right?) or larger (as much as $600$), but this is quite unlikely. Betting that you will see roughly $350$ dots on $100$ rolls is more profitable than betting on roughly $12$. $\endgroup$ Apr 27, 2012 at 11:11
  • $\begingroup$ The expectation of a nominal variable is the vector of probabilities of each possible outcome - see the multinomial distribution - en.wikipedia.org/wiki/Multinomial_distribution $\endgroup$
    – Macro
    Apr 27, 2012 at 11:15
  • 1
    $\begingroup$ Also, while the value of a die roll is discrete, it is not nominal. Count variables don't fit Stevens scheme very well see [this blog post of mine] (statisticalanalysisconsulting.com/…) $\endgroup$
    – Peter Flom
    Apr 27, 2012 at 11:15
  • $\begingroup$ Regarding coins, if the outcome is "number of heads" then there is an expected value. $\endgroup$
    – Peter Flom
    Apr 27, 2012 at 11:16
  • $\begingroup$ Sorry. That last $12$ in my previous comment should have been a $120$ but I noticed the typo too late to fix it. $\endgroup$ Apr 27, 2012 at 12:32

1 Answer 1


There are several different measures of "location" or "central tendency". Expected value is the most popular one, but there are others -- median, mode, geometric mean, etc.

While all measures of central tendency are, in some ways, similar, it is important to remember that they actually measure different things. Here are the interpretations.

  1. Expected value (or mean). Expected value is useful for calculating the total when you have a large number of observations. Say you own a company with a large number ($N$) of employees. The expected salary is $E[X]$. The total salary that you have to pay out is $N E[X]$.
  2. Median. Median tells you about the typical observation. If you want to know what the typical person earns, look at the median salary, not the expected salary.
  3. Mode. Mode tells you the most likely outcome. If you are applying for jobs in a particular field, the modal salary is what you will most probably earn, not the expected salary.

There is confusion because, with symmetric distributions, which are common, mean, median, and mode are numerically equal. So people just think of the mean and not the median or mode. But, depending on the problem, what you are actually interested in could be the mode, even if it is numerically equal to the mean.

Now, let's look at your specific questions.

  • Suppose each time you throw a die, you earn the amount of money that comes up. After throwing it a lot of times, you will earn $N E[X]$. The "fair price" that someone could charge you for making these throws is $N E[X]$.
  • If you assign head and tail to numbers (think: amounts that you will earn or pay), then there certainly is an expected value. For $H = 0$, $T = 1$, $E[X] = 0.5$.
  • $\begingroup$ A symmetrical distribution might have an even number of modes (e.g. 2) in which case the mode won't equal the mean or the median. A simple example has 0 and 1 with probability 0.5 in which case the mean and median are 0.5, but that's not the mode too. These distributions may be uncommon but they are possible. For that matter, exactly symmetric distributions are uncommon too. $\endgroup$
    – Nick Cox
    Jul 4, 2023 at 15:44

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