Mixing
The probability of objects of the same class being in a row.
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Suppose I have 8 blocks of three different colors: red, yellow, and blue. Suppose there are 4 red ones, 3 yellow ones, and 1 blue one. If I randomly line them up, what is the probability that there will be 4 red ones in a row? 3 red ones? 2?
I tried the following:
- 4 red in a row: $left(4! cdot 4choose 4 cdot 5right)/8! = 0.30%$
- 3 red in a row: $left(3! cdot 4choose 3 cdot 6right)/8! = 0.36%$
- 2 red in a row: $left(2! cdot 4choose 2 cdot 7right)/8! = 0.21%$
My reasoning was as follows. In the 4 red in a row case, for example, there are:
- $4!$ ways to order the group of 4 red blocks in a row
- $4choose 4$ ways of choosing 4 red blocks from 4 red blocks (which is trivial in this case but follows the general form)
- $5$ possible positions of 4 consecutive blocks in a group of 8
- $8!$ total possible block orderings
But the probability of 2 red in a row should certainly not be greater than the probability of 3 red in a row, so I must be doing something wrong. Does anyone know how to handle this problem?
The general version would be something like: Suppose there are $n$ objects of $m$ different classes. If these objects are randomly ordered, what is the probability of $k$ objects of the same class being in a row?
probability combinatorics
asked Jul 15 at 21:08
user577374
111
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up vote
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Suppose I have 8 blocks of three different colors: red, yellow, and blue. Suppose there are 4 red ones, 3 yellow ones, and 1 blue one. If I randomly line them up, what is the probability that there will be 4 red ones in a row? 3 red ones? 2?
I tried the following:
- 4 red in a row: $left(4! cdot 4choose 4 cdot 5right)/8! = 0.30%$
- 3 red in a row: $left(3! cdot 4choose 3 cdot 6right)/8! = 0.36%$
- 2 red in a row: $left(2! cdot 4choose 2 cdot 7right)/8! = 0.21%$
My reasoning was as follows. In the 4 red in a row case, for example, there are:
- $4!$ ways to order the group of 4 red blocks in a row
- $4choose 4$ ways of choosing 4 red blocks from 4 red blocks (which is trivial in this case but follows the general form)
- $5$ possible positions of 4 consecutive blocks in a group of 8
- $8!$ total possible block orderings
But the probability of 2 red in a row should certainly not be greater than the probability of 3 red in a row, so I must be doing something wrong. Does anyone know how to handle this problem?
The general version would be something like: Suppose there are $n$ objects of $m$ different classes. If these objects are randomly ordered, what is the probability of $k$ objects of the same class being in a row?
probability combinatorics
asked Jul 15 at 21:08
user577374
111
1
The $m$ different classes don't enter into it. You effectively have only $2$ different classes -- objects that you want to see in a row, and other objects. (Unless by "$k$ objects of the same class" you mean any class -- in that case you should clarify that, since in the first paragraph you were only interested in one particular class.)
– joriki
Jul 15 at 21:18
@Shaun Thanks for the tip, I'll add some more context and show effort in my edits. I'm new here, so I'm getting used to the norms
– user577374
Jul 17 at 2:50
For 2 reds in a row, do you mean only one case of exactly 2 reds in a row ?
– true blue anil
Jul 17 at 6:43
@trueblueanil No, I mean any ordering that has 2 reds in a row. (So it could have more reds in a row, too.)
– user577374
Jul 18 at 6:51
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
Suppose I have 8 blocks of three different colors: red, yellow, and blue. Suppose there are 4 red ones, 3 yellow ones, and 1 blue one. If I randomly line them up, what is the probability that there will be 4 red ones in a row? 3 red ones? 2?
I tried the following:
- 4 red in a row: $left(4! cdot 4choose 4 cdot 5right)/8! = 0.30%$
- 3 red in a row: $left(3! cdot 4choose 3 cdot 6right)/8! = 0.36%$
- 2 red in a row: $left(2! cdot 4choose 2 cdot 7right)/8! = 0.21%$
My reasoning was as follows. In the 4 red in a row case, for example, there are:
- $4!$ ways to order the group of 4 red blocks in a row
- $4choose 4$ ways of choosing 4 red blocks from 4 red blocks (which is trivial in this case but follows the general form)
- $5$ possible positions of 4 consecutive blocks in a group of 8
- $8!$ total possible block orderings
But the probability of 2 red in a row should certainly not be greater than the probability of 3 red in a row, so I must be doing something wrong. Does anyone know how to handle this problem?
The general version would be something like: Suppose there are $n$ objects of $m$ different classes. If these objects are randomly ordered, what is the probability of $k$ objects of the same class being in a row?
probability combinatorics
asked Jul 15 at 21:08
user577374
111
Suppose I have 8 blocks of three different colors: red, yellow, and blue. Suppose there are 4 red ones, 3 yellow ones, and 1 blue one. If I randomly line them up, what is the probability that there will be 4 red ones in a row? 3 red ones? 2?
I tried the following:
- 4 red in a row: $left(4! cdot 4choose 4 cdot 5right)/8! = 0.30%$
- 3 red in a row: $left(3! cdot 4choose 3 cdot 6right)/8! = 0.36%$
- 2 red in a row: $left(2! cdot 4choose 2 cdot 7right)/8! = 0.21%$
My reasoning was as follows. In the 4 red in a row case, for example, there are:
- $4!$ ways to order the group of 4 red blocks in a row
- $4choose 4$ ways of choosing 4 red blocks from 4 red blocks (which is trivial in this case but follows the general form)
- $5$ possible positions of 4 consecutive blocks in a group of 8
- $8!$ total possible block orderings
But the probability of 2 red in a row should certainly not be greater than the probability of 3 red in a row, so I must be doing something wrong. Does anyone know how to handle this problem?
The general version would be something like: Suppose there are $n$ objects of $m$ different classes. If these objects are randomly ordered, what is the probability of $k$ objects of the same class being in a row?
probability combinatorics
asked Jul 15 at 21:08
user577374
111
edited Jul 17 at 3:04
asked Jul 15 at 21:08
user577374
111
asked Jul 15 at 21:08
user577374
111
asked Jul 15 at 21:08
user577374
111
111
1
The $m$ different classes don't enter into it. You effectively have only $2$ different classes -- objects that you want to see in a row, and other objects. (Unless by "$k$ objects of the same class" you mean any class -- in that case you should clarify that, since in the first paragraph you were only interested in one particular class.)
– joriki
Jul 15 at 21:18
@Shaun Thanks for the tip, I'll add some more context and show effort in my edits. I'm new here, so I'm getting used to the norms
– user577374
Jul 17 at 2:50
For 2 reds in a row, do you mean only one case of exactly 2 reds in a row ?
– true blue anil
Jul 17 at 6:43
@trueblueanil No, I mean any ordering that has 2 reds in a row. (So it could have more reds in a row, too.)
– user577374
Jul 18 at 6:51
add a comment |Â
1
The $m$ different classes don't enter into it. You effectively have only $2$ different classes -- objects that you want to see in a row, and other objects. (Unless by "$k$ objects of the same class" you mean any class -- in that case you should clarify that, since in the first paragraph you were only interested in one particular class.)
– joriki
Jul 15 at 21:18
@Shaun Thanks for the tip, I'll add some more context and show effort in my edits. I'm new here, so I'm getting used to the norms
– user577374
Jul 17 at 2:50
For 2 reds in a row, do you mean only one case of exactly 2 reds in a row ?
– true blue anil
Jul 17 at 6:43
@trueblueanil No, I mean any ordering that has 2 reds in a row. (So it could have more reds in a row, too.)
– user577374
Jul 18 at 6:51
1
1
The $m$ different classes don't enter into it. You effectively have only $2$ different classes -- objects that you want to see in a row, and other objects. (Unless by "$k$ objects of the same class" you mean any class -- in that case you should clarify that, since in the first paragraph you were only interested in one particular class.)
– joriki
Jul 15 at 21:18
The $m$ different classes don't enter into it. You effectively have only $2$ different classes -- objects that you want to see in a row, and other objects. (Unless by "$k$ objects of the same class" you mean any class -- in that case you should clarify that, since in the first paragraph you were only interested in one particular class.)
– joriki
Jul 15 at 21:18
@Shaun Thanks for the tip, I'll add some more context and show effort in my edits. I'm new here, so I'm getting used to the norms
– user577374
Jul 17 at 2:50
@Shaun Thanks for the tip, I'll add some more context and show effort in my edits. I'm new here, so I'm getting used to the norms
– user577374
Jul 17 at 2:50
For 2 reds in a row, do you mean only one case of exactly 2 reds in a row ?
– true blue anil
Jul 17 at 6:43
For 2 reds in a row, do you mean only one case of exactly 2 reds in a row ?
– true blue anil
Jul 17 at 6:43
@trueblueanil No, I mean any ordering that has 2 reds in a row. (So it could have more reds in a row, too.)
– user577374
Jul 18 at 6:51
@trueblueanil No, I mean any ordering that has 2 reds in a row. (So it could have more reds in a row, too.)
– user577374
Jul 18 at 6:51
add a comment |Â
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1
The $m$ different classes don't enter into it. You effectively have only $2$ different classes -- objects that you want to see in a row, and other objects. (Unless by "$k$ objects of the same class" you mean any class -- in that case you should clarify that, since in the first paragraph you were only interested in one particular class.)
– joriki
Jul 15 at 21:18
@Shaun Thanks for the tip, I'll add some more context and show effort in my edits. I'm new here, so I'm getting used to the norms
– user577374
Jul 17 at 2:50
For 2 reds in a row, do you mean only one case of exactly 2 reds in a row ?
– true blue anil
Jul 17 at 6:43
@trueblueanil No, I mean any ordering that has 2 reds in a row. (So it could have more reds in a row, too.)
– user577374
Jul 18 at 6:51