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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

A shoelace is lying on the floor, and attached you can see its shadow. If I pull it, what is the probability that it will produce a knot?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

Hi anna_gg

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

All I can tell is that there are 7 intersections, so there are 2^7 =128 possible ways the shoelace can cross itself. Of these, we should keep only the ones that are "over/under/over" and "under/over/under", which will produce a knot. But I cannot figure out how many these are.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

Um, have you seen my answer?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

anonimnystefy wrote:

Um, have you seen my answer?

Yes, but I do not understand how you calculate it

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

Well, I looked at the two parts separately, because they are virtually independent. For one, there are 8 possibilities for one and 16 for the other. Then I just combine them.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

anonimnystefy wrote:

Well, I looked at the two parts separately, because they are virtually independent. For one, there are 8 possibilities for one and 16 for the other. Then I just combine them.

So you mean that in the right part there are only 3/16 possibilities that a knot is produced? Because at the left part I see there are 2/8, that is 4/16.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,532

No, both probabilities are 1/4.

Here lies the reader who will never open this book. He is forever dead.

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

anonimnystefy wrote:

No, both probabilities are 1/4.

So, how do you get 7/16? Am confused

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

Got it. You don't add the two sections, you multiply them

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

How did you get 7 / 16 by multiplying?

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 113

bobbym wrote:

How did you get 7 / 16 by multiplying?

The probability for both parts NOT to produce a knot is 3/4 for each (since the probability to produce a knot is 1/4). Therefore the total probability NOT to have a knot is 3/4*3/4 = 9/16, thus giving us probability to have a knot 1-9/16 = 7/16.

It wasn't me who solved it, I just unraveled anonimnystefy's solution

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,776

Hi;

Okay, thanks for the answer.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**bohr****Member**- Registered: 2013-08-18
- Posts: 1

I understand that the left part has a probability of 1/4 to produce a knot, since only two of the eight possible ways produce a knot (over/under/over and under/over/under). However, how the right part has also 1/4 probability; You mean that from the 16 possible ways, 4 of them produce a knot;

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