Mixing
Number of descendants at the Nth generation. Absurd result… :(
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Are we all the descendants of Charlemagne?
The assumption seems logical but I was wondering if I could calculate the probability, or maybe just the number of descendants at the Nth generation. But it‘s complicated. So here‘s a simplified version:
A group of 4 people: A, B, C and D.
Each generation, two couples are formed (there are 3 possibilities: AB/CD, AC/BD and AD/BC)
These couples have children together (between 0 and 4).
The next generation is also 4 members strong. Each member of the new generation has an equal probability of being the child of the first couple or of the second.
The same process is repeated.
What is the probability of being a descendant of A at the Nth generation?
If I break it up, I would say that at:
Generation 0 (G0) there is only A, so a 25% chance.
Generation 1 (G1): every child has an equal probability of being the child of A or of the other couple. So a 50% chance.
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
....BUT.........
There is a 6.25% probability that NO person in G1 is the descendant of A. In that case, the probability of there not being ANY descendant of A in the Nth generation, MUST be superior to 6.25%!! How come then are we finding it to be 0%??!!
What is wrong here??!!
Thanks a lot in advance!
probability
asked yesterday
Eek
132
add a comment |Â
up vote
2
down vote
favorite
Are we all the descendants of Charlemagne?
The assumption seems logical but I was wondering if I could calculate the probability, or maybe just the number of descendants at the Nth generation. But it‘s complicated. So here‘s a simplified version:
A group of 4 people: A, B, C and D.
Each generation, two couples are formed (there are 3 possibilities: AB/CD, AC/BD and AD/BC)
These couples have children together (between 0 and 4).
The next generation is also 4 members strong. Each member of the new generation has an equal probability of being the child of the first couple or of the second.
The same process is repeated.
What is the probability of being a descendant of A at the Nth generation?
If I break it up, I would say that at:
Generation 0 (G0) there is only A, so a 25% chance.
Generation 1 (G1): every child has an equal probability of being the child of A or of the other couple. So a 50% chance.
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
....BUT.........
There is a 6.25% probability that NO person in G1 is the descendant of A. In that case, the probability of there not being ANY descendant of A in the Nth generation, MUST be superior to 6.25%!! How come then are we finding it to be 0%??!!
What is wrong here??!!
Thanks a lot in advance!
probability
asked yesterday
Eek
132
Sometimes cousins marry each other, reducing your number of distinct ancestors. Indeed all humans alive today are cousins, some more distant than others
– Henry
yesterday
According to Wikipedia, Charlemagne had eighteen children (some with his wives and others with other women) and many of his descendents are clearly alive today, for example through the Habsburgs, Capetians and Plantagenets.
– Henry
yesterday
I think the problem may lie in the statement "if you go on this way, it‘s easy to see the probability rises to 100%..." Are you sure? What happens if you try calculating it for n=3 and beyond?
– alcana
yesterday
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Are we all the descendants of Charlemagne?
The assumption seems logical but I was wondering if I could calculate the probability, or maybe just the number of descendants at the Nth generation. But it‘s complicated. So here‘s a simplified version:
A group of 4 people: A, B, C and D.
Each generation, two couples are formed (there are 3 possibilities: AB/CD, AC/BD and AD/BC)
These couples have children together (between 0 and 4).
The next generation is also 4 members strong. Each member of the new generation has an equal probability of being the child of the first couple or of the second.
The same process is repeated.
What is the probability of being a descendant of A at the Nth generation?
If I break it up, I would say that at:
Generation 0 (G0) there is only A, so a 25% chance.
Generation 1 (G1): every child has an equal probability of being the child of A or of the other couple. So a 50% chance.
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
....BUT.........
There is a 6.25% probability that NO person in G1 is the descendant of A. In that case, the probability of there not being ANY descendant of A in the Nth generation, MUST be superior to 6.25%!! How come then are we finding it to be 0%??!!
What is wrong here??!!
Thanks a lot in advance!
probability
asked yesterday
Eek
132
Are we all the descendants of Charlemagne?
The assumption seems logical but I was wondering if I could calculate the probability, or maybe just the number of descendants at the Nth generation. But it‘s complicated. So here‘s a simplified version:
A group of 4 people: A, B, C and D.
Each generation, two couples are formed (there are 3 possibilities: AB/CD, AC/BD and AD/BC)
These couples have children together (between 0 and 4).
The next generation is also 4 members strong. Each member of the new generation has an equal probability of being the child of the first couple or of the second.
The same process is repeated.
What is the probability of being a descendant of A at the Nth generation?
If I break it up, I would say that at:
Generation 0 (G0) there is only A, so a 25% chance.
Generation 1 (G1): every child has an equal probability of being the child of A or of the other couple. So a 50% chance.
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
....BUT.........
There is a 6.25% probability that NO person in G1 is the descendant of A. In that case, the probability of there not being ANY descendant of A in the Nth generation, MUST be superior to 6.25%!! How come then are we finding it to be 0%??!!
What is wrong here??!!
Thanks a lot in advance!
probability
asked yesterday
Eek
132
asked yesterday
Eek
132
asked yesterday
Eek
132
asked yesterday
Eek
132
132
Sometimes cousins marry each other, reducing your number of distinct ancestors. Indeed all humans alive today are cousins, some more distant than others
– Henry
yesterday
According to Wikipedia, Charlemagne had eighteen children (some with his wives and others with other women) and many of his descendents are clearly alive today, for example through the Habsburgs, Capetians and Plantagenets.
– Henry
yesterday
I think the problem may lie in the statement "if you go on this way, it‘s easy to see the probability rises to 100%..." Are you sure? What happens if you try calculating it for n=3 and beyond?
– alcana
yesterday
add a comment |Â
Sometimes cousins marry each other, reducing your number of distinct ancestors. Indeed all humans alive today are cousins, some more distant than others
– Henry
yesterday
According to Wikipedia, Charlemagne had eighteen children (some with his wives and others with other women) and many of his descendents are clearly alive today, for example through the Habsburgs, Capetians and Plantagenets.
– Henry
yesterday
I think the problem may lie in the statement "if you go on this way, it‘s easy to see the probability rises to 100%..." Are you sure? What happens if you try calculating it for n=3 and beyond?
– alcana
yesterday
Sometimes cousins marry each other, reducing your number of distinct ancestors. Indeed all humans alive today are cousins, some more distant than others
– Henry
yesterday
Sometimes cousins marry each other, reducing your number of distinct ancestors. Indeed all humans alive today are cousins, some more distant than others
– Henry
yesterday
According to Wikipedia, Charlemagne had eighteen children (some with his wives and others with other women) and many of his descendents are clearly alive today, for example through the Habsburgs, Capetians and Plantagenets.
– Henry
yesterday
According to Wikipedia, Charlemagne had eighteen children (some with his wives and others with other women) and many of his descendents are clearly alive today, for example through the Habsburgs, Capetians and Plantagenets.
– Henry
yesterday
I think the problem may lie in the statement "if you go on this way, it‘s easy to see the probability rises to 100%..." Are you sure? What happens if you try calculating it for n=3 and beyond?
– alcana
yesterday
I think the problem may lie in the statement "if you go on this way, it‘s easy to see the probability rises to 100%..." Are you sure? What happens if you try calculating it for n=3 and beyond?
– alcana
yesterday
add a comment |Â
3 Answers
3
active
oldest
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up vote
2
down vote
accepted
Joriki's analysis looks convincing, but skips somewhat quickly past what is wrong with your analysis:
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
So far so good. Each person in G1 has randomly chosen either the A-nonA couple or the nonA-nonA couple as his parents, so the "A-ness" of two randomly chosen persons in G1 is independent.
Without this independence your "25% chance" wouldn't necessarily hold.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
Here it goes wrong if you repeat your analysis from G2 because "mother not descended from A" and "father not descended from A" are no longer independent -- in particular your cannot multiply their probabilities to find the probability of your ancestors being A-free.
On the contrary, in late generations it becomes overwhelmingly likely that either everyone descends from A or nobody does. Thus with high probability, two randomly chosen persons have the same A-ness, which means that the A-ness of your mother and father will be highly correlated.
answered yesterday
Henning Makholm
225k16289516
1
Thank you! I understand now what the problem was! Ok! Thanks a lot for your lightning quick answers! You are really great!
– Eek
yesterday
add a comment |Â
up vote
2
down vote
You don't specify how the couples are chosen; I'll assume that in each generation one of the three possible combinations is independently and uniformly chosen.
The problem with your calculation is that you treat all the probabilities as independent and only keep track of the probability for individuals to be descendants of $A$, whereas you need the entire probability distribution in each generation in order to derive it for the next generation.
In the first generation of descendants, you have a binomial distribution for the number of descendants of $A$:
beginarrayc
n&0&1&2&3&4\hline
p&frac116&frac416&frac616&frac416&frac116
endarray
Now couples are formed. If there are $0$ descendants of $A$, there will be $0$ couples whose children will be descendants of $A$. If there is $1$ descendant, there will be $1$ such couple. If there are $3$ or $4$ descendants, there will be $2$ such couples. The case of $2$ descendants is a bit more complicated: With probability $frac13$ the two descendants end up in the same couple, and then there's only one couple descended from $A$, and with probability $frac23$ the two descendants end up in different couples, which are then both descended from $A$. Thus the probability distribution for the number of couples descended from $A$ is
beginarrayc
m&0&1&2\hline
p&frac116&frac616&frac916
endarray
Now if there are $0$ couples descended from $A$, no-one in the next generation will be descended from $A$, and if there are $2$, everyone will; whereas if there is $1$, we get the binomial distribution again. So in the next generation, the probability distribution for the number of people descended from $A$ is
beginarrayc
n&0&1&2&3&4\hline
p&frac11128½128&frac18128½128&frac75128
endarray
The expected value is $frac916cdot1+frac616cdotfrac12=frac34$, so up to this point your probability for individuals to be descendants of $A$ is correct. But the distribution of the number of couples descended from $A$ in this generation now comes out as
beginarrayc
m&0&1&2\hline
p&frac11128&frac18128&frac99128
endarray
So the expected number of descendants of $A$ in the next generation will be $frac99128cdot1+frac18128cdotfrac12=frac2732$, not nearly $100%$.
We can express the process using matrices to see how quickly one of the two equilibria is reached and how likely they are. We have the following update rule for the vector $p$ of the probability distribution for the number of couples descended from $A$:
$$
p'=pmatrix
1&0&0&0&0\
0&1&frac13&0&0\
0&0&frac23&1&1
pmatrix
1&frac116&0\
0&frac416&0\
0&frac616&0\
0&frac416&0\
0&frac116&1\
=pmatrix
1&frac116&0\
0&frac616&0\
0&frac916&1
p;.
$$
The matrix on the right has two eigenvalues $1$ corresponding to the equilibrium distributions where everyone or no one is a descendant of $A$. The third eigenvalue is $frac616$, with eigenvector $(-1,10,-9)^top$. Your initial distribution $(0,1,0)^top$ decomposes as
$$
pmatrix0\1\0=frac110pmatrix1\0\0+frac110pmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
so after applying the matrix $j$ times this becomes
$$
frac110pmatrix1\0\0+frac110left(frac38right)^jpmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
and the limit for $jtoinfty$ is
$$
pmatrixfrac110\0\frac910;.
$$
So in the long run, the probability that everyone will be descended from $A$ is $90%$ and the probability that no one will be descended from $A$ is $10%$.
answered yesterday
joriki
164k10179328
1
Thanks a lot!!! That was quickly answered and impressive!! Thanks a lot!!
– Eek
yesterday
add a comment |Â
up vote
0
down vote
Turn it around. I have two parents, four grandparents, eight great grandparents and so on. If we go back $30$ generations I would have $2^30 approx 1$ billion ancestors. $600$ years ago there were not a billion people on earth, so many of those ancestors must be the same people. If you assume much mixing at all, I must be descended from everybody alive in the year $1200$.
There are two holes in this analysis. One, some of the people did not have any children, so I am not descended from them. Two, there could be isolated groups who do not mix with the general population. A European of $800$ could do the same calculation and claim to be descended from everybody alive in $400$ or some such year. As far as we know there was no mixing with the inhabitants of the Americas during that period and probably none with some other remote areas.
answered yesterday
Ross Millikan
275k21183348
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Joriki's analysis looks convincing, but skips somewhat quickly past what is wrong with your analysis:
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
So far so good. Each person in G1 has randomly chosen either the A-nonA couple or the nonA-nonA couple as his parents, so the "A-ness" of two randomly chosen persons in G1 is independent.
Without this independence your "25% chance" wouldn't necessarily hold.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
Here it goes wrong if you repeat your analysis from G2 because "mother not descended from A" and "father not descended from A" are no longer independent -- in particular your cannot multiply their probabilities to find the probability of your ancestors being A-free.
On the contrary, in late generations it becomes overwhelmingly likely that either everyone descends from A or nobody does. Thus with high probability, two randomly chosen persons have the same A-ness, which means that the A-ness of your mother and father will be highly correlated.
answered yesterday
Henning Makholm
225k16289516
1
Thank you! I understand now what the problem was! Ok! Thanks a lot for your lightning quick answers! You are really great!
– Eek
yesterday
add a comment |Â
up vote
2
down vote
accepted
Joriki's analysis looks convincing, but skips somewhat quickly past what is wrong with your analysis:
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
So far so good. Each person in G1 has randomly chosen either the A-nonA couple or the nonA-nonA couple as his parents, so the "A-ness" of two randomly chosen persons in G1 is independent.
Without this independence your "25% chance" wouldn't necessarily hold.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
Here it goes wrong if you repeat your analysis from G2 because "mother not descended from A" and "father not descended from A" are no longer independent -- in particular your cannot multiply their probabilities to find the probability of your ancestors being A-free.
On the contrary, in late generations it becomes overwhelmingly likely that either everyone descends from A or nobody does. Thus with high probability, two randomly chosen persons have the same A-ness, which means that the A-ness of your mother and father will be highly correlated.
answered yesterday
Henning Makholm
225k16289516
1
Thank you! I understand now what the problem was! Ok! Thanks a lot for your lightning quick answers! You are really great!
– Eek
yesterday
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Joriki's analysis looks convincing, but skips somewhat quickly past what is wrong with your analysis:
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
So far so good. Each person in G1 has randomly chosen either the A-nonA couple or the nonA-nonA couple as his parents, so the "A-ness" of two randomly chosen persons in G1 is independent.
Without this independence your "25% chance" wouldn't necessarily hold.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
Here it goes wrong if you repeat your analysis from G2 because "mother not descended from A" and "father not descended from A" are no longer independent -- in particular your cannot multiply their probabilities to find the probability of your ancestors being A-free.
On the contrary, in late generations it becomes overwhelmingly likely that either everyone descends from A or nobody does. Thus with high probability, two randomly chosen persons have the same A-ness, which means that the A-ness of your mother and father will be highly correlated.
answered yesterday
Henning Makholm
225k16289516
Joriki's analysis looks convincing, but skips somewhat quickly past what is wrong with your analysis:
Generation 2 (G2): every member of the group of G1 has a 50% chance of not being a descendant of A. When I pair two of them randomly together, they have a 25% chance of neither of them being a descendant of A. So each member of G2 has a 75% chance of being a descendant of A.
So far so good. Each person in G1 has randomly chosen either the A-nonA couple or the nonA-nonA couple as his parents, so the "A-ness" of two randomly chosen persons in G1 is independent.
Without this independence your "25% chance" wouldn't necessarily hold.
Generation N (Gn): if you go on this way, it‘s easy to see the probability rises to 100%...
Here it goes wrong if you repeat your analysis from G2 because "mother not descended from A" and "father not descended from A" are no longer independent -- in particular your cannot multiply their probabilities to find the probability of your ancestors being A-free.
On the contrary, in late generations it becomes overwhelmingly likely that either everyone descends from A or nobody does. Thus with high probability, two randomly chosen persons have the same A-ness, which means that the A-ness of your mother and father will be highly correlated.
answered yesterday
Henning Makholm
225k16289516
answered yesterday
Henning Makholm
225k16289516
answered yesterday
Henning Makholm
225k16289516
answered yesterday
Henning Makholm
225k16289516
225k16289516
1
Thank you! I understand now what the problem was! Ok! Thanks a lot for your lightning quick answers! You are really great!
– Eek
yesterday
add a comment |Â
1
Thank you! I understand now what the problem was! Ok! Thanks a lot for your lightning quick answers! You are really great!
– Eek
yesterday
1
1
Thank you! I understand now what the problem was! Ok! Thanks a lot for your lightning quick answers! You are really great!
– Eek
yesterday
Thank you! I understand now what the problem was! Ok! Thanks a lot for your lightning quick answers! You are really great!
– Eek
yesterday
add a comment |Â
up vote
2
down vote
You don't specify how the couples are chosen; I'll assume that in each generation one of the three possible combinations is independently and uniformly chosen.
The problem with your calculation is that you treat all the probabilities as independent and only keep track of the probability for individuals to be descendants of $A$, whereas you need the entire probability distribution in each generation in order to derive it for the next generation.
In the first generation of descendants, you have a binomial distribution for the number of descendants of $A$:
beginarrayc
n&0&1&2&3&4\hline
p&frac116&frac416&frac616&frac416&frac116
endarray
Now couples are formed. If there are $0$ descendants of $A$, there will be $0$ couples whose children will be descendants of $A$. If there is $1$ descendant, there will be $1$ such couple. If there are $3$ or $4$ descendants, there will be $2$ such couples. The case of $2$ descendants is a bit more complicated: With probability $frac13$ the two descendants end up in the same couple, and then there's only one couple descended from $A$, and with probability $frac23$ the two descendants end up in different couples, which are then both descended from $A$. Thus the probability distribution for the number of couples descended from $A$ is
beginarrayc
m&0&1&2\hline
p&frac116&frac616&frac916
endarray
Now if there are $0$ couples descended from $A$, no-one in the next generation will be descended from $A$, and if there are $2$, everyone will; whereas if there is $1$, we get the binomial distribution again. So in the next generation, the probability distribution for the number of people descended from $A$ is
beginarrayc
n&0&1&2&3&4\hline
p&frac11128½128&frac18128½128&frac75128
endarray
The expected value is $frac916cdot1+frac616cdotfrac12=frac34$, so up to this point your probability for individuals to be descendants of $A$ is correct. But the distribution of the number of couples descended from $A$ in this generation now comes out as
beginarrayc
m&0&1&2\hline
p&frac11128&frac18128&frac99128
endarray
So the expected number of descendants of $A$ in the next generation will be $frac99128cdot1+frac18128cdotfrac12=frac2732$, not nearly $100%$.
We can express the process using matrices to see how quickly one of the two equilibria is reached and how likely they are. We have the following update rule for the vector $p$ of the probability distribution for the number of couples descended from $A$:
$$
p'=pmatrix
1&0&0&0&0\
0&1&frac13&0&0\
0&0&frac23&1&1
pmatrix
1&frac116&0\
0&frac416&0\
0&frac616&0\
0&frac416&0\
0&frac116&1\
=pmatrix
1&frac116&0\
0&frac616&0\
0&frac916&1
p;.
$$
The matrix on the right has two eigenvalues $1$ corresponding to the equilibrium distributions where everyone or no one is a descendant of $A$. The third eigenvalue is $frac616$, with eigenvector $(-1,10,-9)^top$. Your initial distribution $(0,1,0)^top$ decomposes as
$$
pmatrix0\1\0=frac110pmatrix1\0\0+frac110pmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
so after applying the matrix $j$ times this becomes
$$
frac110pmatrix1\0\0+frac110left(frac38right)^jpmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
and the limit for $jtoinfty$ is
$$
pmatrixfrac110\0\frac910;.
$$
So in the long run, the probability that everyone will be descended from $A$ is $90%$ and the probability that no one will be descended from $A$ is $10%$.
answered yesterday
joriki
164k10179328
1
Thanks a lot!!! That was quickly answered and impressive!! Thanks a lot!!
– Eek
yesterday
add a comment |Â
up vote
2
down vote
You don't specify how the couples are chosen; I'll assume that in each generation one of the three possible combinations is independently and uniformly chosen.
The problem with your calculation is that you treat all the probabilities as independent and only keep track of the probability for individuals to be descendants of $A$, whereas you need the entire probability distribution in each generation in order to derive it for the next generation.
In the first generation of descendants, you have a binomial distribution for the number of descendants of $A$:
beginarrayc
n&0&1&2&3&4\hline
p&frac116&frac416&frac616&frac416&frac116
endarray
Now couples are formed. If there are $0$ descendants of $A$, there will be $0$ couples whose children will be descendants of $A$. If there is $1$ descendant, there will be $1$ such couple. If there are $3$ or $4$ descendants, there will be $2$ such couples. The case of $2$ descendants is a bit more complicated: With probability $frac13$ the two descendants end up in the same couple, and then there's only one couple descended from $A$, and with probability $frac23$ the two descendants end up in different couples, which are then both descended from $A$. Thus the probability distribution for the number of couples descended from $A$ is
beginarrayc
m&0&1&2\hline
p&frac116&frac616&frac916
endarray
Now if there are $0$ couples descended from $A$, no-one in the next generation will be descended from $A$, and if there are $2$, everyone will; whereas if there is $1$, we get the binomial distribution again. So in the next generation, the probability distribution for the number of people descended from $A$ is
beginarrayc
n&0&1&2&3&4\hline
p&frac11128½128&frac18128½128&frac75128
endarray
The expected value is $frac916cdot1+frac616cdotfrac12=frac34$, so up to this point your probability for individuals to be descendants of $A$ is correct. But the distribution of the number of couples descended from $A$ in this generation now comes out as
beginarrayc
m&0&1&2\hline
p&frac11128&frac18128&frac99128
endarray
So the expected number of descendants of $A$ in the next generation will be $frac99128cdot1+frac18128cdotfrac12=frac2732$, not nearly $100%$.
We can express the process using matrices to see how quickly one of the two equilibria is reached and how likely they are. We have the following update rule for the vector $p$ of the probability distribution for the number of couples descended from $A$:
$$
p'=pmatrix
1&0&0&0&0\
0&1&frac13&0&0\
0&0&frac23&1&1
pmatrix
1&frac116&0\
0&frac416&0\
0&frac616&0\
0&frac416&0\
0&frac116&1\
=pmatrix
1&frac116&0\
0&frac616&0\
0&frac916&1
p;.
$$
The matrix on the right has two eigenvalues $1$ corresponding to the equilibrium distributions where everyone or no one is a descendant of $A$. The third eigenvalue is $frac616$, with eigenvector $(-1,10,-9)^top$. Your initial distribution $(0,1,0)^top$ decomposes as
$$
pmatrix0\1\0=frac110pmatrix1\0\0+frac110pmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
so after applying the matrix $j$ times this becomes
$$
frac110pmatrix1\0\0+frac110left(frac38right)^jpmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
and the limit for $jtoinfty$ is
$$
pmatrixfrac110\0\frac910;.
$$
So in the long run, the probability that everyone will be descended from $A$ is $90%$ and the probability that no one will be descended from $A$ is $10%$.
answered yesterday
joriki
164k10179328
1
Thanks a lot!!! That was quickly answered and impressive!! Thanks a lot!!
– Eek
yesterday
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You don't specify how the couples are chosen; I'll assume that in each generation one of the three possible combinations is independently and uniformly chosen.
The problem with your calculation is that you treat all the probabilities as independent and only keep track of the probability for individuals to be descendants of $A$, whereas you need the entire probability distribution in each generation in order to derive it for the next generation.
In the first generation of descendants, you have a binomial distribution for the number of descendants of $A$:
beginarrayc
n&0&1&2&3&4\hline
p&frac116&frac416&frac616&frac416&frac116
endarray
Now couples are formed. If there are $0$ descendants of $A$, there will be $0$ couples whose children will be descendants of $A$. If there is $1$ descendant, there will be $1$ such couple. If there are $3$ or $4$ descendants, there will be $2$ such couples. The case of $2$ descendants is a bit more complicated: With probability $frac13$ the two descendants end up in the same couple, and then there's only one couple descended from $A$, and with probability $frac23$ the two descendants end up in different couples, which are then both descended from $A$. Thus the probability distribution for the number of couples descended from $A$ is
beginarrayc
m&0&1&2\hline
p&frac116&frac616&frac916
endarray
Now if there are $0$ couples descended from $A$, no-one in the next generation will be descended from $A$, and if there are $2$, everyone will; whereas if there is $1$, we get the binomial distribution again. So in the next generation, the probability distribution for the number of people descended from $A$ is
beginarrayc
n&0&1&2&3&4\hline
p&frac11128½128&frac18128½128&frac75128
endarray
The expected value is $frac916cdot1+frac616cdotfrac12=frac34$, so up to this point your probability for individuals to be descendants of $A$ is correct. But the distribution of the number of couples descended from $A$ in this generation now comes out as
beginarrayc
m&0&1&2\hline
p&frac11128&frac18128&frac99128
endarray
So the expected number of descendants of $A$ in the next generation will be $frac99128cdot1+frac18128cdotfrac12=frac2732$, not nearly $100%$.
We can express the process using matrices to see how quickly one of the two equilibria is reached and how likely they are. We have the following update rule for the vector $p$ of the probability distribution for the number of couples descended from $A$:
$$
p'=pmatrix
1&0&0&0&0\
0&1&frac13&0&0\
0&0&frac23&1&1
pmatrix
1&frac116&0\
0&frac416&0\
0&frac616&0\
0&frac416&0\
0&frac116&1\
=pmatrix
1&frac116&0\
0&frac616&0\
0&frac916&1
p;.
$$
The matrix on the right has two eigenvalues $1$ corresponding to the equilibrium distributions where everyone or no one is a descendant of $A$. The third eigenvalue is $frac616$, with eigenvector $(-1,10,-9)^top$. Your initial distribution $(0,1,0)^top$ decomposes as
$$
pmatrix0\1\0=frac110pmatrix1\0\0+frac110pmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
so after applying the matrix $j$ times this becomes
$$
frac110pmatrix1\0\0+frac110left(frac38right)^jpmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
and the limit for $jtoinfty$ is
$$
pmatrixfrac110\0\frac910;.
$$
So in the long run, the probability that everyone will be descended from $A$ is $90%$ and the probability that no one will be descended from $A$ is $10%$.
answered yesterday
joriki
164k10179328
You don't specify how the couples are chosen; I'll assume that in each generation one of the three possible combinations is independently and uniformly chosen.
The problem with your calculation is that you treat all the probabilities as independent and only keep track of the probability for individuals to be descendants of $A$, whereas you need the entire probability distribution in each generation in order to derive it for the next generation.
In the first generation of descendants, you have a binomial distribution for the number of descendants of $A$:
beginarrayc
n&0&1&2&3&4\hline
p&frac116&frac416&frac616&frac416&frac116
endarray
Now couples are formed. If there are $0$ descendants of $A$, there will be $0$ couples whose children will be descendants of $A$. If there is $1$ descendant, there will be $1$ such couple. If there are $3$ or $4$ descendants, there will be $2$ such couples. The case of $2$ descendants is a bit more complicated: With probability $frac13$ the two descendants end up in the same couple, and then there's only one couple descended from $A$, and with probability $frac23$ the two descendants end up in different couples, which are then both descended from $A$. Thus the probability distribution for the number of couples descended from $A$ is
beginarrayc
m&0&1&2\hline
p&frac116&frac616&frac916
endarray
Now if there are $0$ couples descended from $A$, no-one in the next generation will be descended from $A$, and if there are $2$, everyone will; whereas if there is $1$, we get the binomial distribution again. So in the next generation, the probability distribution for the number of people descended from $A$ is
beginarrayc
n&0&1&2&3&4\hline
p&frac11128½128&frac18128½128&frac75128
endarray
The expected value is $frac916cdot1+frac616cdotfrac12=frac34$, so up to this point your probability for individuals to be descendants of $A$ is correct. But the distribution of the number of couples descended from $A$ in this generation now comes out as
beginarrayc
m&0&1&2\hline
p&frac11128&frac18128&frac99128
endarray
So the expected number of descendants of $A$ in the next generation will be $frac99128cdot1+frac18128cdotfrac12=frac2732$, not nearly $100%$.
We can express the process using matrices to see how quickly one of the two equilibria is reached and how likely they are. We have the following update rule for the vector $p$ of the probability distribution for the number of couples descended from $A$:
$$
p'=pmatrix
1&0&0&0&0\
0&1&frac13&0&0\
0&0&frac23&1&1
pmatrix
1&frac116&0\
0&frac416&0\
0&frac616&0\
0&frac416&0\
0&frac116&1\
=pmatrix
1&frac116&0\
0&frac616&0\
0&frac916&1
p;.
$$
The matrix on the right has two eigenvalues $1$ corresponding to the equilibrium distributions where everyone or no one is a descendant of $A$. The third eigenvalue is $frac616$, with eigenvector $(-1,10,-9)^top$. Your initial distribution $(0,1,0)^top$ decomposes as
$$
pmatrix0\1\0=frac110pmatrix1\0\0+frac110pmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
so after applying the matrix $j$ times this becomes
$$
frac110pmatrix1\0\0+frac110left(frac38right)^jpmatrix-1\10\-9+frac910pmatrix0\0\1;,
$$
and the limit for $jtoinfty$ is
$$
pmatrixfrac110\0\frac910;.
$$
So in the long run, the probability that everyone will be descended from $A$ is $90%$ and the probability that no one will be descended from $A$ is $10%$.
answered yesterday
joriki
164k10179328
edited yesterday
answered yesterday
joriki
164k10179328
answered yesterday
joriki
164k10179328
answered yesterday
joriki
164k10179328
164k10179328
1
Thanks a lot!!! That was quickly answered and impressive!! Thanks a lot!!
– Eek
yesterday
add a comment |Â
1
Thanks a lot!!! That was quickly answered and impressive!! Thanks a lot!!
– Eek
yesterday
1
1
Thanks a lot!!! That was quickly answered and impressive!! Thanks a lot!!
– Eek
yesterday
Thanks a lot!!! That was quickly answered and impressive!! Thanks a lot!!
– Eek
yesterday
add a comment |Â
up vote
0
down vote
Turn it around. I have two parents, four grandparents, eight great grandparents and so on. If we go back $30$ generations I would have $2^30 approx 1$ billion ancestors. $600$ years ago there were not a billion people on earth, so many of those ancestors must be the same people. If you assume much mixing at all, I must be descended from everybody alive in the year $1200$.
There are two holes in this analysis. One, some of the people did not have any children, so I am not descended from them. Two, there could be isolated groups who do not mix with the general population. A European of $800$ could do the same calculation and claim to be descended from everybody alive in $400$ or some such year. As far as we know there was no mixing with the inhabitants of the Americas during that period and probably none with some other remote areas.
answered yesterday
Ross Millikan
275k21183348
add a comment |Â
up vote
0
down vote
Turn it around. I have two parents, four grandparents, eight great grandparents and so on. If we go back $30$ generations I would have $2^30 approx 1$ billion ancestors. $600$ years ago there were not a billion people on earth, so many of those ancestors must be the same people. If you assume much mixing at all, I must be descended from everybody alive in the year $1200$.
There are two holes in this analysis. One, some of the people did not have any children, so I am not descended from them. Two, there could be isolated groups who do not mix with the general population. A European of $800$ could do the same calculation and claim to be descended from everybody alive in $400$ or some such year. As far as we know there was no mixing with the inhabitants of the Americas during that period and probably none with some other remote areas.
answered yesterday
Ross Millikan
275k21183348
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Turn it around. I have two parents, four grandparents, eight great grandparents and so on. If we go back $30$ generations I would have $2^30 approx 1$ billion ancestors. $600$ years ago there were not a billion people on earth, so many of those ancestors must be the same people. If you assume much mixing at all, I must be descended from everybody alive in the year $1200$.
There are two holes in this analysis. One, some of the people did not have any children, so I am not descended from them. Two, there could be isolated groups who do not mix with the general population. A European of $800$ could do the same calculation and claim to be descended from everybody alive in $400$ or some such year. As far as we know there was no mixing with the inhabitants of the Americas during that period and probably none with some other remote areas.
answered yesterday
Ross Millikan
275k21183348
Turn it around. I have two parents, four grandparents, eight great grandparents and so on. If we go back $30$ generations I would have $2^30 approx 1$ billion ancestors. $600$ years ago there were not a billion people on earth, so many of those ancestors must be the same people. If you assume much mixing at all, I must be descended from everybody alive in the year $1200$.
There are two holes in this analysis. One, some of the people did not have any children, so I am not descended from them. Two, there could be isolated groups who do not mix with the general population. A European of $800$ could do the same calculation and claim to be descended from everybody alive in $400$ or some such year. As far as we know there was no mixing with the inhabitants of the Americas during that period and probably none with some other remote areas.
answered yesterday
Ross Millikan
275k21183348
answered yesterday
Ross Millikan
275k21183348
answered yesterday
Ross Millikan
275k21183348
answered yesterday
Ross Millikan
275k21183348
275k21183348
add a comment |Â
add a comment |Â
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Sometimes cousins marry each other, reducing your number of distinct ancestors. Indeed all humans alive today are cousins, some more distant than others
– Henry
yesterday
According to Wikipedia, Charlemagne had eighteen children (some with his wives and others with other women) and many of his descendents are clearly alive today, for example through the Habsburgs, Capetians and Plantagenets.
– Henry
yesterday
I think the problem may lie in the statement "if you go on this way, it‘s easy to see the probability rises to 100%..." Are you sure? What happens if you try calculating it for n=3 and beyond?
– alcana
yesterday