What's the probability of guessing on an outcome from a random distribution?
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Problem:
Given a random distribution A and random variable X~A, X$in${1,2,3}, with f.x. probabilities $P_{textrm{A}}(X = 1) = P_{textrm{A}}(X = 2) = 0.1$ and $P_{textrm{A}}(X = 3) = 0.8$. Let's say I'm sampling from this distribution, and also throwing a fair (non-biased) 3-sided dice with numbers 1, 2 and 3. What is the probability of the number on the dice and the sampling from the distribution agree?
My solution
From intuition I would say that the probability is:
$P = sum_{i=1}^3 P_{textrm{Dice}}(i) * P_{textrm{A}}(X = i)$
The problem
This gives us $P = sum_{i=1}^3 frac{1}{3} * P_{A}(X = i) = frac{1}{3} sum_{i=1}^3 P_{A}(X = i) = frac{1}{3} 1 = frac{1}{3}$
So the distribution we are sampling from doesn't matter..., the answer is always 1/3. I guess that might be ok, but I'm not so sure. Am I on the right track? :)
probability probability-distributions
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
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|
show 1 more comment
$begingroup$
Problem:
Given a random distribution A and random variable X~A, X$in${1,2,3}, with f.x. probabilities $P_{textrm{A}}(X = 1) = P_{textrm{A}}(X = 2) = 0.1$ and $P_{textrm{A}}(X = 3) = 0.8$. Let's say I'm sampling from this distribution, and also throwing a fair (non-biased) 3-sided dice with numbers 1, 2 and 3. What is the probability of the number on the dice and the sampling from the distribution agree?
My solution
From intuition I would say that the probability is:
$P = sum_{i=1}^3 P_{textrm{Dice}}(i) * P_{textrm{A}}(X = i)$
The problem
This gives us $P = sum_{i=1}^3 frac{1}{3} * P_{A}(X = i) = frac{1}{3} sum_{i=1}^3 P_{A}(X = i) = frac{1}{3} 1 = frac{1}{3}$
So the distribution we are sampling from doesn't matter..., the answer is always 1/3. I guess that might be ok, but I'm not so sure. Am I on the right track? :)
probability probability-distributions
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
$endgroup$
$begingroup$
Answer and methodology are both good. As you see, there's nothing special about the $3$...your observation generalizes.
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– lulu
Jan 25 at 16:26
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The important facts here are that (1) the die roll and the random variable $X$ are independent, (2) they have the same possible outcomes, and (3) the die roll is uniformly distributed. As long as those three conditions are satisfied, then the probability of them matching will be $1/n$ where $n$ is the number of possible outcomes of the die roll. (More precisely, the possible values of $X$ have to be a subset of the possible outcomes of the die roll, but they don't necessarily have to be equal.)
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– kccu
Jan 25 at 16:42
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Thank you @kccu for a detailed answer. When you say that the values of X have to be a subset of the values of the die, but don't have to be equal. Do you mean that the die could f.x. be a 6 sided die, but the values of X could be unchanged (1,2 and 3)?
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– Frimann Bjornsson
Jan 25 at 20:08
$begingroup$
@FrimannBjornsson That's right. Because you would still have $$P(text{match})=sum_{i=1}^6 P_{text{Die}}(i) cdot P_A(X=i)=sum_{i=1}^6 frac{1}{6}P_A(X=i) = frac{1}{6}sum_{i=1}^6 P_A(X=i) = frac{1}{6}.$$
$endgroup$
– kccu
Jan 25 at 20:10
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I don't think this will generalize nicely to a biased die, because the important part was that we could factor $P_{text{Die}}(i)$ out of the sum since it was constant, so did not depend on $i$.
$endgroup$
– kccu
Jan 25 at 20:12
|
show 1 more comment
$begingroup$
Problem:
Given a random distribution A and random variable X~A, X$in${1,2,3}, with f.x. probabilities $P_{textrm{A}}(X = 1) = P_{textrm{A}}(X = 2) = 0.1$ and $P_{textrm{A}}(X = 3) = 0.8$. Let's say I'm sampling from this distribution, and also throwing a fair (non-biased) 3-sided dice with numbers 1, 2 and 3. What is the probability of the number on the dice and the sampling from the distribution agree?
My solution
From intuition I would say that the probability is:
$P = sum_{i=1}^3 P_{textrm{Dice}}(i) * P_{textrm{A}}(X = i)$
The problem
This gives us $P = sum_{i=1}^3 frac{1}{3} * P_{A}(X = i) = frac{1}{3} sum_{i=1}^3 P_{A}(X = i) = frac{1}{3} 1 = frac{1}{3}$
So the distribution we are sampling from doesn't matter..., the answer is always 1/3. I guess that might be ok, but I'm not so sure. Am I on the right track? :)
probability probability-distributions
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
$endgroup$
Problem:
Given a random distribution A and random variable X~A, X$in${1,2,3}, with f.x. probabilities $P_{textrm{A}}(X = 1) = P_{textrm{A}}(X = 2) = 0.1$ and $P_{textrm{A}}(X = 3) = 0.8$. Let's say I'm sampling from this distribution, and also throwing a fair (non-biased) 3-sided dice with numbers 1, 2 and 3. What is the probability of the number on the dice and the sampling from the distribution agree?
My solution
From intuition I would say that the probability is:
$P = sum_{i=1}^3 P_{textrm{Dice}}(i) * P_{textrm{A}}(X = i)$
The problem
This gives us $P = sum_{i=1}^3 frac{1}{3} * P_{A}(X = i) = frac{1}{3} sum_{i=1}^3 P_{A}(X = i) = frac{1}{3} 1 = frac{1}{3}$
So the distribution we are sampling from doesn't matter..., the answer is always 1/3. I guess that might be ok, but I'm not so sure. Am I on the right track? :)
probability probability-distributions
probability probability-distributions
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
asked Jan 25 at 16:23
Frimann BjornssonFrimann Bjornsson
31
31
$begingroup$
Answer and methodology are both good. As you see, there's nothing special about the $3$...your observation generalizes.
$endgroup$
– lulu
Jan 25 at 16:26
$begingroup$
The important facts here are that (1) the die roll and the random variable $X$ are independent, (2) they have the same possible outcomes, and (3) the die roll is uniformly distributed. As long as those three conditions are satisfied, then the probability of them matching will be $1/n$ where $n$ is the number of possible outcomes of the die roll. (More precisely, the possible values of $X$ have to be a subset of the possible outcomes of the die roll, but they don't necessarily have to be equal.)
$endgroup$
– kccu
Jan 25 at 16:42
$begingroup$
Thank you @kccu for a detailed answer. When you say that the values of X have to be a subset of the values of the die, but don't have to be equal. Do you mean that the die could f.x. be a 6 sided die, but the values of X could be unchanged (1,2 and 3)?
$endgroup$
– Frimann Bjornsson
Jan 25 at 20:08
$begingroup$
@FrimannBjornsson That's right. Because you would still have $$P(text{match})=sum_{i=1}^6 P_{text{Die}}(i) cdot P_A(X=i)=sum_{i=1}^6 frac{1}{6}P_A(X=i) = frac{1}{6}sum_{i=1}^6 P_A(X=i) = frac{1}{6}.$$
$endgroup$
– kccu
Jan 25 at 20:10
$begingroup$
I don't think this will generalize nicely to a biased die, because the important part was that we could factor $P_{text{Die}}(i)$ out of the sum since it was constant, so did not depend on $i$.
$endgroup$
– kccu
Jan 25 at 20:12
|
show 1 more comment
$begingroup$
Answer and methodology are both good. As you see, there's nothing special about the $3$...your observation generalizes.
$endgroup$
– lulu
Jan 25 at 16:26
$begingroup$
The important facts here are that (1) the die roll and the random variable $X$ are independent, (2) they have the same possible outcomes, and (3) the die roll is uniformly distributed. As long as those three conditions are satisfied, then the probability of them matching will be $1/n$ where $n$ is the number of possible outcomes of the die roll. (More precisely, the possible values of $X$ have to be a subset of the possible outcomes of the die roll, but they don't necessarily have to be equal.)
$endgroup$
– kccu
Jan 25 at 16:42
$begingroup$
Thank you @kccu for a detailed answer. When you say that the values of X have to be a subset of the values of the die, but don't have to be equal. Do you mean that the die could f.x. be a 6 sided die, but the values of X could be unchanged (1,2 and 3)?
$endgroup$
– Frimann Bjornsson
Jan 25 at 20:08
$begingroup$
@FrimannBjornsson That's right. Because you would still have $$P(text{match})=sum_{i=1}^6 P_{text{Die}}(i) cdot P_A(X=i)=sum_{i=1}^6 frac{1}{6}P_A(X=i) = frac{1}{6}sum_{i=1}^6 P_A(X=i) = frac{1}{6}.$$
$endgroup$
– kccu
Jan 25 at 20:10
$begingroup$
I don't think this will generalize nicely to a biased die, because the important part was that we could factor $P_{text{Die}}(i)$ out of the sum since it was constant, so did not depend on $i$.
$endgroup$
– kccu
Jan 25 at 20:12
$begingroup$
Answer and methodology are both good. As you see, there's nothing special about the $3$...your observation generalizes.
$endgroup$
– lulu
Jan 25 at 16:26
$begingroup$
Answer and methodology are both good. As you see, there's nothing special about the $3$...your observation generalizes.
$endgroup$
– lulu
Jan 25 at 16:26
$begingroup$
The important facts here are that (1) the die roll and the random variable $X$ are independent, (2) they have the same possible outcomes, and (3) the die roll is uniformly distributed. As long as those three conditions are satisfied, then the probability of them matching will be $1/n$ where $n$ is the number of possible outcomes of the die roll. (More precisely, the possible values of $X$ have to be a subset of the possible outcomes of the die roll, but they don't necessarily have to be equal.)
$endgroup$
– kccu
Jan 25 at 16:42
$begingroup$
The important facts here are that (1) the die roll and the random variable $X$ are independent, (2) they have the same possible outcomes, and (3) the die roll is uniformly distributed. As long as those three conditions are satisfied, then the probability of them matching will be $1/n$ where $n$ is the number of possible outcomes of the die roll. (More precisely, the possible values of $X$ have to be a subset of the possible outcomes of the die roll, but they don't necessarily have to be equal.)
$endgroup$
– kccu
Jan 25 at 16:42
$begingroup$
Thank you @kccu for a detailed answer. When you say that the values of X have to be a subset of the values of the die, but don't have to be equal. Do you mean that the die could f.x. be a 6 sided die, but the values of X could be unchanged (1,2 and 3)?
$endgroup$
– Frimann Bjornsson
Jan 25 at 20:08
$begingroup$
Thank you @kccu for a detailed answer. When you say that the values of X have to be a subset of the values of the die, but don't have to be equal. Do you mean that the die could f.x. be a 6 sided die, but the values of X could be unchanged (1,2 and 3)?
$endgroup$
– Frimann Bjornsson
Jan 25 at 20:08
$begingroup$
@FrimannBjornsson That's right. Because you would still have $$P(text{match})=sum_{i=1}^6 P_{text{Die}}(i) cdot P_A(X=i)=sum_{i=1}^6 frac{1}{6}P_A(X=i) = frac{1}{6}sum_{i=1}^6 P_A(X=i) = frac{1}{6}.$$
$endgroup$
– kccu
Jan 25 at 20:10
$begingroup$
@FrimannBjornsson That's right. Because you would still have $$P(text{match})=sum_{i=1}^6 P_{text{Die}}(i) cdot P_A(X=i)=sum_{i=1}^6 frac{1}{6}P_A(X=i) = frac{1}{6}sum_{i=1}^6 P_A(X=i) = frac{1}{6}.$$
$endgroup$
– kccu
Jan 25 at 20:10
$begingroup$
I don't think this will generalize nicely to a biased die, because the important part was that we could factor $P_{text{Die}}(i)$ out of the sum since it was constant, so did not depend on $i$.
$endgroup$
– kccu
Jan 25 at 20:12
$begingroup$
I don't think this will generalize nicely to a biased die, because the important part was that we could factor $P_{text{Die}}(i)$ out of the sum since it was constant, so did not depend on $i$.
$endgroup$
– kccu
Jan 25 at 20:12
|
show 1 more comment
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$begingroup$
Answer and methodology are both good. As you see, there's nothing special about the $3$...your observation generalizes.
$endgroup$
– lulu
Jan 25 at 16:26
$begingroup$
The important facts here are that (1) the die roll and the random variable $X$ are independent, (2) they have the same possible outcomes, and (3) the die roll is uniformly distributed. As long as those three conditions are satisfied, then the probability of them matching will be $1/n$ where $n$ is the number of possible outcomes of the die roll. (More precisely, the possible values of $X$ have to be a subset of the possible outcomes of the die roll, but they don't necessarily have to be equal.)
$endgroup$
– kccu
Jan 25 at 16:42
$begingroup$
Thank you @kccu for a detailed answer. When you say that the values of X have to be a subset of the values of the die, but don't have to be equal. Do you mean that the die could f.x. be a 6 sided die, but the values of X could be unchanged (1,2 and 3)?
$endgroup$
– Frimann Bjornsson
Jan 25 at 20:08
$begingroup$
@FrimannBjornsson That's right. Because you would still have $$P(text{match})=sum_{i=1}^6 P_{text{Die}}(i) cdot P_A(X=i)=sum_{i=1}^6 frac{1}{6}P_A(X=i) = frac{1}{6}sum_{i=1}^6 P_A(X=i) = frac{1}{6}.$$
$endgroup$
– kccu
Jan 25 at 20:10
$begingroup$
I don't think this will generalize nicely to a biased die, because the important part was that we could factor $P_{text{Die}}(i)$ out of the sum since it was constant, so did not depend on $i$.
$endgroup$
– kccu
Jan 25 at 20:12