Probability of Repeated Coin Toss Results on the Nth Toss | Coin Toss Homework

[ArcanaNoir]

Homework Statement

A coin is tossed until the same result appears twice in a row. Find the probability that this event occurs on the nth toss.

(the answer is $\frac{1}{2^{n-1}}$)

Homework Equations

The Attempt at a Solution

I made a tree and first only considered the results of the 2, 3, 4..ect toss that did not have the same result twice in a row. I kept getting 1/2. When I included ALL the results on the 3, 4, 5 etc toss, I did not get something equivalent to the answer.

 

[Dickfore]

In the first n - 1 tosses, what must the distribution of outcomes be?

 

[ArcanaNoir]

In the first n - 1 tosses, what must the distribution of outcomes be?

1/2 heads, 1/2 tails?

 

[Dickfore]

That is true, but not sufficient. Can you have:

HHHTTT

for example?

EDIT:
Actually, if n - 1 is odd, it isn't even true.

 

[ArcanaNoir]

Of course

 

[Dickfore]

Of course

Of course what?

 

[ArcanaNoir]

Of course you can have HHHTTT

 

[ArcanaNoir]

but if you toss unti lyou get the same twice in a row, maybe you can't have hhhttt

 

[Dickfore]

So, can you or can't you? Think before you type.

 

[ArcanaNoir]

Okay how about I don't know? When I tried both ways I still got the wrong numbers.

 

[Dickfore]

Lol. So, let's say that in the 7th toss you got the same outcome as in the 6th toss, but never before that. What are the possible ways of tossing the coin (in all previous trials)?

 

[ArcanaNoir]

HTHTHTT or THTHTHH

 

[Dickfore]

True. What is the probability that any of these events will occur (use conditional probability)?

 

[ArcanaNoir]

(2/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) ?

 

[Dickfore]

Why do you have 2/2 in the first factor?

 

[ArcanaNoir]

Cause we could start either heads or tails.

 

[Dickfore]

Cause we could start either heads or tails.

Oh, so you went one step beyond my question and evaluated the TOTAL probability of the two union of the two favorable events. Ok, so if you count the powers of 2 in the denominator, you will get 1/2⁶.

Can you generalize this pattern if we change 7 in our example with a general integer n? What do you get?

 

[ArcanaNoir]

I got, it right? I easily see how (2/2)(1/2)(1/2)(1/2)(1/2)(1/2)(1/2) = $\frac{1}{2^{7-1}}$

 

[Dickfore]

Yes, so the point is that if you knew that you had a double toss in the nth trial and you know what you had, you can go back and trace your whole tossing history. But, you can have HH or TT as a double toss, so this increases your favorable outcomes by a factor of 2.

Note: It is not that easy to do it with a dice, for example.

 

[ArcanaNoir]

thank you very much for the help!

 

[Dickfore]

i didn't do anything. I was just chatting with you. You did all the work.

 

[ArcanaNoir]

But you got me in the right direction. I just stared at it otherwise.

So, what's the probability that this event occurs on an even toss? Answer is 2/3

I see that it's can't happen on the first (an odd) toss, and then after that it's almost like 50/50. But how do I find 2/3?

 

[Dickfore]

So, what's the probability that this event occurs on an even toss? Answer is 2/3

No, this is not true. It does not matter if its an even or odd toss. The point is that (n-1)-st is the same as n-th and they alternate if you go backwards after that. So, for n = 6 (even) you would have:

HTHTHH

THTHTT

 

[ArcanaNoir]

What isn't true? That occurring on an even toss is 2/3? This is what the book says.

 

[Dickfore]

What isn't true? That occurring on an even toss is 2/3? This is what the book says.

OOOhhhhh, on ANY even toss.


Yes, well that is the probability of occurring on the 3rd (it cannot on the 1st cause you didn't make 2 tosses yet), 5th, 7th, etc.

Using your formula you need to evaluate the geometric series:

$$\frac{1}{2^{3 - 1}} + \frac{1}{2^{5 - 1}} + \frac{1}{2^{7 - 1}} + \ldots$$

If you know how to sum these, then you will get the answer from the book.

 

[ArcanaNoir]

So the probability that it occurs on an EVEN toss is $$\frac{1}{2^{1}}+\frac{1}{2^{3}} + \frac{1}{2^{5}} + \frac{1}{2^{7}} + \ldots$$

So... how DO I sum that?

 

[Dickfore]

It says in your avatar that you have and Undergrad in Mathematics. Surely you know how to sum geometric series. That is something you learn in Senior High School.

 

[ArcanaNoir]

You're cruel. So I forgot! Sheesh. $$\frac{1}{a-r}$$ or somthing. Fine, I'll look it up.

 

[Dickfore]

Fine, I'll look it up.

Yes, this is the right attitude. Always! BTW, what you had posted is incorrect.

 

[ArcanaNoir]

Yes, I see that... *trudges up the stairs to get stewie's calculus*

 

[ArcanaNoir]

Cool. Got it. Thanks again!