Mixing
Runs Application Probability Problem Feller
Clash Royale CLAN TAG#URR8PPP
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Taken from a problem in Feller's Probability Theory Vol. 1 Chapter 2 Section 5:
Suppose that an observation yielded the following arrangement of empty and occupied seats along a lunch counter: EOEEOEEEOEEEOEOE. There are $11$ runs here, and Feller argues that the probability of eleven runs would be $0.0578...$ and so therefore it is unlikely that this arrangement is due to chance, and this excess of runs (with five occupied and eleven empty seats it is impossible to get more than eleven runs) points to intentional mixing (people not wanting to sit next to each other if possible). Feller grabs this number out the air, so does anyone know how he calculates $0.0578..$ and how one would do this in a more general case.
Thank you.
probability combinatorics statistics
asked Aug 5 at 23:42
Daniele1234
716215
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up vote
1
down vote
favorite
Taken from a problem in Feller's Probability Theory Vol. 1 Chapter 2 Section 5:
Suppose that an observation yielded the following arrangement of empty and occupied seats along a lunch counter: EOEEOEEEOEEEOEOE. There are $11$ runs here, and Feller argues that the probability of eleven runs would be $0.0578...$ and so therefore it is unlikely that this arrangement is due to chance, and this excess of runs (with five occupied and eleven empty seats it is impossible to get more than eleven runs) points to intentional mixing (people not wanting to sit next to each other if possible). Feller grabs this number out the air, so does anyone know how he calculates $0.0578..$ and how one would do this in a more general case.
Thank you.
probability combinatorics statistics
asked Aug 5 at 23:42
Daniele1234
716215
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Taken from a problem in Feller's Probability Theory Vol. 1 Chapter 2 Section 5:
Suppose that an observation yielded the following arrangement of empty and occupied seats along a lunch counter: EOEEOEEEOEEEOEOE. There are $11$ runs here, and Feller argues that the probability of eleven runs would be $0.0578...$ and so therefore it is unlikely that this arrangement is due to chance, and this excess of runs (with five occupied and eleven empty seats it is impossible to get more than eleven runs) points to intentional mixing (people not wanting to sit next to each other if possible). Feller grabs this number out the air, so does anyone know how he calculates $0.0578..$ and how one would do this in a more general case.
Thank you.
probability combinatorics statistics
asked Aug 5 at 23:42
Daniele1234
716215
Taken from a problem in Feller's Probability Theory Vol. 1 Chapter 2 Section 5:
Suppose that an observation yielded the following arrangement of empty and occupied seats along a lunch counter: EOEEOEEEOEEEOEOE. There are $11$ runs here, and Feller argues that the probability of eleven runs would be $0.0578...$ and so therefore it is unlikely that this arrangement is due to chance, and this excess of runs (with five occupied and eleven empty seats it is impossible to get more than eleven runs) points to intentional mixing (people not wanting to sit next to each other if possible). Feller grabs this number out the air, so does anyone know how he calculates $0.0578..$ and how one would do this in a more general case.
Thank you.
probability combinatorics statistics
asked Aug 5 at 23:42
Daniele1234
716215
asked Aug 5 at 23:42
Daniele1234
716215
asked Aug 5 at 23:42
Daniele1234
716215
asked Aug 5 at 23:42
Daniele1234
716215
716215
add a comment |Â
add a comment |Â
1 Answer
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up vote
1
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I think it is possible that Feller's calculator was off by 0.0001, since I am getting the answer to be 0.0577 after rounding. (It is 3/52).
So, how do we go about calculating this? This is a simple counting argument: Assuming a length 16 string with 6 Es and 10 Os, how do we count the probability of getting a string that has 11 "runs" (by a run, we refer to a contiguous sequence of Es or Os)?
We quickly note that (like OP noted) that 11 is indeed the maximum, and the only way that that can happen is if we start and with an E, and between each E there is at least one E. This reduces the problem to a stars and bars problem, since any string we have must be of the form:
E $B_1$ O E $B_2$ O E $B_3$ O E $B_4$ O E $B_5$ O E $B_6$ where $B_i$ contains some number of Es and the sum of all Es in all those buckets is 5 (we used 6 of them to guarantee that we have 11 runs). Now, by using a stars and bars argument, you can count the number of such strings is $binom105$ (to count, just note that permutations of 5 stars and 5 bars corresponds to filling 6 buckets with 5 Es).
Now, just count the total number of options, which is $binom165$, and thus the ratio is indeed $fracbinom105binom165$ which is approximately 0.0577.
answered Aug 6 at 0:05
E-A
1,8101312
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
I think it is possible that Feller's calculator was off by 0.0001, since I am getting the answer to be 0.0577 after rounding. (It is 3/52).
So, how do we go about calculating this? This is a simple counting argument: Assuming a length 16 string with 6 Es and 10 Os, how do we count the probability of getting a string that has 11 "runs" (by a run, we refer to a contiguous sequence of Es or Os)?
We quickly note that (like OP noted) that 11 is indeed the maximum, and the only way that that can happen is if we start and with an E, and between each E there is at least one E. This reduces the problem to a stars and bars problem, since any string we have must be of the form:
E $B_1$ O E $B_2$ O E $B_3$ O E $B_4$ O E $B_5$ O E $B_6$ where $B_i$ contains some number of Es and the sum of all Es in all those buckets is 5 (we used 6 of them to guarantee that we have 11 runs). Now, by using a stars and bars argument, you can count the number of such strings is $binom105$ (to count, just note that permutations of 5 stars and 5 bars corresponds to filling 6 buckets with 5 Es).
Now, just count the total number of options, which is $binom165$, and thus the ratio is indeed $fracbinom105binom165$ which is approximately 0.0577.
answered Aug 6 at 0:05
E-A
1,8101312
add a comment |Â
up vote
1
down vote
accepted
I think it is possible that Feller's calculator was off by 0.0001, since I am getting the answer to be 0.0577 after rounding. (It is 3/52).
So, how do we go about calculating this? This is a simple counting argument: Assuming a length 16 string with 6 Es and 10 Os, how do we count the probability of getting a string that has 11 "runs" (by a run, we refer to a contiguous sequence of Es or Os)?
We quickly note that (like OP noted) that 11 is indeed the maximum, and the only way that that can happen is if we start and with an E, and between each E there is at least one E. This reduces the problem to a stars and bars problem, since any string we have must be of the form:
E $B_1$ O E $B_2$ O E $B_3$ O E $B_4$ O E $B_5$ O E $B_6$ where $B_i$ contains some number of Es and the sum of all Es in all those buckets is 5 (we used 6 of them to guarantee that we have 11 runs). Now, by using a stars and bars argument, you can count the number of such strings is $binom105$ (to count, just note that permutations of 5 stars and 5 bars corresponds to filling 6 buckets with 5 Es).
Now, just count the total number of options, which is $binom165$, and thus the ratio is indeed $fracbinom105binom165$ which is approximately 0.0577.
answered Aug 6 at 0:05
E-A
1,8101312
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
I think it is possible that Feller's calculator was off by 0.0001, since I am getting the answer to be 0.0577 after rounding. (It is 3/52).
So, how do we go about calculating this? This is a simple counting argument: Assuming a length 16 string with 6 Es and 10 Os, how do we count the probability of getting a string that has 11 "runs" (by a run, we refer to a contiguous sequence of Es or Os)?
We quickly note that (like OP noted) that 11 is indeed the maximum, and the only way that that can happen is if we start and with an E, and between each E there is at least one E. This reduces the problem to a stars and bars problem, since any string we have must be of the form:
E $B_1$ O E $B_2$ O E $B_3$ O E $B_4$ O E $B_5$ O E $B_6$ where $B_i$ contains some number of Es and the sum of all Es in all those buckets is 5 (we used 6 of them to guarantee that we have 11 runs). Now, by using a stars and bars argument, you can count the number of such strings is $binom105$ (to count, just note that permutations of 5 stars and 5 bars corresponds to filling 6 buckets with 5 Es).
Now, just count the total number of options, which is $binom165$, and thus the ratio is indeed $fracbinom105binom165$ which is approximately 0.0577.
answered Aug 6 at 0:05
E-A
1,8101312
I think it is possible that Feller's calculator was off by 0.0001, since I am getting the answer to be 0.0577 after rounding. (It is 3/52).
So, how do we go about calculating this? This is a simple counting argument: Assuming a length 16 string with 6 Es and 10 Os, how do we count the probability of getting a string that has 11 "runs" (by a run, we refer to a contiguous sequence of Es or Os)?
We quickly note that (like OP noted) that 11 is indeed the maximum, and the only way that that can happen is if we start and with an E, and between each E there is at least one E. This reduces the problem to a stars and bars problem, since any string we have must be of the form:
E $B_1$ O E $B_2$ O E $B_3$ O E $B_4$ O E $B_5$ O E $B_6$ where $B_i$ contains some number of Es and the sum of all Es in all those buckets is 5 (we used 6 of them to guarantee that we have 11 runs). Now, by using a stars and bars argument, you can count the number of such strings is $binom105$ (to count, just note that permutations of 5 stars and 5 bars corresponds to filling 6 buckets with 5 Es).
Now, just count the total number of options, which is $binom165$, and thus the ratio is indeed $fracbinom105binom165$ which is approximately 0.0577.
answered Aug 6 at 0:05
E-A
1,8101312
answered Aug 6 at 0:05
E-A
1,8101312
answered Aug 6 at 0:05
E-A
1,8101312
answered Aug 6 at 0:05
E-A
1,8101312
1,8101312
add a comment |Â
add a comment |Â
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