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I am trying to use the formula for the birthday paradox as a reference to figure out an equation that represents the probability of a fingerprint match. Here's the equation for probability of a matching birthday.

$$ p(n) = 1-\frac{364}{365}^{\frac{n(n-1)}{2}} $$

Where n is the number of people in the room.

There are obviously a few things that are different for fingerprint probability.

  1. The available different fingerprints is theoretically infinite but I am going to go off of the assumption Apple made that there is a 1 in 50,000 chance of a match.
  2. Each person has more than one fingerprint (10 total) that can be used
  3. The secured device can have one or more fingerprints registered as secure (up to 10) which will also increase probability of a match

I tried to adapt this equation myself and ended up with this. Fr represents the number of registered fingers, n the amount of people in the room (multiplied by 10 to include all fingers), and Pr is the match probability (1/50000)

$$ 1-Pr^{\frac{Fr*((n*10)-Fr)}{2}} $$

This equation doesn't work for a couple reasons though (there's probably more I'm missing too)

  1. It doesn't remove the registered fingerprints from a match per phone. e.g. if two people, each with their own phone, have registered 1 finger each then the total available fingerprints for a match is 19 per phone.
  2. The equation assumes each person has a phone with registered fingerprints (I'm OK with that assumption)

When I work out this equation with 50 people each registering 3 fingerprints I get

$$ 1-\frac{49999}{50000}^{\frac{150*((50*10)-150)}{2}} $$

Which gives me 52.9% chance of a match which seems way too high. Can someone help me figure out what I'm doing wrong?

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    $\begingroup$ I'm pretty sure your equation for the birthday paradox is wrong. It seems to assume that the chance for all pairs of people sharing a birthday are independent, but if we are given that $A$ shares a birthday with neither $B$ nor $C$, then the chance for $B$ and $C$ to share a birthday is higher (an intuitive way to see this is to pretend for a moment that there's only $2$ days in a year). Note that your formula would give a non-zero probability that among $367$ people, no two share a birthday! $\endgroup$
    – MartianInvader
    Commented Jan 9, 2014 at 22:42
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    $\begingroup$ @MartianInvader. It $is$ wrong. The correct formula, assuming exactly $365$ days in a year,is $$1-\prod_{j=1}^n\frac {365-(j-1)}{365}.$$ $\endgroup$
    – DanielWainfleet
    Commented Aug 29, 2017 at 16:28
  • $\begingroup$ I think birthday paradox might not apply here. Because in the BP there are only 365 days to be selected and elimination of one person having a particular birthday decrease the total available days. This is why @DanielWainfleet's formula is the correct one. However, I doubt that there are only 50k fingerprints in the world, but it is more like a false positive probability and nothing to do with the number of the registered fingerprints. (I read the link you gave but I think the phrasing there is unclear). $\endgroup$
    – obareey
    Commented Oct 12, 2021 at 13:23

1 Answer 1

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The exponent $\frac {n(n-1)}2$ in the paradox is the number of pairs of people, which is the number of tries to find a match. You have to define what are the pairs you are looking for. It looks like you are asking the chance of a random match between one of $150$ registered fingerprints (3 each from 50 people) and any finger of the same people (500). In that case your calculation is correct and is in line with the birthday paradox. Once you have as many pairs that might match as items, the chance of a match is rather high. You have $37425$ pairs, which is not far from $50000$. In the usual birthday paradox there are $253$ pairs compared to $365$ days to get a match chance of $50\%$

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  • $\begingroup$ You are correct in what I am trying to match. The only difference being I would want to exclude each finger from matching its own registration. Technically a person could have two of the same fingerprint so I only want to exclude 150 fingers from the match (hence the -150) in the example above. I guess I really wanted someone else to verify what I had looked right because it didn't seem right to me. $\endgroup$
    – Justin Garrison
    Commented Jan 22, 2014 at 16:58
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    $\begingroup$ But the formula for the birthday paradox is wrong? (See the comments under the question.) $\endgroup$
    – joriki
    Commented Sep 17, 2018 at 22:12

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