With <em>t</em> = 2.5 and 98 degrees of freedom, <em>r</em> = 0.245. With the same <em>t</em> and 18 degrees of freedom, <em>r</em> = 0.508. The conversion that produces them approximates nothing: it is <strong>algebra</strong>, the same quantity written two ways. And that sets it apart from the conversion from <em>d</em> to <em>r</em>, whose textbook 4 is really an approximation that depends on how many people you have in each group.
r = t / √(t² + df) r² = t² / (t² + df) t = r · √df / √(1 − r²)
t = 2.5: df = 18 → r = 0.508 df = 98 → r = 0.245 df = 998 → r = 0.079
Take your dependent variable and regress it on the group coded 0/1. The t of the coefficient is your t, and the R² of that model is exactly t²/(t²+df). There are not two quantities that resemble each other: there is one, written in two notations. That is why the conversion carries no assumption of its own. The assumptions are the ones you already accepted when running the test — independence, equal variances, normality of residuals — and the r adds none on top.
Compare it with d to r. That one also ends in a point-biserial, and there is no latent distribution involved; what happens there is something else: the formula everybody copies, r = d/√(d²+4), is not exact. The correct term is N(N−2)/(n₁·n₂), which with equal groups is 4(n−1)/n. With 20 per group that is 3.8 and not 4, and with d = 0.50 the published r falls 0.006 short; with 200 per group the difference drops to 0.0006 and stops mattering. The figure depends on the effect: with the same 20 per group, a d of 0.20 falls 0.003 short and one of 1.00, 0.009. Here there is nothing equivalent to declare: it is enough to say which t and which df you used.
The way back does not check what it seems to. If you undo the conversion with t = r·√df/√(1−r²) using the same df you entered, you always recover your t, even if that df was wrong: it is the same equation rearranged, not a second measurement. With df = 20 typed in where 18 belonged, t = 2.00 gives r = 0.408 and the way back happily returns 2.000000. Where it does help is when both the r and the t come from the paper and you supply the df: if your df does not reproduce the published t, your df is not the test’s. The p is preserved, but with a condition: the same df and both two-tailed. The r is always tested two-tailed, so if your t was one-tailed its p was half and you will not reproduce it. With df = 30, a t of 1.70 has a one-tailed p of .050 — significant — and the corresponding r of .296 comes out with p = .099.
With t = 2.5, the degrees of freedom decide the result. With df = 18 you get r = 0.508; with df = 98, 0.245; with df = 998, 0.079. One statistic, three effect sizes a reader would classify as large, between small and medium, and negligible. It is not a quirk of the formula: the t measures evidence against the null and the r measures magnitude, and with a thousand people very little magnitude is needed to accumulate evidence.
Hence the practical rule. A t without its df lets you compute nothing, and an r without its N neither: you cannot rebuild the interval or the weight it would get in a meta-analysis. Always publish the three together, even if the journal only asks for two.
The square reads directly as a proportion of variance. Those three correlations are 25.8%, 6.0% and 0.6%. It is worth looking at before writing the discussion: an effect that sounds respectable in the r metric sometimes explains 6% of the variability, and that sentence changes the tone of the whole paragraph.
The point-biserial is not the correlation you would have obtained with a continuous predictor. Splitting in two compresses: with 50/50 groups the point-biserial is about 0.80 times the underlying biserial, with 80/20 about 0.70 and with 90/10 about 0.58. The factor is φ(zp)/√(p·q) and depends only on the split, but it exists only if you accept that under the dichotomy there is a normal variable. It is worth saying out loud, because it is precisely the assumption the t → r conversion does not need: without latent normality there is no biserial to estimate and no factor to compute, even though the point-biserial remains exact. What changes with the split is not the exactness: it is the quantity being pointed at.
The consequence is for tables and for meta-analysis. If you put point-biserial correlations from trials and Pearson correlations between two continuous variables in the same column, the group split has slipped in as a moderator without you declaring it. Either model it as such, or separate the families, or say in methods that you lived with it. What you cannot do is keep quiet and average.
A nuance that gets forgotten: if the group is experimental, the compression is not a defect to correct. There is no latent continuum down there to recover — the assignment really is 0/1 — and the biserial correction only makes sense when you yourself dichotomized something continuous. In that case, the problem is not the conversion: it is the dichotomization, and no formula returns the information you already threw away.
All thirteen effect size conversions, grouped by what each one is for, are in the index.
If you use it in your thesis or in a paper, cite it so that readers of your Methods section know how the calculation was done.
Ciudad-Fernández, V. (2026). t to point-biserial r converter [Web application]. Ciudad Science. Retrieved October 14, 2026, from https://analisisdedatospsicologia.com/en/convert/t-to-r
For your Methods section: The point-biserial correlation was obtained from the t statistic and its degrees of freedom, using the Ciudad Science t to point-biserial r converter (Ciudad-Fernández, 2026).
@misc{ciudadfernandez2026convtar,
author = {Ciudad-Fern{\'a}ndez, V{\'i}ctor},
title = {{t to point-biserial r converter}},
year = {2026},
publisher = {Ciudad Science},
howpublished = {Web application},
url = {https://analisisdedatospsicologia.com/en/convert/t-to-r},
urldate = {2026-10-14},
note = {Accessed 2026-10-14},
}The test’s: n₁ + n₂ − 2 with two independent groups. Using the total N instead inflates the denominator and shrinks the r. With t = 2.00 and N = 20, the correct df (18) give 0.426 and the N gives 0.408. If the paper reports the df, copy them as they are.
With equal group sizes Welch’s statistic and Student’s coincide to the last decimal; only the df change. Use df = n₁ + n₂ − 2 and the r you get is the exact point-biserial. If you enter Welch’s fractional df, which are smaller, the r goes up and stops being exact.
A number comes out, but it is not the correlation between group and dependent variable: it inherits the correlation between the two measurements. With a pre-post that correlates 0.80, that r sits above the one from a between-subjects design with the same effect. For pre-post, report d_z and say so explicitly.
The clean route is to compute the interval of the test with the noncentral t and convert its two limits with this same formula, which, being exact, respects the ends. Fisher’s z with variance 1/(N−3) is an approximation designed for the bivariate normal; with point-biserial and unequal groups its coverage gets worse.