What Is the Probability of Finding 10 Consecutive Shoes with 5 Left and 5 Right?

[LCSphysicist]

Homework Statement

> 20 shoes, from 10 pairs of shoes, are lined randomly. What is the
> probability that there is a set of 10 consecutive shoes with 5
> left shoes and 5 right shoes?

Relevant Equations

.

> 20 shoes, from 10 pairs of shoes, are lined randomly. What is the
> probability that there is a set of 10 consecutive shoes with 5 left shoes
> and 5 right shoes?

I thought that would be a good idea to imagine 10 shoes as one unique object, as follow:
Instead enumerate 20 objects, let's separate 10 shoes and think them as another object (call it X), so that there is 11 in total.

There is 11! ways to permute each objects. The problem is to know how to count X.

So i am stuck in how to count |X| right. I thought that we could first choose $C_{10,5} = 10!/(5!5!)$ right shoes, $C_{10,5}$ left shoes. Now, the number of ways that satisfy our conditions is $C_{10,5}*C_{10,5}*\frac{{10!}*11!}{1}$, so the prob is $C_{10,5}*C_{10,5}*\frac{{10!}*11!}{20!}$. But this is greater than 1 (...), i have no idea what to do now.

I would appreciate to receive an answer that specifies where is my error :.

 

[sysprog]

Please specify your answer in probablistic terms $-$ e.g. what is the percentage of likelihood, or at what value between: '0 = certainly false'; and '1 = certainly true' is your yea or nay going?

 

[LCSphysicist]

Please specify your answer in probablistic terms $-$ e.g. what is the percentage of likelihood, or where in the confidence interval between: '0 = certainly false'; and '1 = certainly true' is your yea or nay going?

Sorry, i don't understand what you mean with yea nay (I am not a english native speaker, so i am searching about it right now... never head before XD)
But, if i understood, i think it is previously assumed in probabilistic problems (pure probabilistic problems) that 0 and 1 have the meaning you pointed. I assumed that since nothing was said about it.

 

[sysprog]

I apologize @LCSphysicist $-$ I think that 'yea' and 'nay' mean 'I proclaim that yes' and 'I proclaim that no', respectively, and also at the same time mean 'yes' and 'no' $\dots$

 

[LCSphysicist]

Oh i understood but wrote wrong . SO i proclaim that yes, 1 is certainty, while 0 is impossible**,anyway, thank you for the information ;)

So what do you think? I can't see any error in my solution :/

 

[sysprog]

@LCSphysicist I would like you to please say a number for what the probability is regarding the podiality (analogous to chirality) left-or-right character of the shoes specified in the problem.

The question says:

What is the
probability that there is a set of 10 consecutive shoes with 5
left shoes and 5 right shoes?​

The correct answer to that question has to include a probabilty number. You wrote some beautiful stuff. Please show an attempt at the answer.

 

[PeroK]

Homework Statement:: > 20 shoes, from 10 pairs of shoes, are lined randomly. What is the
> probability that there is a set of 10 consecutive shoes with 5
> left shoes and 5 right shoes?
Relevant Equations:: .

> 20 shoes, from 10 pairs of shoes, are lined randomly. What is the
> probability that there is a set of 10 consecutive shoes with 5 left shoes
> and 5 right shoes?

Just to make sure we all understand and agree on the question.

First we look at the first 10 shoes and if we have 5 left and 5 right, then that meets the criterion. That is a successful outcome.

If not, we look at the second to eleventh shoes. And if that sequence has 5 left and 5 right, then we have a successful outcome.

If not, we do the same for the third to twelfth shoes and so on until we get to the eleventh to twentieth shoes. And, if any of these sequences has 5 left and 5 right then we have a successful outcome.

We want to calculate the probability of a successful outcome.

 

[PeroK]

Homework Statement:: > 20 shoes, from 10 pairs of shoes, are lined randomly. What is the
> probability that there is a set of 10 consecutive shoes with 5
> left shoes and 5 right shoes?

What I would do is this. Try to find an example of an unsuccessful outcome, to get an idea of what it might look like. And, 20 shoes is a lot, so perhaps start with 4 shoes, then 8 shoes etc. See what unsuccessful outcomes look like.

 

[sysprog]

@PeroK Your 'outcomes' approach reminds me of the 'Scarne on Cards' explanation of how to determine poker hand odds $-$ I think that although it can be tough on some people to be disabused of their fantasaical notions, the approach that you recommend could save some people some money $\dots$

 

[vela]

I thought that would be a good idea to imagine 10 shoes as one unique object, as follow:
Instead enumerate 20 objects, let's separate 10 shoes and think them as another object (call it X), so that there is 11 in total.

There is 11! ways to permute each objects. The problem is to know how to count X.

It's not completely clear to me what you meant here, so let me describe what I think you meant.

Object X is an arrangement of 5 left shoes followed by 5 right shoes, i.e. X = (L1, L2, L3, L4, L5, R1, R2, R3, R4, R5). That leaves 10 more shoes, which I'll call S1 through S10. The collection {X, S1, ..., S10} can be arranged in 11! ways.

Now, the number of ways that satisfy our conditions is $C_{10,5}*C_{10,5}*\frac{{10!}*11!}{1}$

I assume the 11! is the same as the 11! above. I'm not seeing where the 10! comes from or what the 1 in the denominator corresponds to. Could you elaborate on your logic here?

so the prob is $C_{10,5}*C_{10,5}*\frac{{10!}*11!}{20!}$.

Finally, you divide by 20! because that's how many ways you can permute the 20 shoes.

Is that right?

 

[LCSphysicist]

Ok, i will try to be clearer ;)

We can chose 5 left shoes of a total of 10, and the same to right shoes.
So that we have C10,5 for each case, a total of C10,5*C10,5.
Now that we "choose" our shoes, we can arrange themselves in X, that is, 10 options (from the 10 shoes chosen) for the first position in X, 9 for the second ... 10!

For each arrangement of X, we yet have the opportunity to arrange X + the others shoes, so that we can have X as the first element, or X as the second... 11!

Multiplying everything give us
C10,5*C10,5*10!*11!

And the total number of permutations is 20!

The prob is C10,5*C10,5*10!*11!/20!

Probably the error is that i am over counting something ...

 

[vela]

We can chose 5 left shoes of a total of 10, and the same to right shoes.
So that we have C10,5 for each case, a total of C10,5*C10,5.
Now that we "choose" our shoes, we can arrange themselves in X, that is, 10 options (from the 10 shoes chosen) for the first position in X, 9 for the second ... 10!

But that would include arrangements like L1 R1 L2 R2 L3 R3 L4 R4 L5 R5, right?

 

[LCSphysicist]

But that would include arrangements like L1 R1 L2 R2 L3 R3 L4 R4 L5 R5, right?

Oh, yes. I took time to realized too, but what the question want is to find the possible combinations, from all combination, in which it is always possible to find a (let's call) set of ten consecutive shoes such that there is five left shoes and five right shoes in this set. And so calc the probability. That's how i interpret it (I should mentioned that in the first comment, but i forgot that this was another possible interpretation) (yes is so really ambiguous)

The intuition says it is 1, but i can not prove it.
This question smell pigeonhole principle!

 

[vela]

Oh, yes. I took time to realized too, but what the question want is to find the possible combinations, from all combination, in which it is always possible to find a (let's call) set of ten consecutive shoes such that there is five left shoes and five right shoes in this set. And so calc the probability. That's how i interpret it (I should mentioned that in the first comment, but i forgot that this was another possible interpretation) (yes is so really ambiguous)

OK, I misunderstood. I read more into the problem statement than was there.

 

[PeroK]

OK, I misunderstood. I read more into the problem statement than was there.

The problem was elucidated in post #7. One way to help the OP is to find an arrangement that does not have the required property - just so we can see what sort of arrangement is needed.

 

[vela]

ETA: I'm probably misinterpreting the problem again, so ignore this post if so.

Multiplying everything give us
C10,5*C10,5*10!*11!

And the total number of permutations is 20!

The prob is C10,5*C10,5*10!*11!/20!

Probably the error is that i am over counting something ...

If you compare to the total number of permutations, then the numerator is definitely an over-count. It's probably easiest to see this if we look at a smaller number of shoes, say 2 pairs. Take X = (L1 R1). According to your calculation method, for this X, there'd be 2! ways of permuting the elements of X and 3! ways of arranging X and the remaining shoes:

X L2 R2 (*)
X R2 L2
L2 X R2
R2 X L2
L2 R2 X
R2 L2 X
+ 6 more from reversing the elements of X.

So far so good. Then you would multiply by 4 = C2,1*C2,1, the number of ways you can choose the components of X. One of these four possibilities is X=(L2 R2) for which we have the permutations

X L1 R1
X R1 L1
L1 X R1
R1 X L1
L1 R1 X (*)
R1 L1 X
+ 6 more from permuting the elements of X.

Note that the two permutations marked with (*) are identical.

That said, you can keep your calculation in the numerator, but you need to modify the denominator so you're comparing against the appropriate number of possibilities.

 

[haruspex]

The problem was elucidated in post #4. One way to help the OP is to find an arrangement that does not have the required property - just so we can see what sort of arrangement is needed.

You mean post #7, yes?

 

[haruspex]

'0 = certainly false'; and '1 = certainly true'

In general, no. A probability of zero is described as "almost surely" false; similarly 1 is almost surely true, or true "except on a set of measure zero".
E.g. if you pick a real between 0 and 1, using a uniform distribution, you could pick 0.5, but the probability is zero.
But of course here there is only a finite number of possibilities, so no need for such quibbles.

The intuition says it is 1, but i can not prove it.

Could be, but if so, since there is only a finite number of possibilities, it is not really a probability question at all.

Inspect the first 10. Suppose there are r>5 right shoes. Step along one, so now you are looking at 2 to 11, as in @PeroK's description in post #7. What are the possible numbers of right shoes now?

 

[sysprog]

@haruspex I didn't intend the example description of the confidence interval to be an assertion; I intended merely to ask the OP to please give us a number.

Please specify your answer in probablistic terms − e.g. what is the percentage of likelihood, or where in the confidence interval between: '0 = certainly false'; and '1 = certainly true' is your yea or nay going?

In some 'fuzzy logic' systems, the confidence interval is described as between -1 and +1.

 

[PeroK]

You mean post #7, yes?

Yes, post #7. I've edited it.