How Often Do NFL Starters Share a Birthday?

The birthday paradox says that there's about a 50% chance that at least two people in a group of just 23 share a birthday.

A starting NFL lineup features 23 players (if you include a kicker), so I've collected data from NFL depth charts from the past five years to empirically test how well the birthday paradox holds.

Two starters on the 2023 Kansas City Chiefs shared at least one birthday.


QB	Patrick Mahomes	Sep 17	27
RB	Isiah Pacheco	Mar 2	24
WR	Rashee Rice	Apr 22	23
WR	Marquez Valdes-Scantling	Oct 10	28
TE	Noah Gray	Apr 30	24
TE	Travis Kelce	Oct 5	33
LT	Donovan Smith	Jun 23	30
LG	Joe Thuney	Nov 18	30
C	Creed Humphrey	Jun 28	24
RG	Trey Smith	Jun 16	24
RT	Jawaan Taylor	Nov 25	25
LDE	George Karlaftis	Apr 3	22
LDT	Chris Jones	Jul 3	29
RDT	Derrick Nnadi	May 9	27
RDE	Mike Danna	Dec 1	25
LLB	Willie Gay	Feb 15	25
MLB	Nick Bolton	Mar 10	23
RLB	Leo Chenal	Oct 26	22
LCB	Trent McDuffie	Sep 13	22
RCB	L'Jarius Sneed	Jan 21	26
SS	Justin Reid	Feb 15	26
FS	Bryan Cook	Sep 7	23
K	Harrison Butker	Jul 14	28

Data: Pro Football Reference, Sleeper

Altogether, 83 of 160 rosters (50.7%) match the birthday paradox's assertion—which is even higher than the expected percentage.

If you're curious to dig into the details, here's an approach to understanding the birthday paradox.

Rather than calculate the chances of choosing 23 people that share one or more birthdays, it's easier to calculate the chances of choosing 23 people that share zero birthdays (and then subtract from 100%). This is allowed because the two events are mutually exclusive and exhaustive.

Mutually exclusive and exhaustive events.

50.7%

100% - 49.3%

0 shared birthdays among 23 people

49.3%

100% - 50.7%

≥ 1 shared birthdays among 23 people

the birthday paradoxChicago Bears' regular season record by wins and lossesMarcus Mariota's postseason scoring by touchdowns

Now consider the probability of choosing each person's birthday without repeating a previous birthday, ignoring leap years.

The first person's birthday can be any day of the year.

\frac{365}{365}

The second person's birthday can be any of the remaining 364 (365 − 1) days in the year.

\frac{365}{365}, \frac{364}{365}

And so on, until the 23^rd person's birthday has 343 (365 − 22) remaining days in the year to not repeat one of the previous 22 birthdays.

\frac{365}{365}, \frac{364}{365}, \dots, \frac{343}{365}

We need all 23 events to take place, so we multiply the individual probabilities together to produce a single probability.

\frac{365}{365} \times \frac{364}{365} \times \dots \times \frac{343}{365} ≅ 0.493

Finally, we subtract from 100% to get the probability of the original problem.

1 - 0.493 = 0.507