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Shannon Entropy of a fair dice
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The formula for Shannon entropy is as follows,
$$textEntropy(S) = - sum_i p_i log_2 p_i
$$
Thus, a fair six sided dice should have the Entropy,
$$- sum_i=1^6 dfrac16 log_2 dfrac16 = log_2 (6) = 2.5849...$$
However, the Entropy should also correspond to the average number of questions you have to ask in order to know the outcome (as exampled in this guide under the headline Information Theory).
Now, trying to construct a decision tree to describe the average number of questions we have to ask in order to know the outcome of a dice, this seems to be the optimal one:
Decision tree for dice
Looking at the average number of questions in the image, there are 3 questions in 4/6 cases in 2 questions in 2/6 cases. Thus the Entropy should be:
$$dfrac46 times 3 + dfrac26 times 2 = 2.6666...$$
So, obvisouly the result for the Entropy isn't the same in the two calculations. How come?
probability-theory information-theory entropy decision-trees
asked 2 hours ago
Mountain_sheep
162
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2
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The formula for Shannon entropy is as follows,
$$textEntropy(S) = - sum_i p_i log_2 p_i
$$
Thus, a fair six sided dice should have the Entropy,
$$- sum_i=1^6 dfrac16 log_2 dfrac16 = log_2 (6) = 2.5849...$$
However, the Entropy should also correspond to the average number of questions you have to ask in order to know the outcome (as exampled in this guide under the headline Information Theory).
Now, trying to construct a decision tree to describe the average number of questions we have to ask in order to know the outcome of a dice, this seems to be the optimal one:
Decision tree for dice
Looking at the average number of questions in the image, there are 3 questions in 4/6 cases in 2 questions in 2/6 cases. Thus the Entropy should be:
$$dfrac46 times 3 + dfrac26 times 2 = 2.6666...$$
So, obvisouly the result for the Entropy isn't the same in the two calculations. How come?
probability-theory information-theory entropy decision-trees
asked 2 hours ago
Mountain_sheep
162
Take base 6 entropy, and $H_6(X)=1$, which is the same as max depth of your 6-ary decision tree which decides your outcome. The discrepancy comes form the base of the log. you are trying to encode 6 different outcomes in base 2, so you are going to have to use bit-strings of length at least 3. There is clearly some waste here, since there are $2^3 =8$ unique "encodings" on 3 bits, but you only need to encode 6 elements. In general, the entropy bound gets tighter as the number of elements to encode grows, as the first answer already mentioned.
– mm8511
48 mins ago
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up vote
2
down vote
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up vote
2
down vote
favorite
The formula for Shannon entropy is as follows,
$$textEntropy(S) = - sum_i p_i log_2 p_i
$$
Thus, a fair six sided dice should have the Entropy,
$$- sum_i=1^6 dfrac16 log_2 dfrac16 = log_2 (6) = 2.5849...$$
However, the Entropy should also correspond to the average number of questions you have to ask in order to know the outcome (as exampled in this guide under the headline Information Theory).
Now, trying to construct a decision tree to describe the average number of questions we have to ask in order to know the outcome of a dice, this seems to be the optimal one:
Decision tree for dice
Looking at the average number of questions in the image, there are 3 questions in 4/6 cases in 2 questions in 2/6 cases. Thus the Entropy should be:
$$dfrac46 times 3 + dfrac26 times 2 = 2.6666...$$
So, obvisouly the result for the Entropy isn't the same in the two calculations. How come?
probability-theory information-theory entropy decision-trees
asked 2 hours ago
Mountain_sheep
162
The formula for Shannon entropy is as follows,
$$textEntropy(S) = - sum_i p_i log_2 p_i
$$
Thus, a fair six sided dice should have the Entropy,
$$- sum_i=1^6 dfrac16 log_2 dfrac16 = log_2 (6) = 2.5849...$$
However, the Entropy should also correspond to the average number of questions you have to ask in order to know the outcome (as exampled in this guide under the headline Information Theory).
Now, trying to construct a decision tree to describe the average number of questions we have to ask in order to know the outcome of a dice, this seems to be the optimal one:
Decision tree for dice
Looking at the average number of questions in the image, there are 3 questions in 4/6 cases in 2 questions in 2/6 cases. Thus the Entropy should be:
$$dfrac46 times 3 + dfrac26 times 2 = 2.6666...$$
So, obvisouly the result for the Entropy isn't the same in the two calculations. How come?
probability-theory information-theory entropy decision-trees
probability-theory information-theory entropy decision-trees
asked 2 hours ago
Mountain_sheep
162
asked 2 hours ago
Mountain_sheep
162
edited 1 hour ago
asked 2 hours ago
Mountain_sheep
162
asked 2 hours ago
Mountain_sheep
162
asked 2 hours ago
Mountain_sheep
162
162
Take base 6 entropy, and $H_6(X)=1$, which is the same as max depth of your 6-ary decision tree which decides your outcome. The discrepancy comes form the base of the log. you are trying to encode 6 different outcomes in base 2, so you are going to have to use bit-strings of length at least 3. There is clearly some waste here, since there are $2^3 =8$ unique "encodings" on 3 bits, but you only need to encode 6 elements. In general, the entropy bound gets tighter as the number of elements to encode grows, as the first answer already mentioned.
– mm8511
48 mins ago
add a comment |Â
Take base 6 entropy, and $H_6(X)=1$, which is the same as max depth of your 6-ary decision tree which decides your outcome. The discrepancy comes form the base of the log. you are trying to encode 6 different outcomes in base 2, so you are going to have to use bit-strings of length at least 3. There is clearly some waste here, since there are $2^3 =8$ unique "encodings" on 3 bits, but you only need to encode 6 elements. In general, the entropy bound gets tighter as the number of elements to encode grows, as the first answer already mentioned.
– mm8511
48 mins ago
Take base 6 entropy, and $H_6(X)=1$, which is the same as max depth of your 6-ary decision tree which decides your outcome. The discrepancy comes form the base of the log. you are trying to encode 6 different outcomes in base 2, so you are going to have to use bit-strings of length at least 3. There is clearly some waste here, since there are $2^3 =8$ unique "encodings" on 3 bits, but you only need to encode 6 elements. In general, the entropy bound gets tighter as the number of elements to encode grows, as the first answer already mentioned.
– mm8511
48 mins ago
Take base 6 entropy, and $H_6(X)=1$, which is the same as max depth of your 6-ary decision tree which decides your outcome. The discrepancy comes form the base of the log. you are trying to encode 6 different outcomes in base 2, so you are going to have to use bit-strings of length at least 3. There is clearly some waste here, since there are $2^3 =8$ unique "encodings" on 3 bits, but you only need to encode 6 elements. In general, the entropy bound gets tighter as the number of elements to encode grows, as the first answer already mentioned.
– mm8511
48 mins ago
add a comment |Â
2 Answers
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In your setting, the Shannon entropy is "just" a lower bound for an entropy of any decision tree (including optimal ones). These don't have to coincide. To get closer to what the Shannon entropy is, imagine an optimal decision tree identifying outcomes of throwing a dice $N$ times with some large $N$ (assuming independence). The larger $N$ is, the smaller (yet nonnegative) is the difference between the "averaged" (i.e. divided by $N$) entropy of this "compound" decision tree and the Shannon entropy of the dice. (It resembles a background of arithmetic coding).
answered 1 hour ago
metamorphy
9178
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3
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To recover entropy, you have to consider a sequence of dice throws, and ask how many questions per roll you need in an optimal strategy, in the limit that the number of rolls goes to infinity. Note that each question can cover all the rolls, for example for two rolls, you could ask at some point: “Are the results in $16,21,22,23$?†(where the first digit denotes the first throw, and the second digit denotes the second throw).
I'm too lazy to do it for 36 possibilities, therefore here a simpler example: Consider that each roll gives only one of three results with equal probability. Then the entropy is about $1.58496$.
For one toss, the optimal strategy is simply to ask “was it $1$?†followed by â€Âwas it $2$?â€Â, which on average gives $5/3 = 1.66$ questions.
For two tosses, an optimal strategy would be to first ask “was it one of $11,12,13,21$?†(where the first digit gives the result of the first toss, and the second digit the result of the second toss). If the answer is “yesâ€Â, then use two questions to single out one of the for results. Otherwise, ask “was the first toss a $2$?â€Â, if yes then it was one of $22$ or $23$, and one question is sufficient to determine that. In the remaining case you know the first toss was $3$ and know nothing about the second, so you employ the one-toss strategy to determine the second toss.
This strategy needs on average $29/9=3.2222$ questions, or $1.61111$ questions per toss. Which is already much better, and indeed only $1.65,%$ worse that the value given by the entropy.
Note that the average number of questions of the single-toss optimal strategy can differ dramatically from the entropy. For this, consider the toss of a biased coin. The entropy of this can be made arbitrary low by making the coin sufficiently biased. But obviously there's no way you can get the result of a coin toss with less than one question.
answered 55 mins ago
celtschk
28.8k75497
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
In your setting, the Shannon entropy is "just" a lower bound for an entropy of any decision tree (including optimal ones). These don't have to coincide. To get closer to what the Shannon entropy is, imagine an optimal decision tree identifying outcomes of throwing a dice $N$ times with some large $N$ (assuming independence). The larger $N$ is, the smaller (yet nonnegative) is the difference between the "averaged" (i.e. divided by $N$) entropy of this "compound" decision tree and the Shannon entropy of the dice. (It resembles a background of arithmetic coding).
answered 1 hour ago
metamorphy
9178
add a comment |Â
up vote
4
down vote
In your setting, the Shannon entropy is "just" a lower bound for an entropy of any decision tree (including optimal ones). These don't have to coincide. To get closer to what the Shannon entropy is, imagine an optimal decision tree identifying outcomes of throwing a dice $N$ times with some large $N$ (assuming independence). The larger $N$ is, the smaller (yet nonnegative) is the difference between the "averaged" (i.e. divided by $N$) entropy of this "compound" decision tree and the Shannon entropy of the dice. (It resembles a background of arithmetic coding).
answered 1 hour ago
metamorphy
9178
add a comment |Â
up vote
4
down vote
up vote
4
down vote
In your setting, the Shannon entropy is "just" a lower bound for an entropy of any decision tree (including optimal ones). These don't have to coincide. To get closer to what the Shannon entropy is, imagine an optimal decision tree identifying outcomes of throwing a dice $N$ times with some large $N$ (assuming independence). The larger $N$ is, the smaller (yet nonnegative) is the difference between the "averaged" (i.e. divided by $N$) entropy of this "compound" decision tree and the Shannon entropy of the dice. (It resembles a background of arithmetic coding).
answered 1 hour ago
metamorphy
9178
In your setting, the Shannon entropy is "just" a lower bound for an entropy of any decision tree (including optimal ones). These don't have to coincide. To get closer to what the Shannon entropy is, imagine an optimal decision tree identifying outcomes of throwing a dice $N$ times with some large $N$ (assuming independence). The larger $N$ is, the smaller (yet nonnegative) is the difference between the "averaged" (i.e. divided by $N$) entropy of this "compound" decision tree and the Shannon entropy of the dice. (It resembles a background of arithmetic coding).
answered 1 hour ago
metamorphy
9178
edited 1 min ago
answered 1 hour ago
metamorphy
9178
answered 1 hour ago
metamorphy
9178
answered 1 hour ago
metamorphy
9178
9178
add a comment |Â
add a comment |Â
up vote
3
down vote
To recover entropy, you have to consider a sequence of dice throws, and ask how many questions per roll you need in an optimal strategy, in the limit that the number of rolls goes to infinity. Note that each question can cover all the rolls, for example for two rolls, you could ask at some point: “Are the results in $16,21,22,23$?†(where the first digit denotes the first throw, and the second digit denotes the second throw).
I'm too lazy to do it for 36 possibilities, therefore here a simpler example: Consider that each roll gives only one of three results with equal probability. Then the entropy is about $1.58496$.
For one toss, the optimal strategy is simply to ask “was it $1$?†followed by â€Âwas it $2$?â€Â, which on average gives $5/3 = 1.66$ questions.
For two tosses, an optimal strategy would be to first ask “was it one of $11,12,13,21$?†(where the first digit gives the result of the first toss, and the second digit the result of the second toss). If the answer is “yesâ€Â, then use two questions to single out one of the for results. Otherwise, ask “was the first toss a $2$?â€Â, if yes then it was one of $22$ or $23$, and one question is sufficient to determine that. In the remaining case you know the first toss was $3$ and know nothing about the second, so you employ the one-toss strategy to determine the second toss.
This strategy needs on average $29/9=3.2222$ questions, or $1.61111$ questions per toss. Which is already much better, and indeed only $1.65,%$ worse that the value given by the entropy.
Note that the average number of questions of the single-toss optimal strategy can differ dramatically from the entropy. For this, consider the toss of a biased coin. The entropy of this can be made arbitrary low by making the coin sufficiently biased. But obviously there's no way you can get the result of a coin toss with less than one question.
answered 55 mins ago
celtschk
28.8k75497
add a comment |Â
up vote
3
down vote
To recover entropy, you have to consider a sequence of dice throws, and ask how many questions per roll you need in an optimal strategy, in the limit that the number of rolls goes to infinity. Note that each question can cover all the rolls, for example for two rolls, you could ask at some point: “Are the results in $16,21,22,23$?†(where the first digit denotes the first throw, and the second digit denotes the second throw).
I'm too lazy to do it for 36 possibilities, therefore here a simpler example: Consider that each roll gives only one of three results with equal probability. Then the entropy is about $1.58496$.
For one toss, the optimal strategy is simply to ask “was it $1$?†followed by â€Âwas it $2$?â€Â, which on average gives $5/3 = 1.66$ questions.
For two tosses, an optimal strategy would be to first ask “was it one of $11,12,13,21$?†(where the first digit gives the result of the first toss, and the second digit the result of the second toss). If the answer is “yesâ€Â, then use two questions to single out one of the for results. Otherwise, ask “was the first toss a $2$?â€Â, if yes then it was one of $22$ or $23$, and one question is sufficient to determine that. In the remaining case you know the first toss was $3$ and know nothing about the second, so you employ the one-toss strategy to determine the second toss.
This strategy needs on average $29/9=3.2222$ questions, or $1.61111$ questions per toss. Which is already much better, and indeed only $1.65,%$ worse that the value given by the entropy.
Note that the average number of questions of the single-toss optimal strategy can differ dramatically from the entropy. For this, consider the toss of a biased coin. The entropy of this can be made arbitrary low by making the coin sufficiently biased. But obviously there's no way you can get the result of a coin toss with less than one question.
answered 55 mins ago
celtschk
28.8k75497
add a comment |Â
up vote
3
down vote
up vote
3
down vote
To recover entropy, you have to consider a sequence of dice throws, and ask how many questions per roll you need in an optimal strategy, in the limit that the number of rolls goes to infinity. Note that each question can cover all the rolls, for example for two rolls, you could ask at some point: “Are the results in $16,21,22,23$?†(where the first digit denotes the first throw, and the second digit denotes the second throw).
I'm too lazy to do it for 36 possibilities, therefore here a simpler example: Consider that each roll gives only one of three results with equal probability. Then the entropy is about $1.58496$.
For one toss, the optimal strategy is simply to ask “was it $1$?†followed by â€Âwas it $2$?â€Â, which on average gives $5/3 = 1.66$ questions.
For two tosses, an optimal strategy would be to first ask “was it one of $11,12,13,21$?†(where the first digit gives the result of the first toss, and the second digit the result of the second toss). If the answer is “yesâ€Â, then use two questions to single out one of the for results. Otherwise, ask “was the first toss a $2$?â€Â, if yes then it was one of $22$ or $23$, and one question is sufficient to determine that. In the remaining case you know the first toss was $3$ and know nothing about the second, so you employ the one-toss strategy to determine the second toss.
This strategy needs on average $29/9=3.2222$ questions, or $1.61111$ questions per toss. Which is already much better, and indeed only $1.65,%$ worse that the value given by the entropy.
Note that the average number of questions of the single-toss optimal strategy can differ dramatically from the entropy. For this, consider the toss of a biased coin. The entropy of this can be made arbitrary low by making the coin sufficiently biased. But obviously there's no way you can get the result of a coin toss with less than one question.
answered 55 mins ago
celtschk
28.8k75497
To recover entropy, you have to consider a sequence of dice throws, and ask how many questions per roll you need in an optimal strategy, in the limit that the number of rolls goes to infinity. Note that each question can cover all the rolls, for example for two rolls, you could ask at some point: “Are the results in $16,21,22,23$?†(where the first digit denotes the first throw, and the second digit denotes the second throw).
I'm too lazy to do it for 36 possibilities, therefore here a simpler example: Consider that each roll gives only one of three results with equal probability. Then the entropy is about $1.58496$.
For one toss, the optimal strategy is simply to ask “was it $1$?†followed by â€Âwas it $2$?â€Â, which on average gives $5/3 = 1.66$ questions.
For two tosses, an optimal strategy would be to first ask “was it one of $11,12,13,21$?†(where the first digit gives the result of the first toss, and the second digit the result of the second toss). If the answer is “yesâ€Â, then use two questions to single out one of the for results. Otherwise, ask “was the first toss a $2$?â€Â, if yes then it was one of $22$ or $23$, and one question is sufficient to determine that. In the remaining case you know the first toss was $3$ and know nothing about the second, so you employ the one-toss strategy to determine the second toss.
This strategy needs on average $29/9=3.2222$ questions, or $1.61111$ questions per toss. Which is already much better, and indeed only $1.65,%$ worse that the value given by the entropy.
Note that the average number of questions of the single-toss optimal strategy can differ dramatically from the entropy. For this, consider the toss of a biased coin. The entropy of this can be made arbitrary low by making the coin sufficiently biased. But obviously there's no way you can get the result of a coin toss with less than one question.
answered 55 mins ago
celtschk
28.8k75497
edited 46 mins ago
answered 55 mins ago
celtschk
28.8k75497
answered 55 mins ago
celtschk
28.8k75497
answered 55 mins ago
celtschk
28.8k75497
28.8k75497
add a comment |Â
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Take base 6 entropy, and $H_6(X)=1$, which is the same as max depth of your 6-ary decision tree which decides your outcome. The discrepancy comes form the base of the log. you are trying to encode 6 different outcomes in base 2, so you are going to have to use bit-strings of length at least 3. There is clearly some waste here, since there are $2^3 =8$ unique "encodings" on 3 bits, but you only need to encode 6 elements. In general, the entropy bound gets tighter as the number of elements to encode grows, as the first answer already mentioned.
– mm8511
48 mins ago