# Coin toss

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• June 4th 2010, 03:16 PM
BertieWheen
Coin toss
This is a question aimed at people newer to statistics/probabilities. I made this question as a test to myself because I haven't done stats in AGES.

A fair coin is thrown fairly 10, 20, 50, and 100 times.

What is the probability that the same side of the coin will not occur twice in a row.

Give the probability for each one.

(Or, if you're lazy, simply give a formula for finding the probabilities :)

EDIT: In answer to people's questions, this is to find out what people come up with, and maybe help somebody new to statistics who wants an easy question.
• June 4th 2010, 03:21 PM
undefined
Quote:

Originally Posted by BertieWheen
This is a question aimed at people newer to statistics/probabilities. I made this question as a test to myself because I haven't done stats in AGES.

A fair coin is thrown fairly 10, 20, 50, and 100 times.

What is the probability that the same side of the coin will not occur twice in a row.

Give the probability for each one.

(Or, if you're lazy, simply give a formula for finding the probabilities :)

Edit: never mind, my answer was not well thought out.
• June 4th 2010, 03:24 PM
ANDS!
Are you asking for help, or is this a "can you solve this problem" problem. If you are asking for help:

Spoiler:
The specific sequence of coin tosses you are asking for is one out of an "n" number of permutations of coin tosses. Thus the probability that you do not throw the same coin on subsequent coin tosses is 1/n (whatever n happens to be for each of those numbers of coin tosses).
• June 4th 2010, 03:25 PM
undefined
Quote:

Originally Posted by BertieWheen
This is a question aimed at people newer to statistics/probabilities. I made this question as a test to myself because I haven't done stats in AGES.

A fair coin is thrown fairly 10, 20, 50, and 100 times.

What is the probability that the same side of the coin will not occur twice in a row.

Give the probability for each one.

(Or, if you're lazy, simply give a formula for finding the probabilities :)

So you already know the answer and just want to see what people come up with? I'll put my answer in a spoiler box then.

Spoiler:

In order for the same side of the coin not to occur twice in a row, we must have alternating heads and tails, either starting with a head, or starting with a tail.

So let n = the number of fair tosses, n greater than or equal to 1.

Let p(n) be the probability asked for.

p(1) = 1.

p(n) = 2(0.5)^n = (1/2)^(n-1), for n > 1
(Edited because my first reply was not well thought out. Same answer though, as it turns out.)

Edit 2: whoops. Meant to edit my first post rather than make a new post. Browser's back button didn't behave as I expected.
• June 4th 2010, 03:30 PM
Plato
Toss a coin 10 times there are $2^{10}$ outcomes.
But there are only two in which two identical outcomes do not happen: $HTHTHTHTHT~\&~THTHTHTHTH$.
• June 4th 2010, 10:33 PM
BertieWheen
Here I will put the "results" (Wink) a.k.a if people get it right or not.

ANDS! - Incorrect but close, although it was rather vague (didn't give a formula for finding $n$)
Plato - Correct =D

What Plato said applied into a formula:

Spoiler:
Plato's formula is $2$ in $2^n$, or $1$ in $(2^n)/2$, where n = the number of fair tosses.

So if there are 10 tosses we simply substitute n = 10 into the formula like so, and then simplify:

$p = 2$ in $2^n$
$p = 2$ in http://www.mathhelpforum.com/math-he...2522daba-1.gif
$p = 2$ in $1024$
$p = 1$ in $512$
or $p = \frac {1}{512}$

Alternatively we could do the following:

$p = 1$ in $(2^n)/2$
$p = 1$ in $(2^{10})/2$
$p = 1$ in $1024/2$
$p = 1$ in $512$
or $p = \frac {1}{512}$

undefined's formulas:

Spoiler:
The first one is $p = 2(0.5)^n$, where n is the number of tosses.
For example:
$p = 2(0.5)^n$
$p = 2(0.5)^n$
$n = 10$
$p = 2(0.5)^{10}$
$p = 2(\frac {1}{1024})$
$p = \frac {1}{512}$

The second formula is $0.5$^ $(n-1)$, and for this formula $n > 1$, meaning you are tossing the coin at least twice, because otherwise it come's out with $0.5^0$, which we really don't need in a problem like this. An example would be:

$p = 0.5^{(n-1)}$
$n = 10$
$p = 0.5^9$
$p = \frac {1}{512}$

Also ANDS!, here's my explanation for why your answer is not correct:

Quote:

Originally Posted by ANDS!
Are you asking for help, or is this a "can you solve this problem" problem. If you are asking for help:

Spoiler:
The specific sequence of coin tosses you are asking for is one out of an "n" number of permutations of coin tosses. Thus the probability that you do not throw the same coin on subsequent coin tosses is 1/n (whatever n happens to be for each of those numbers of coin tosses).

Spoiler:
Instead of $1/n$, it is either $2/n$, or $1/(n/2)$ - if that makes sense.

This is because out of all the possible permutations (n amount), there are $2$ correct ones, not $1$.

For instance, like Plato said, if there were 10 tosses, HTHTHTHTHT and THTHTHTHTH would both be correct permutations that fit the given pattern (The same side not coming up more than once in a row)
• June 4th 2010, 10:45 PM
ANDS!
Why are you dividing by 2? That would mean that you're excluding half of the possible permutations that you can get from tossing "n" number of times.
• June 4th 2010, 11:00 PM
undefined
Well I think this post has been up long enough that we don't need spoiler boxes anymore.

I didn't notice on my last post that the case n = 1 conforms to the formula for case n > 1.

So, using p(n) as defined in my last post, for all integers $n \geq 1$ we have $p(n)=\frac{1}{2^{n-1}}$.
• June 4th 2010, 11:26 PM
BertieWheen
Quote:

Originally Posted by ANDS!
Why are you dividing by 2? That would mean that you're excluding half of the possible permutations that you can get from tossing "n" number of times.

What I said was:
Quote:

Instead of http://www.mathhelpforum.com/math-he...e801e4d9-1.gif, it is either http://www.mathhelpforum.com/math-he...a20217a7-1.gif, or http://www.mathhelpforum.com/math-he...2433d178-1.gif - if that makes sense.

This is because out of all the possible permutations (n amount), there are http://www.mathhelpforum.com/math-he...cc14862c-1.gif correct ones, not http://www.mathhelpforum.com/math-he...6f75849b-1.gif.

For instance, like Plato said, if there were 10 tosses, HTHTHTHTHT and THTHTHTHTH would both be correct permutations that fit the given pattern (The same side not coming up more than once in a row)
I'll try and put it another way.

You put the formula as $1/n$, where $n$ is the number of possible permutations, however the correct formula would be $2/n$, because out of all the possible permutations, there are 2 that are correct.

So:
p = 2/n
However we can change that so it is
p = 1/(n/2)
Treat the fraction as an equation, so for instance it would be
2 = n
1 = n/2
See?

Note: If you don't believe me, here's an example of why 2/n = 1/(n/2)

If n = 4, then:
2/4 = 1/(4/2)
2/4 = 1/2
• June 4th 2010, 11:30 PM
BertieWheen
Quote:

Originally Posted by undefined
Well I think this post has been up long enough that we don't need spoiler boxes anymore.

I didn't notice on my last post that the case n = 1 conforms to the formula for case n > 1.

So, using p(n) as defined in my last post, for all integers $n \geq 1$ we have $p(n)=\frac{1}{2^{n-1}}$.

Surely if n = 1 then:

$p(n)=\frac{1}{2^{n-1}}$

$p(n)=\frac{1}{2^{1-1}}$

$p(n)=\frac{1}{2^0}$

$p(n)=\frac{1}{1}$

$p(n)=1$

Oh wait, you were right never mind (Giggle)
• June 4th 2010, 11:42 PM
undefined
It should be noted that there is incorrect usage of the word "permutations" happening on this thread. A permutation refers to ways in which some elements can be re-arranged while not changing the elements themselves.

Thus THTH is a permutation of HTHT and TTHH, but not of TTTT.

Typically permutations are over a set of distinct elements, such as {a, b, c, d}, for which the number of permutations is $4!$.

Sometimes permutations are over a set of not necessarily distinct elements, i.e., a multiset, such as {a,a,b,b,b,c,c,c,c,d,d}, for which the number of permutations is $\binom{11}{2,3,4,2}=\frac{11!}{(2!)^2(3!)(4!)}$.

Anyway, we are not talking about permutations but about possible sequences of coin flips.
• June 4th 2010, 11:45 PM
BertieWheen
What I meant by permutations is the possible outcomes from flipping a coin $x$ times, such as for 10 flips there would be THTHTHTHTH, THHTHTHTTT, HTHTTHTHHT etc, which I saw and see to be correct
• June 4th 2010, 11:49 PM
undefined
Quote:

Originally Posted by BertieWheen
What I meant by permutations is the possible outcomes from flipping a coin $x$ times, such as for 10 flips there would be THTHTHTHTH, THHTHTHTTT, HTHTTHTHHT etc, which I saw and see to be correct

What I mean to convey is that the word "permutation" is already in use, and means something else. Of course we can define any word to mean anything we like, but this can cause confusion because some people will think the word means what is written in the dictionary, rather than what you say it means.
• June 4th 2010, 11:53 PM
BertieWheen
Quote:

Originally Posted by undefined
What I mean to convey is that the word "permutation" is already in use, and means something else. Of course we can define any word to mean anything we like, but this can cause confusion because some people will think the word means what is written in the dictionary, rather than what you say it means.

Haha, I was just going on the definition I was told, although clearly I misunderstood.

Thanks for the heads up, I'll look it up later (Smile)

NB. I really appreciate posts like that because it helps with my English, and I'm 15 and doing my GCSEs so I need to be learning :)
• June 4th 2010, 11:53 PM
ANDS!
You're right. My "1" should have been a "2" - I read Heads/Tails when the distinction wasn't made. Consider it me giving you an opportunity to flex your algebraic skills.
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