Distributing random marbles among five personsProving a variable as discrete random (confusing question)k persons N trials biased coin gameAn urn contains 12 red marbles and 10 blue marbles.Start with 10 dices, and after each round of throwing, we remove the one with number 6, and repeat process.Probability of drawing 'm' same colored marbles out of 'k' boxesCalculating probability of rolling a die exact number of times in aConfusion about distribution of marbles among 5 persons. Example by Kahneman and TverskyProbability of drawing x unique marbles from a bag of size yProbability Question for 2-Sided Coin (Verification)Fair game with marbles
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Distributing random marbles among five persons
Proving a variable as discrete random (confusing question)k persons N trials biased coin gameAn urn contains 12 red marbles and 10 blue marbles.Start with 10 dices, and after each round of throwing, we remove the one with number 6, and repeat process.Probability of drawing 'm' same colored marbles out of 'k' boxesCalculating probability of rolling a die exact number of times in aConfusion about distribution of marbles among 5 persons. Example by Kahneman and TverskyProbability of drawing x unique marbles from a bag of size yProbability Question for 2-Sided Coin (Verification)Fair game with marbles
$begingroup$
On each round of a game, 20 marbles are distributed at random among five children: A, B, C, D and E.
- A: 4, B: 4, C: 5, D: 4 and E: 3
- A: 4, B: 4, C: 4, D: 4 and E: 4
In many rounds of the game, will there be more results of type I or type II?
Can we say that the probability of both scenarios will be same as $frac1(4 from 20)$?
probability probability-distributions
asked Mar 14 at 7:24
Ali J.Ali J.
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
On each round of a game, 20 marbles are distributed at random among five children: A, B, C, D and E.
- A: 4, B: 4, C: 5, D: 4 and E: 3
- A: 4, B: 4, C: 4, D: 4 and E: 4
In many rounds of the game, will there be more results of type I or type II?
Can we say that the probability of both scenarios will be same as $frac1(4 from 20)$?
probability probability-distributions
asked Mar 14 at 7:24
Ali J.Ali J.
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
What exactly counts as a type 1 result? For example, would 54443 count? In any case, probabilities will NOT be the same
$endgroup$
– asdf
Mar 14 at 7:31
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
Yes, but the same kid getting all the marbles is not as probable as each getting 4
$endgroup$
– asdf
Mar 14 at 8:23
add a comment |
$begingroup$
On each round of a game, 20 marbles are distributed at random among five children: A, B, C, D and E.
- A: 4, B: 4, C: 5, D: 4 and E: 3
- A: 4, B: 4, C: 4, D: 4 and E: 4
In many rounds of the game, will there be more results of type I or type II?
Can we say that the probability of both scenarios will be same as $frac1(4 from 20)$?
probability probability-distributions
asked Mar 14 at 7:24
Ali J.Ali J.
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
On each round of a game, 20 marbles are distributed at random among five children: A, B, C, D and E.
- A: 4, B: 4, C: 5, D: 4 and E: 3
- A: 4, B: 4, C: 4, D: 4 and E: 4
In many rounds of the game, will there be more results of type I or type II?
Can we say that the probability of both scenarios will be same as $frac1(4 from 20)$?
probability probability-distributions
probability probability-distributions
asked Mar 14 at 7:24
Ali J.Ali J.
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 14 at 7:24
Ali J.Ali J.
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 14 at 8:22
Ali J.
asked Mar 14 at 7:24
Ali J.Ali J.
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 14 at 7:24
Ali J.Ali J.
484
asked Mar 14 at 7:24
Ali J.Ali J.
484
484
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ali J. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
What exactly counts as a type 1 result? For example, would 54443 count? In any case, probabilities will NOT be the same
$endgroup$
– asdf
Mar 14 at 7:31
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
Yes, but the same kid getting all the marbles is not as probable as each getting 4
$endgroup$
– asdf
Mar 14 at 8:23
add a comment |
1
$begingroup$
What exactly counts as a type 1 result? For example, would 54443 count? In any case, probabilities will NOT be the same
$endgroup$
– asdf
Mar 14 at 7:31
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
Yes, but the same kid getting all the marbles is not as probable as each getting 4
$endgroup$
– asdf
Mar 14 at 8:23
1
1
$begingroup$
What exactly counts as a type 1 result? For example, would 54443 count? In any case, probabilities will NOT be the same
$endgroup$
– asdf
Mar 14 at 7:31
$begingroup$
What exactly counts as a type 1 result? For example, would 54443 count? In any case, probabilities will NOT be the same
$endgroup$
– asdf
Mar 14 at 7:31
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
Yes, but the same kid getting all the marbles is not as probable as each getting 4
$endgroup$
– asdf
Mar 14 at 8:23
$begingroup$
Yes, but the same kid getting all the marbles is not as probable as each getting 4
$endgroup$
– asdf
Mar 14 at 8:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
No, the probability will not be the same, at least not under the scenario that one would usually call 'distributed at random'.
Just like there is a slightly higher probability to see 50 heads out of 100 than to see 49, there is a slightly higher probability of seeing the second configuration (assuming, per asdf's comment, that for the first, we only count exactly the configuration $(4,4,5,4,3)$ and don't also include symmetry-related configurations like $(5,4,4,4,3)$).
The probability of the different configurations here follows a multinomial distribution, but we can just count them up manually. We have five buckets to choose from for each of the twenty objects, so there are $5^20$ total possible allocations. For the first configuration, we choose $4$ for bucket A, $4$ of the remaining for bucket B, five of the remaining for bucket C and so on. So that's $$ 20 choose 416 choose 412 choose 57 choose 43 choose 3$$ different allocations that have the configuration $(4,4,5,4,3).$ Dividing this by $5^20$ gives approximately $0.00256$ or $0.256%.$
On the other hand the total number for configuration $(4,4,4,4,4)$ is$$ 20 choose 416 choose 412 choose 48 choose 44 choose 4$$ which, divided by $5^20$ gives approximately $0.00320,$ or $0.320%,$ which is slightly higher, in accord with our intuition from the coin flip example.
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
@AliJ. But if there were only two kids and 20 marbles, then it's the same thing as coin flips. Would you say the probability of (10,10) was the same as $(11,9)$? If so, you're saying that the probability of getting $10/20$ heads is the same as getting $11/20$ heads. Worse, would you say $(10,10)$ is the same as $(20,0)$? Cause then you'd be saying the probability of getting $10/20$ heads is the same as getting $20/20,$ something that should surely feel intuitively wrong.
$endgroup$
– spaceisdarkgreen
Mar 14 at 8:28
add a comment |
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$begingroup$
No, the probability will not be the same, at least not under the scenario that one would usually call 'distributed at random'.
Just like there is a slightly higher probability to see 50 heads out of 100 than to see 49, there is a slightly higher probability of seeing the second configuration (assuming, per asdf's comment, that for the first, we only count exactly the configuration $(4,4,5,4,3)$ and don't also include symmetry-related configurations like $(5,4,4,4,3)$).
The probability of the different configurations here follows a multinomial distribution, but we can just count them up manually. We have five buckets to choose from for each of the twenty objects, so there are $5^20$ total possible allocations. For the first configuration, we choose $4$ for bucket A, $4$ of the remaining for bucket B, five of the remaining for bucket C and so on. So that's $$ 20 choose 416 choose 412 choose 57 choose 43 choose 3$$ different allocations that have the configuration $(4,4,5,4,3).$ Dividing this by $5^20$ gives approximately $0.00256$ or $0.256%.$
On the other hand the total number for configuration $(4,4,4,4,4)$ is$$ 20 choose 416 choose 412 choose 48 choose 44 choose 4$$ which, divided by $5^20$ gives approximately $0.00320,$ or $0.320%,$ which is slightly higher, in accord with our intuition from the coin flip example.
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
@AliJ. But if there were only two kids and 20 marbles, then it's the same thing as coin flips. Would you say the probability of (10,10) was the same as $(11,9)$? If so, you're saying that the probability of getting $10/20$ heads is the same as getting $11/20$ heads. Worse, would you say $(10,10)$ is the same as $(20,0)$? Cause then you'd be saying the probability of getting $10/20$ heads is the same as getting $20/20,$ something that should surely feel intuitively wrong.
$endgroup$
– spaceisdarkgreen
Mar 14 at 8:28
add a comment |
$begingroup$
No, the probability will not be the same, at least not under the scenario that one would usually call 'distributed at random'.
Just like there is a slightly higher probability to see 50 heads out of 100 than to see 49, there is a slightly higher probability of seeing the second configuration (assuming, per asdf's comment, that for the first, we only count exactly the configuration $(4,4,5,4,3)$ and don't also include symmetry-related configurations like $(5,4,4,4,3)$).
The probability of the different configurations here follows a multinomial distribution, but we can just count them up manually. We have five buckets to choose from for each of the twenty objects, so there are $5^20$ total possible allocations. For the first configuration, we choose $4$ for bucket A, $4$ of the remaining for bucket B, five of the remaining for bucket C and so on. So that's $$ 20 choose 416 choose 412 choose 57 choose 43 choose 3$$ different allocations that have the configuration $(4,4,5,4,3).$ Dividing this by $5^20$ gives approximately $0.00256$ or $0.256%.$
On the other hand the total number for configuration $(4,4,4,4,4)$ is$$ 20 choose 416 choose 412 choose 48 choose 44 choose 4$$ which, divided by $5^20$ gives approximately $0.00320,$ or $0.320%,$ which is slightly higher, in accord with our intuition from the coin flip example.
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
@AliJ. But if there were only two kids and 20 marbles, then it's the same thing as coin flips. Would you say the probability of (10,10) was the same as $(11,9)$? If so, you're saying that the probability of getting $10/20$ heads is the same as getting $11/20$ heads. Worse, would you say $(10,10)$ is the same as $(20,0)$? Cause then you'd be saying the probability of getting $10/20$ heads is the same as getting $20/20,$ something that should surely feel intuitively wrong.
$endgroup$
– spaceisdarkgreen
Mar 14 at 8:28
add a comment |
$begingroup$
No, the probability will not be the same, at least not under the scenario that one would usually call 'distributed at random'.
Just like there is a slightly higher probability to see 50 heads out of 100 than to see 49, there is a slightly higher probability of seeing the second configuration (assuming, per asdf's comment, that for the first, we only count exactly the configuration $(4,4,5,4,3)$ and don't also include symmetry-related configurations like $(5,4,4,4,3)$).
The probability of the different configurations here follows a multinomial distribution, but we can just count them up manually. We have five buckets to choose from for each of the twenty objects, so there are $5^20$ total possible allocations. For the first configuration, we choose $4$ for bucket A, $4$ of the remaining for bucket B, five of the remaining for bucket C and so on. So that's $$ 20 choose 416 choose 412 choose 57 choose 43 choose 3$$ different allocations that have the configuration $(4,4,5,4,3).$ Dividing this by $5^20$ gives approximately $0.00256$ or $0.256%.$
On the other hand the total number for configuration $(4,4,4,4,4)$ is$$ 20 choose 416 choose 412 choose 48 choose 44 choose 4$$ which, divided by $5^20$ gives approximately $0.00320,$ or $0.320%,$ which is slightly higher, in accord with our intuition from the coin flip example.
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
$endgroup$
No, the probability will not be the same, at least not under the scenario that one would usually call 'distributed at random'.
Just like there is a slightly higher probability to see 50 heads out of 100 than to see 49, there is a slightly higher probability of seeing the second configuration (assuming, per asdf's comment, that for the first, we only count exactly the configuration $(4,4,5,4,3)$ and don't also include symmetry-related configurations like $(5,4,4,4,3)$).
The probability of the different configurations here follows a multinomial distribution, but we can just count them up manually. We have five buckets to choose from for each of the twenty objects, so there are $5^20$ total possible allocations. For the first configuration, we choose $4$ for bucket A, $4$ of the remaining for bucket B, five of the remaining for bucket C and so on. So that's $$ 20 choose 416 choose 412 choose 57 choose 43 choose 3$$ different allocations that have the configuration $(4,4,5,4,3).$ Dividing this by $5^20$ gives approximately $0.00256$ or $0.256%.$
On the other hand the total number for configuration $(4,4,4,4,4)$ is$$ 20 choose 416 choose 412 choose 48 choose 44 choose 4$$ which, divided by $5^20$ gives approximately $0.00320,$ or $0.320%,$ which is slightly higher, in accord with our intuition from the coin flip example.
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
answered Mar 14 at 8:01
spaceisdarkgreenspaceisdarkgreen
33.5k21753
33.5k21753
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
@AliJ. But if there were only two kids and 20 marbles, then it's the same thing as coin flips. Would you say the probability of (10,10) was the same as $(11,9)$? If so, you're saying that the probability of getting $10/20$ heads is the same as getting $11/20$ heads. Worse, would you say $(10,10)$ is the same as $(20,0)$? Cause then you'd be saying the probability of getting $10/20$ heads is the same as getting $20/20,$ something that should surely feel intuitively wrong.
$endgroup$
– spaceisdarkgreen
Mar 14 at 8:28
add a comment |
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
@AliJ. But if there were only two kids and 20 marbles, then it's the same thing as coin flips. Would you say the probability of (10,10) was the same as $(11,9)$? If so, you're saying that the probability of getting $10/20$ heads is the same as getting $11/20$ heads. Worse, would you say $(10,10)$ is the same as $(20,0)$? Cause then you'd be saying the probability of getting $10/20$ heads is the same as getting $20/20,$ something that should surely feel intuitively wrong.
$endgroup$
– spaceisdarkgreen
Mar 14 at 8:28
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
@AliJ. But if there were only two kids and 20 marbles, then it's the same thing as coin flips. Would you say the probability of (10,10) was the same as $(11,9)$? If so, you're saying that the probability of getting $10/20$ heads is the same as getting $11/20$ heads. Worse, would you say $(10,10)$ is the same as $(20,0)$? Cause then you'd be saying the probability of getting $10/20$ heads is the same as getting $20/20,$ something that should surely feel intuitively wrong.
$endgroup$
– spaceisdarkgreen
Mar 14 at 8:28
$begingroup$
@AliJ. But if there were only two kids and 20 marbles, then it's the same thing as coin flips. Would you say the probability of (10,10) was the same as $(11,9)$? If so, you're saying that the probability of getting $10/20$ heads is the same as getting $11/20$ heads. Worse, would you say $(10,10)$ is the same as $(20,0)$? Cause then you'd be saying the probability of getting $10/20$ heads is the same as getting $20/20,$ something that should surely feel intuitively wrong.
$endgroup$
– spaceisdarkgreen
Mar 14 at 8:28
add a comment |
Ali J. is a new contributor. Be nice, and check out our Code of Conduct.
Ali J. is a new contributor. Be nice, and check out our Code of Conduct.
Ali J. is a new contributor. Be nice, and check out our Code of Conduct.
Ali J. is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
What exactly counts as a type 1 result? For example, would 54443 count? In any case, probabilities will NOT be the same
$endgroup$
– asdf
Mar 14 at 7:31
$begingroup$
Why I said equal probability? Because the marbles are not difference and there are 20 same marbles.
$endgroup$
– Ali J.
Mar 14 at 8:21
$begingroup$
Yes, but the same kid getting all the marbles is not as probable as each getting 4
$endgroup$
– asdf
Mar 14 at 8:23