I have recently encountered the remark that if a logit model's pseudo $R^2$ is lower than $0.5$ the result is completely worthless because a coin toss is a better model. Is this interpretation correct? If not, what is wrong with it?

$R^2 = \frac{\text{#Correct}}{\text{Total count}}$

  • 2
    $\begingroup$ This would depend on your definition of your pseudo R^2. For example, are you using a Count R^2? Efron's? McFadden's? The choice affects the interpretation. $\endgroup$
    – A. Webb
    Mar 27, 2015 at 14:57
  • $\begingroup$ That would be the plain count $R^2$. $\endgroup$
    – Constantin
    Mar 27, 2015 at 15:06
  • $\begingroup$ Plain count R^2 is number correct out of total. So if your model gets the answer right less than 50% of the time, then I'd agree with that interpretation if you are basing decisions off of rounding the outcomes to binary responses, e.g. p < 0.5 versus p>=0.5. However, the model provides more information than a coin flip in that perhaps near p = 0.5 your decision is a third response, e.g. that the results are indefinite. In some cases this would improve overall results. $\endgroup$
    – A. Webb
    Mar 27, 2015 at 15:11

1 Answer 1


(Percent correct)/(Total count) is usually termed the Correct Classification Rate. This is not one of the pseudo-R-squared indicators, and it's generally considered an inferior way of assessing model fit because it simplifies so much; it doesn't take into account the differences in predicted probability from observation to observation.

The pseudo-R-squared can be calculated in several different ways, as noted by @A. Webb and @kjetil b halvorsen, but by any of those methods, a result not just of 0.50 but even of, say, 0.03 will, in a sample of a few hundred or a thousand, reflect a model that is much more informative and/or a better guide to decision-making than a simple coin flip. This can be seen concretely by comparing the two distributions of predicted probabilities the model generates: one for observations with a "1" on the dependent variable and one for those with a "0." The predicted probabilities for the "1"s will be noticeably shifted right relative to those for the "0"s. An ROC curve, too, will mark out noticeably more area for this model than it would for a null model based on coin flips.


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