Consider four subjects (A,B,C,D) competing 12 times in a game with 5 possible scores (5,6,7,8,9):

A) Median: 7  Mean: 6.92  Mode: 5
    5:#####              avgL||avgR
    6:                    6.9||8.5 (+3.2)
    7:##                555557789999
    8:#                      ^^

B) Median: 7  Mean: 7.08  Mode: 7
    5:#                  avgL||avgR
    6:###                 6.2||8.0 (+1.8)
    7:####              566677778899

C) Median: 7  Mean: 7.00  Mode: 7
    5:                   avgL||avgR
    6:                    7.0||7.0 (±0.0)
    7:############      777777777777

D) Median: 7  Mean: 6.83  Mode: 7
    5:##                 avgL||avgR
    6:##                  6.0||7.7 (+1.7)
    7:######            556677777799

In this case: Who won?

  • Going by the median, they're all equal
  • A has the lowest mean and the lowest mode, but the highest average right half of the scores
  • B has the second-highest mean and the most normal distribution
  • C has the most consistent score
  • D has the highest mean, but the second-lowest skew toward the higher scores

Would it make sense to average out or otherwise combine a few metrics?

  • 4
    $\begingroup$ The winner is whoever has the highest score of the metric of your choosing. In case of the median then everyone (or no one) is a winner. In case of the mean it's D. avgL and avgR are very arbitrary. $\endgroup$ – user2974951 Dec 16 '20 at 6:40
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    $\begingroup$ Why do you think that the median or mean do not answer your question, about who is the winner? Why do you think they are inappropriate? Does the shape of the distribution bother you? $\endgroup$ – user2974951 Dec 16 '20 at 6:44
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    $\begingroup$ This is a subjective question. The average score (D wins) makes the most sense to me because of the simplicity. Btw, I like the way you present the data. $\endgroup$ – Kota Mori Dec 16 '20 at 7:03
  • $\begingroup$ For right skewed data one often has mode < median < mean. So maybe it makes sense just to use the median. These three measures of centrality have very different definitions and thus different appropriate uses in practice, so it would be difficult to explain what criterion for 'best' is expressed by an average of mean, median, and mode. [For left-skewed data its often mode > median > mean; for roughly symmetrical unimodal data (think normal) the three may be about equal.] $\endgroup$ – BruceET Dec 16 '20 at 7:19
  • $\begingroup$ After some additional thought, it seems I should have mentioned (as did @NickCox) that with data from only 12 games played by each participant, you may have ties values for the 'mode'. This strengthens @FrankHarrel's advice against averaging mean, median, and mode. // Perhaps more important, it seems that your ultimate objective is to find a way to say which participant is would be 'best' in a match up between any two. I have posted an Addendum showing one way that might be answered. $\endgroup$ – BruceET Dec 18 '20 at 2:34

Three examples:

Right-skewed data. Salaries within an organization may be right-skewed. Mode is the most common salary, mean is total payroll divided by number of employees considered. Any one of mode, median, mean might be the one best way to describe or summarize the data.

The summary of the data below shows the median and mean of hypothetical data.

set.seed(2020);  x = rgamma(500, 3, .01)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  20.69  175.49  268.61  294.28  384.20  964.41 

Roughly, the midpoint (250) of the tallest histogram bar might be used as the 'mode' of the sample. A better way to estimate the mode of the population distribution might be to find where (234) the kernel density estimator (KDE) of the sample is a maximum.

It is not clear what attribute of the data the average (about 266) of the mean, median, and mode might describe. As mentioned in my comment, this is not far from the median, which does have a clear meaning.

hist(x, prob=T, col="skyblue2", main="500 Salaries")
 lines(density(x), col="red", lwd=2)

enter image description here

mx = max(density(x)$y); mx
[1] 0.002711106
mode = mean(density(x)$x[density(x)$y==mx]);  mode
[1] 233.6744

[Note: For this gamma population distribution the mean, median, and mode are 300, 267.4, and 200, respectively.]

Symmetrical, unimodal data. For data from a normal population, the mean, median, and mode may be nearly the same (all about 100 for the sample below). Perhaps averages are most often used to describe the center of such a distribution; but for various purposes the mode or median might be used as well, and if they are all nearly the same, it doesn't much matter.

set.seed(1215);  y = rnorm(500, 100, 15)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  49.59   91.30   99.69  100.44  109.07  146.51 

hist(y, prob=T, col="skyblue2", main="500 Test Scores")
 lines(density(y), col="red", lwd=2)

enter image description here

Left-skewed data. Here the mean (17.1) tends to be smallest and the mode (18.5) largest. Depending on the type of device under study, any one of these three measures of centrality might be most useful.

set.seed(1159);  v = 24*rbeta(200, 10, 4)
  Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  10.23   15.30   17.50   17.13   19.13   22.13 

hist(v, prob=T, col="skyblue2", main="200 Voltages")
 lines(density(v), col="red", lwd=2)

enter image description here

mx = max(density(v)$y); mx
[1] 0.1479789
mode = mean(density(v)$x[density(v)$y==mx]);  mode
[1] 18.51482

[Note: For this population distribution, a multiple of a beta distribution, the mean, median, and mode are 17.14, 17.39, and 18.0, respectively.]

Addendum. Of two participants, which is best? One answer is to try to predict which of the two would win if the participants faced each other in a match:

Consider A vs. D. On the 12 games for which you have data A averaged 6.92 and D averaged 6.83.

# A vs D
A = c(5,5,5,5,5,7,7,8,9,9,9,9)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  5.000   5.000   7.000   6.917   9.000   9.000 
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  5.000   6.000   7.000   6.833   7.000   9.000 

It seems reasonable to ask whether A is likely to win in a match with D. By choosing performances on the 12 earlier games at random, we can simulate a thousand matches between A and D. It seems that A beats or ties D about 63% of the time.

a = sample(A, 1000, rep=T)
d = sample(D, 1000, rep=T)
mean(a >= d)
[1] 0.628

Playing around with a few other such hypothetical tournaments of 1000 games, I found that the player with the higher average does best. But it is possible to make up player histories of 12 games where this is is not the case when averages over the 12 games are nearly equal. I will let you play around with similar simulated tournaments. If this criterion for 'best' appeals to you, you might search this site and google for information on stochastic domination.

Here is one more hypothetical tournament---for B vs. C, both of whom have averages near 7 for the 12 games you report. However, B has more games with high scores, which seems to give B an advantage: B wins or ties about 67% of the time.

B = c(5,6,6,6,7,7,7,7,8,8,9,9)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  5.000   6.000   7.000   7.083   8.000   9.000 
C = rep(7, 12)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
      7       7       7       7       7       7 

# B vs C
b = sample(B, 1000, rep=T)
c = sample(C, 1000, rep=T)
mean(b >= c)
[1] 0.673
  • 5
    $\begingroup$ Executive summary, perhaps: If measures agree, it doesn't much matter in practice what you use, except that on principle you might prefer one. If measures disagree, they're capturing different aspects of the distribution, and quoting two or three might be informative. $\endgroup$ – Nick Cox Dec 16 '20 at 9:45
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    $\begingroup$ And (not explicitly noted above) if knowing population parameters is an issue: for a given sample size the mean is often most accurately estimated and the mode least accurately estimated. $\endgroup$ – BruceET Dec 16 '20 at 9:51
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    $\begingroup$ In this particular case, I would worry first that there are ties for mode given the discreteness of the variable. $\endgroup$ – Nick Cox Dec 16 '20 at 10:03
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    $\begingroup$ @Frank Harrell The general rule seems sound, but still to me leaves some scope for exceptions. The midrange is the average of the minimum and maximum. It is, or was, quite often used as a summary, e.g. in reporting from min-max thermometers observed daily. It is a practical choice whenever those extremes are the only available measurements. Daily range of temperatures is, however, unusual because all the temperatures between extremes are known to have been experienced in the previous 24 hours. $\endgroup$ – Nick Cox Dec 16 '20 at 15:45
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    $\begingroup$ @Vaelus I doubt that Bruce ET is in a better position than the OP to know what is best for the problem. You're always at liberty to post an answer yourself if dissatisfied. $\endgroup$ – Nick Cox Dec 16 '20 at 17:22

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