Mixing
What is the probability that the sent codeword will be eventually decoded correctly
$begingroup$
I seem to be stuck on the following problem:
Let $C={000,111,222}$ be a ternary code which is sent over a symmetric channel with symbol-error probability $p$.
We use an error detection system, so that if the received word is not a codeword in $C$, we request a retransmission.
The codeword $000in C$ is sent. What is the probability that we eventually decode the codeword correctly, perhaps after several retransmissions?
Here is the progress I have made so far:
The probability that we decode correctly upon receiving the first transmission is the probability that no errors occur, i.e. $(1-p)^3$. The probability that we decode correctly upon receiving the second transmission given that the first transmission was rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)$, where the $(3p-3p^2+frac{3}{4}p^3)$ term comes from the fact that the probability that a codeword is rejected is one minus the probability that no errors occurred during the first transmission plus the probability that $000$ was changed to $111$ or $222$ as a result of errors introduced during the first transmission, which is: $(1-p)^3-frac{1}{4}p^3=1-3p+3p^2-frac{3}{4}p^3$. Likewise, the probability that the codeword is decoded correctly upon receiving the third transmission given that the previous two were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^2$ and so on. My initial idea was to simply extrapolate this to the general case; that the probability that the codeword was decoded correctly upon receiving the $(n+1)$'th transmission given that the previous $n$ transmissions were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n$, and then taking the sum from zero to infinity, which in my mind would yield: $sum_{n=0}^{infty}(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3sum_{n=0}^{infty}(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(frac{1}{1+frac{1}{4}(frac{p}{1-p})^3})$. Needless to say, I immediately recognised the fallaciouness of this argument as the events are not independent.
Now I find myself in a position in which I believe that the problem does admit a simple solution, which would be immediately recognised by a mind more lucid than mine. All help and input would, as always, be highly appreciated.
probability coding-theory
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
$endgroup$
add a comment |
$begingroup$
I seem to be stuck on the following problem:
Let $C={000,111,222}$ be a ternary code which is sent over a symmetric channel with symbol-error probability $p$.
We use an error detection system, so that if the received word is not a codeword in $C$, we request a retransmission.
The codeword $000in C$ is sent. What is the probability that we eventually decode the codeword correctly, perhaps after several retransmissions?
Here is the progress I have made so far:
The probability that we decode correctly upon receiving the first transmission is the probability that no errors occur, i.e. $(1-p)^3$. The probability that we decode correctly upon receiving the second transmission given that the first transmission was rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)$, where the $(3p-3p^2+frac{3}{4}p^3)$ term comes from the fact that the probability that a codeword is rejected is one minus the probability that no errors occurred during the first transmission plus the probability that $000$ was changed to $111$ or $222$ as a result of errors introduced during the first transmission, which is: $(1-p)^3-frac{1}{4}p^3=1-3p+3p^2-frac{3}{4}p^3$. Likewise, the probability that the codeword is decoded correctly upon receiving the third transmission given that the previous two were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^2$ and so on. My initial idea was to simply extrapolate this to the general case; that the probability that the codeword was decoded correctly upon receiving the $(n+1)$'th transmission given that the previous $n$ transmissions were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n$, and then taking the sum from zero to infinity, which in my mind would yield: $sum_{n=0}^{infty}(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3sum_{n=0}^{infty}(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(frac{1}{1+frac{1}{4}(frac{p}{1-p})^3})$. Needless to say, I immediately recognised the fallaciouness of this argument as the events are not independent.
Now I find myself in a position in which I believe that the problem does admit a simple solution, which would be immediately recognised by a mind more lucid than mine. All help and input would, as always, be highly appreciated.
probability coding-theory
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
$endgroup$
add a comment |
$begingroup$
I seem to be stuck on the following problem:
Let $C={000,111,222}$ be a ternary code which is sent over a symmetric channel with symbol-error probability $p$.
We use an error detection system, so that if the received word is not a codeword in $C$, we request a retransmission.
The codeword $000in C$ is sent. What is the probability that we eventually decode the codeword correctly, perhaps after several retransmissions?
Here is the progress I have made so far:
The probability that we decode correctly upon receiving the first transmission is the probability that no errors occur, i.e. $(1-p)^3$. The probability that we decode correctly upon receiving the second transmission given that the first transmission was rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)$, where the $(3p-3p^2+frac{3}{4}p^3)$ term comes from the fact that the probability that a codeword is rejected is one minus the probability that no errors occurred during the first transmission plus the probability that $000$ was changed to $111$ or $222$ as a result of errors introduced during the first transmission, which is: $(1-p)^3-frac{1}{4}p^3=1-3p+3p^2-frac{3}{4}p^3$. Likewise, the probability that the codeword is decoded correctly upon receiving the third transmission given that the previous two were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^2$ and so on. My initial idea was to simply extrapolate this to the general case; that the probability that the codeword was decoded correctly upon receiving the $(n+1)$'th transmission given that the previous $n$ transmissions were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n$, and then taking the sum from zero to infinity, which in my mind would yield: $sum_{n=0}^{infty}(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3sum_{n=0}^{infty}(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(frac{1}{1+frac{1}{4}(frac{p}{1-p})^3})$. Needless to say, I immediately recognised the fallaciouness of this argument as the events are not independent.
Now I find myself in a position in which I believe that the problem does admit a simple solution, which would be immediately recognised by a mind more lucid than mine. All help and input would, as always, be highly appreciated.
probability coding-theory
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
$endgroup$
I seem to be stuck on the following problem:
Let $C={000,111,222}$ be a ternary code which is sent over a symmetric channel with symbol-error probability $p$.
We use an error detection system, so that if the received word is not a codeword in $C$, we request a retransmission.
The codeword $000in C$ is sent. What is the probability that we eventually decode the codeword correctly, perhaps after several retransmissions?
Here is the progress I have made so far:
The probability that we decode correctly upon receiving the first transmission is the probability that no errors occur, i.e. $(1-p)^3$. The probability that we decode correctly upon receiving the second transmission given that the first transmission was rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)$, where the $(3p-3p^2+frac{3}{4}p^3)$ term comes from the fact that the probability that a codeword is rejected is one minus the probability that no errors occurred during the first transmission plus the probability that $000$ was changed to $111$ or $222$ as a result of errors introduced during the first transmission, which is: $(1-p)^3-frac{1}{4}p^3=1-3p+3p^2-frac{3}{4}p^3$. Likewise, the probability that the codeword is decoded correctly upon receiving the third transmission given that the previous two were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^2$ and so on. My initial idea was to simply extrapolate this to the general case; that the probability that the codeword was decoded correctly upon receiving the $(n+1)$'th transmission given that the previous $n$ transmissions were rejected is $(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n$, and then taking the sum from zero to infinity, which in my mind would yield: $sum_{n=0}^{infty}(1-p)^3(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3sum_{n=0}^{infty}(3p-3p^2+frac{3}{4}p^3)^n=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(1-p)^3(frac{1}{(1-p)^3+frac{1}{4}p^3})=(frac{1}{1+frac{1}{4}(frac{p}{1-p})^3})$. Needless to say, I immediately recognised the fallaciouness of this argument as the events are not independent.
Now I find myself in a position in which I believe that the problem does admit a simple solution, which would be immediately recognised by a mind more lucid than mine. All help and input would, as always, be highly appreciated.
probability coding-theory
probability coding-theory
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
asked Jan 27 at 21:16
Heinrich WagnerHeinrich Wagner
450211
450211
add a comment |
add a comment |
1 Answer
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$begingroup$
The calculation is correct, there is no problem with independece.
Summing probabilities is correct if the events under consideration are disjoint: $P(X cup Y) = P(X)+P(Y)$ if $X$ and $Y$ cannot happen at the same time. In your case, $X$ is the event that we end after $x$ retransmissions and $Y$ that we end after $y$ retransmissions.
Independence comes into play only when we don't have disjointess; the above formula becomes
$$P(X cup Y) = P(X)+P(Y) - P(X cap Y)$$
If we know that $X$ and $Y$ are independent, then we have $P(X cap Y)=P(X)P(Y)$.
But this is not necessary here, disjoint events are even simpler as $P(X cap Y)=0$ since $X$ and $Y$ cannot both happen.
++++++++++++++++++
I'll give another solution that avoids the inifinite sum by some 'clever' argument:
In any given (re)transmission, there are 3 events relevant for the problem:
$E_1$: We decode as "$000$",
$E_2$: We decode as "$000$" or "$111$" or "$222$" (so a word from $C$),
$E_3$: We decode as any other word (so a word not in $C$).
The probability that event 1 happens is correctly calculated in the original post as
$$p_1=P(E_1)=(1-p)^3$$
The probability that event 2 happens is also correctly calculated (with a small clerical sign error) in the original post as
$$p_2=P(E_2)=(1-p)^3 + frac18p^3 + frac18p^3 = (1-p)^3 + frac14p^3 = 1-3p+3p^2-frac34p^3$$
Events 2 and 3 are complementary, so we have
$$p_3=P(E_3)=1-p_2=3p-3p^2+frac34p^3$$
which also appears in the original post.
In case of event 3, we ask for a retransmission and everything we did up to now is irrelevant: The retransmission will decide what happens (maybe involving even more retransmissions).
So the decoded word is finally accepted when we get event 2 in any transmission; if this is also event 1, we decode correctly. Since the previous transmissions are not important now (they only influenced the fact that we got to this transmission, but they did not affect what happened in this transmission), the probability that we decode correctly ($p_{000}$) is that we get event 1 under the condition that we have gotten event 2:
$$p_{000}=P(E_1|E_2) = frac{P(E_1 cap E_2)}{P(E_2)}= frac{P(E_1)}{P(E_2)}=frac{p_1}{p_2}=frac{(1-p)^3}{(1-p)^3 + frac14p^3}$$
answered Jan 27 at 23:19
IngixIngix
5,077159
$endgroup$
$begingroup$
You made it all very clear to me. I especially like your alternative and far more elegant solution. Thank you for your help.
$endgroup$
– Heinrich Wagner
Jan 28 at 14:16
add a comment |
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$begingroup$
The calculation is correct, there is no problem with independece.
Summing probabilities is correct if the events under consideration are disjoint: $P(X cup Y) = P(X)+P(Y)$ if $X$ and $Y$ cannot happen at the same time. In your case, $X$ is the event that we end after $x$ retransmissions and $Y$ that we end after $y$ retransmissions.
Independence comes into play only when we don't have disjointess; the above formula becomes
$$P(X cup Y) = P(X)+P(Y) - P(X cap Y)$$
If we know that $X$ and $Y$ are independent, then we have $P(X cap Y)=P(X)P(Y)$.
But this is not necessary here, disjoint events are even simpler as $P(X cap Y)=0$ since $X$ and $Y$ cannot both happen.
++++++++++++++++++
I'll give another solution that avoids the inifinite sum by some 'clever' argument:
In any given (re)transmission, there are 3 events relevant for the problem:
$E_1$: We decode as "$000$",
$E_2$: We decode as "$000$" or "$111$" or "$222$" (so a word from $C$),
$E_3$: We decode as any other word (so a word not in $C$).
The probability that event 1 happens is correctly calculated in the original post as
$$p_1=P(E_1)=(1-p)^3$$
The probability that event 2 happens is also correctly calculated (with a small clerical sign error) in the original post as
$$p_2=P(E_2)=(1-p)^3 + frac18p^3 + frac18p^3 = (1-p)^3 + frac14p^3 = 1-3p+3p^2-frac34p^3$$
Events 2 and 3 are complementary, so we have
$$p_3=P(E_3)=1-p_2=3p-3p^2+frac34p^3$$
which also appears in the original post.
In case of event 3, we ask for a retransmission and everything we did up to now is irrelevant: The retransmission will decide what happens (maybe involving even more retransmissions).
So the decoded word is finally accepted when we get event 2 in any transmission; if this is also event 1, we decode correctly. Since the previous transmissions are not important now (they only influenced the fact that we got to this transmission, but they did not affect what happened in this transmission), the probability that we decode correctly ($p_{000}$) is that we get event 1 under the condition that we have gotten event 2:
$$p_{000}=P(E_1|E_2) = frac{P(E_1 cap E_2)}{P(E_2)}= frac{P(E_1)}{P(E_2)}=frac{p_1}{p_2}=frac{(1-p)^3}{(1-p)^3 + frac14p^3}$$
answered Jan 27 at 23:19
IngixIngix
5,077159
$endgroup$
$begingroup$
You made it all very clear to me. I especially like your alternative and far more elegant solution. Thank you for your help.
$endgroup$
– Heinrich Wagner
Jan 28 at 14:16
add a comment |
$begingroup$
The calculation is correct, there is no problem with independece.
Summing probabilities is correct if the events under consideration are disjoint: $P(X cup Y) = P(X)+P(Y)$ if $X$ and $Y$ cannot happen at the same time. In your case, $X$ is the event that we end after $x$ retransmissions and $Y$ that we end after $y$ retransmissions.
Independence comes into play only when we don't have disjointess; the above formula becomes
$$P(X cup Y) = P(X)+P(Y) - P(X cap Y)$$
If we know that $X$ and $Y$ are independent, then we have $P(X cap Y)=P(X)P(Y)$.
But this is not necessary here, disjoint events are even simpler as $P(X cap Y)=0$ since $X$ and $Y$ cannot both happen.
++++++++++++++++++
I'll give another solution that avoids the inifinite sum by some 'clever' argument:
In any given (re)transmission, there are 3 events relevant for the problem:
$E_1$: We decode as "$000$",
$E_2$: We decode as "$000$" or "$111$" or "$222$" (so a word from $C$),
$E_3$: We decode as any other word (so a word not in $C$).
The probability that event 1 happens is correctly calculated in the original post as
$$p_1=P(E_1)=(1-p)^3$$
The probability that event 2 happens is also correctly calculated (with a small clerical sign error) in the original post as
$$p_2=P(E_2)=(1-p)^3 + frac18p^3 + frac18p^3 = (1-p)^3 + frac14p^3 = 1-3p+3p^2-frac34p^3$$
Events 2 and 3 are complementary, so we have
$$p_3=P(E_3)=1-p_2=3p-3p^2+frac34p^3$$
which also appears in the original post.
In case of event 3, we ask for a retransmission and everything we did up to now is irrelevant: The retransmission will decide what happens (maybe involving even more retransmissions).
So the decoded word is finally accepted when we get event 2 in any transmission; if this is also event 1, we decode correctly. Since the previous transmissions are not important now (they only influenced the fact that we got to this transmission, but they did not affect what happened in this transmission), the probability that we decode correctly ($p_{000}$) is that we get event 1 under the condition that we have gotten event 2:
$$p_{000}=P(E_1|E_2) = frac{P(E_1 cap E_2)}{P(E_2)}= frac{P(E_1)}{P(E_2)}=frac{p_1}{p_2}=frac{(1-p)^3}{(1-p)^3 + frac14p^3}$$
answered Jan 27 at 23:19
IngixIngix
5,077159
$endgroup$
$begingroup$
You made it all very clear to me. I especially like your alternative and far more elegant solution. Thank you for your help.
$endgroup$
– Heinrich Wagner
Jan 28 at 14:16
add a comment |
$begingroup$
The calculation is correct, there is no problem with independece.
Summing probabilities is correct if the events under consideration are disjoint: $P(X cup Y) = P(X)+P(Y)$ if $X$ and $Y$ cannot happen at the same time. In your case, $X$ is the event that we end after $x$ retransmissions and $Y$ that we end after $y$ retransmissions.
Independence comes into play only when we don't have disjointess; the above formula becomes
$$P(X cup Y) = P(X)+P(Y) - P(X cap Y)$$
If we know that $X$ and $Y$ are independent, then we have $P(X cap Y)=P(X)P(Y)$.
But this is not necessary here, disjoint events are even simpler as $P(X cap Y)=0$ since $X$ and $Y$ cannot both happen.
++++++++++++++++++
I'll give another solution that avoids the inifinite sum by some 'clever' argument:
In any given (re)transmission, there are 3 events relevant for the problem:
$E_1$: We decode as "$000$",
$E_2$: We decode as "$000$" or "$111$" or "$222$" (so a word from $C$),
$E_3$: We decode as any other word (so a word not in $C$).
The probability that event 1 happens is correctly calculated in the original post as
$$p_1=P(E_1)=(1-p)^3$$
The probability that event 2 happens is also correctly calculated (with a small clerical sign error) in the original post as
$$p_2=P(E_2)=(1-p)^3 + frac18p^3 + frac18p^3 = (1-p)^3 + frac14p^3 = 1-3p+3p^2-frac34p^3$$
Events 2 and 3 are complementary, so we have
$$p_3=P(E_3)=1-p_2=3p-3p^2+frac34p^3$$
which also appears in the original post.
In case of event 3, we ask for a retransmission and everything we did up to now is irrelevant: The retransmission will decide what happens (maybe involving even more retransmissions).
So the decoded word is finally accepted when we get event 2 in any transmission; if this is also event 1, we decode correctly. Since the previous transmissions are not important now (they only influenced the fact that we got to this transmission, but they did not affect what happened in this transmission), the probability that we decode correctly ($p_{000}$) is that we get event 1 under the condition that we have gotten event 2:
$$p_{000}=P(E_1|E_2) = frac{P(E_1 cap E_2)}{P(E_2)}= frac{P(E_1)}{P(E_2)}=frac{p_1}{p_2}=frac{(1-p)^3}{(1-p)^3 + frac14p^3}$$
answered Jan 27 at 23:19
IngixIngix
5,077159
$endgroup$
The calculation is correct, there is no problem with independece.
Summing probabilities is correct if the events under consideration are disjoint: $P(X cup Y) = P(X)+P(Y)$ if $X$ and $Y$ cannot happen at the same time. In your case, $X$ is the event that we end after $x$ retransmissions and $Y$ that we end after $y$ retransmissions.
Independence comes into play only when we don't have disjointess; the above formula becomes
$$P(X cup Y) = P(X)+P(Y) - P(X cap Y)$$
If we know that $X$ and $Y$ are independent, then we have $P(X cap Y)=P(X)P(Y)$.
But this is not necessary here, disjoint events are even simpler as $P(X cap Y)=0$ since $X$ and $Y$ cannot both happen.
++++++++++++++++++
I'll give another solution that avoids the inifinite sum by some 'clever' argument:
In any given (re)transmission, there are 3 events relevant for the problem:
$E_1$: We decode as "$000$",
$E_2$: We decode as "$000$" or "$111$" or "$222$" (so a word from $C$),
$E_3$: We decode as any other word (so a word not in $C$).
The probability that event 1 happens is correctly calculated in the original post as
$$p_1=P(E_1)=(1-p)^3$$
The probability that event 2 happens is also correctly calculated (with a small clerical sign error) in the original post as
$$p_2=P(E_2)=(1-p)^3 + frac18p^3 + frac18p^3 = (1-p)^3 + frac14p^3 = 1-3p+3p^2-frac34p^3$$
Events 2 and 3 are complementary, so we have
$$p_3=P(E_3)=1-p_2=3p-3p^2+frac34p^3$$
which also appears in the original post.
In case of event 3, we ask for a retransmission and everything we did up to now is irrelevant: The retransmission will decide what happens (maybe involving even more retransmissions).
So the decoded word is finally accepted when we get event 2 in any transmission; if this is also event 1, we decode correctly. Since the previous transmissions are not important now (they only influenced the fact that we got to this transmission, but they did not affect what happened in this transmission), the probability that we decode correctly ($p_{000}$) is that we get event 1 under the condition that we have gotten event 2:
$$p_{000}=P(E_1|E_2) = frac{P(E_1 cap E_2)}{P(E_2)}= frac{P(E_1)}{P(E_2)}=frac{p_1}{p_2}=frac{(1-p)^3}{(1-p)^3 + frac14p^3}$$
answered Jan 27 at 23:19
IngixIngix
5,077159
answered Jan 27 at 23:19
IngixIngix
5,077159
answered Jan 27 at 23:19
IngixIngix
5,077159
answered Jan 27 at 23:19
IngixIngix
5,077159
5,077159
$begingroup$
You made it all very clear to me. I especially like your alternative and far more elegant solution. Thank you for your help.
$endgroup$
– Heinrich Wagner
Jan 28 at 14:16
add a comment |
$begingroup$
You made it all very clear to me. I especially like your alternative and far more elegant solution. Thank you for your help.
$endgroup$
– Heinrich Wagner
Jan 28 at 14:16
$begingroup$
You made it all very clear to me. I especially like your alternative and far more elegant solution. Thank you for your help.
$endgroup$
– Heinrich Wagner
Jan 28 at 14:16
$begingroup$
You made it all very clear to me. I especially like your alternative and far more elegant solution. Thank you for your help.
$endgroup$
– Heinrich Wagner
Jan 28 at 14:16
add a comment |
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