Fingerprint match probability
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.
- 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.
- Each person has more than one fingerprint (10 total) that can be used
- 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)
- 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.
- 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?
probability paradoxes
asked Jan 9 '14 at 21:00
Rothgar
18814
add a comment |
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.
- 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.
- Each person has more than one fingerprint (10 total) that can be used
- 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)
- 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.
- 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?
probability paradoxes
asked Jan 9 '14 at 21:00
Rothgar
18814
4
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!
– MartianInvader
Jan 9 '14 at 22:42
2
@MartianInvader. It $is$ wrong. The correct formula, assuming exactly $365$ days in a year,is $$1-prod_{j=1}^nfrac {365-(j-1)}{365}.$$
– DanielWainfleet
Aug 29 '17 at 16:28
add a comment |
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.
- 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.
- Each person has more than one fingerprint (10 total) that can be used
- 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)
- 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.
- 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?
probability paradoxes
asked Jan 9 '14 at 21:00
Rothgar
18814
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.
- 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.
- Each person has more than one fingerprint (10 total) that can be used
- 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)
- 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.
- 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?
probability paradoxes
probability paradoxes
asked Jan 9 '14 at 21:00
Rothgar
18814
asked Jan 9 '14 at 21:00
Rothgar
18814
asked Jan 9 '14 at 21:00
Rothgar
18814
asked Jan 9 '14 at 21:00
Rothgar
18814
asked Jan 9 '14 at 21:00
Rothgar
18814
18814
4
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!
– MartianInvader
Jan 9 '14 at 22:42
2
@MartianInvader. It $is$ wrong. The correct formula, assuming exactly $365$ days in a year,is $$1-prod_{j=1}^nfrac {365-(j-1)}{365}.$$
– DanielWainfleet
Aug 29 '17 at 16:28
add a comment |
4
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!
– MartianInvader
Jan 9 '14 at 22:42
2
@MartianInvader. It $is$ wrong. The correct formula, assuming exactly $365$ days in a year,is $$1-prod_{j=1}^nfrac {365-(j-1)}{365}.$$
– DanielWainfleet
Aug 29 '17 at 16:28
4
4
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!
– MartianInvader
Jan 9 '14 at 22:42
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!
– MartianInvader
Jan 9 '14 at 22:42
2
2
@MartianInvader. It $is$ wrong. The correct formula, assuming exactly $365$ days in a year,is $$1-prod_{j=1}^nfrac {365-(j-1)}{365}.$$
– DanielWainfleet
Aug 29 '17 at 16:28
@MartianInvader. It $is$ wrong. The correct formula, assuming exactly $365$ days in a year,is $$1-prod_{j=1}^nfrac {365-(j-1)}{365}.$$
– DanielWainfleet
Aug 29 '17 at 16:28
add a comment |
1 Answer
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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 $50%$
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
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.
– Rothgar
Jan 22 '14 at 16:58
But the formula for the birthday paradox is wrong? (See the comments under the question.)
– joriki
Sep 17 at 22:12
add a comment |
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1 Answer
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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 $50%$
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
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.
– Rothgar
Jan 22 '14 at 16:58
But the formula for the birthday paradox is wrong? (See the comments under the question.)
– joriki
Sep 17 at 22:12
add a comment |
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 $50%$
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
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.
– Rothgar
Jan 22 '14 at 16:58
But the formula for the birthday paradox is wrong? (See the comments under the question.)
– joriki
Sep 17 at 22:12
add a comment |
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 $50%$
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
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 $50%$
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
edited Jan 20 '17 at 16:42
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
answered Jan 9 '14 at 22:13
Ross Millikan
291k23196370
291k23196370
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.
– Rothgar
Jan 22 '14 at 16:58
But the formula for the birthday paradox is wrong? (See the comments under the question.)
– joriki
Sep 17 at 22:12
add a comment |
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.
– Rothgar
Jan 22 '14 at 16:58
But the formula for the birthday paradox is wrong? (See the comments under the question.)
– joriki
Sep 17 at 22:12
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.
– Rothgar
Jan 22 '14 at 16:58
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.
– Rothgar
Jan 22 '14 at 16:58
But the formula for the birthday paradox is wrong? (See the comments under the question.)
– joriki
Sep 17 at 22:12
But the formula for the birthday paradox is wrong? (See the comments under the question.)
– joriki
Sep 17 at 22:12
add a comment |
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4
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!
– MartianInvader
Jan 9 '14 at 22:42
2
@MartianInvader. It $is$ wrong. The correct formula, assuming exactly $365$ days in a year,is $$1-prod_{j=1}^nfrac {365-(j-1)}{365}.$$
– DanielWainfleet
Aug 29 '17 at 16:28