Probability question on fair coin

[ichabodgrant]

Homework Statement

A fair coin is continually flipped. What is the probability that the pattern T,H occurs before the pattern H,H, where T and H respectively denote Tail and Head of a coin?

Homework Equations

Prob. = (n r) (p^r)(1-p)^n-r

The Attempt at a Solution

I am thinking whether the question asks about:

i. prob. of TH vs HT vs TT vs HH
ii. THHHHHHH...or TTHHHHHH... or HTHHHHHH... or HTTHHHHH... or HTTTTTHHH... or ...

If it is the 2nd one, how can we calculate the prob.? It looks like an infinitely long series... or should I let there be n trials? Then I use binomial distribution to find?

 

[haruspex]

Is there any reason why one of the two results is more likely than the other?

 

[ichabodgrant]

I don't quite understand your question...
The prob. of getting a head or a tail is 1/2? Is this what you are asking?

 

[haruspex]

I don't quite understand your question...
The prob. of getting a head or a tail is 1/2? Is this what you are asking?

I'm asking whether you can think of any reason why getting HT is more or less likely than getting TH.

 

[ichabodgrant]

Same?

 

[haruspex]

Same?

Yes.

 

[ichabodgrant]

So simply calculate the first case?

 

[haruspex]

So simply calculate the first case?

What will the two probabilities add up to?

 

[ichabodgrant]

Which two? TH and HT?

0.5*0.5 + 0.5*0.5 = 0.5?

 

[haruspex]

Which two? TH and HT?

0.5*0.5 + 0.5*0.5 = 0.5?

No.
The two events are,

• that TH occurs before HT in an arbitrarily long sequence,
• that HT occurs before TH in an arbitrarily long sequence

What must those two probabilities add up to?

 

[ichabodgrant]

1?

 

[ichabodgrant]

but it seems 0.5*0.5*0.5*... , gets you 0...

 

[haruspex]

1?

Yes.

but it seems 0.5*0.5*0.5*... , gets you 0...

What exactly is the event for which that calculation applies?

 

[ichabodgrant]

Oh Sorry...

I have a typo in the question...
The correct question is "T,H before HH"

 

[ichabodgrant]

Should I indeed use conditional probability?
In case I have a T at the 1st trial, then it already achieves the event that TH appears before HH?

 

[haruspex]

The correct question is "T,H before HH"

Ok, that makes it a lot more interesting.
Assign unknowns to those two probabilities. Consider the first two tosses. There are four situations at that point, equally likely. Consider the probabilities of the two outcomes of interest when continuing from each of those four positions.
See what equations you can extract.

 

[ichabodgrant]

Actually TT, HT,TH all means TH must occur before HH, right?

 

[ichabodgrant]

so if there is no HH, then it is alright?
Let the probability of getting a H be p.

Then the required prob. is 1 - p²?

 

[haruspex]

so if there is no HH, then it is alright?

What do you mean "it's alright"?
There are four equally likely states after two tosses:
HH
HT
TH
TT
In two of those, the outcome is already determined, yes?
Look at the remaining two. Can you predict what the eventual outcome will be for those?

 

[ichabodgrant]

If it is HT, then no matter the next one is T or H, TH already occurs before HH, right?
If it is TT, then also no matter the next one is T or H, TH already occurs before HH, right?
If it is TH, then obviously no matter the next one is T or H, TH already occurs before HH, right?
Just the case if you obtain all H and no T at the first 2 tosses will give you HH before TH..

I am thinking in this way

 

[haruspex]

If it is HT, then no matter the next one is T or H, TH already occurs before HH, right?
If it is TT, then also no matter the next one is T or H, TH already occurs before HH, right?
If it is TH, then obviously no matter the next one is T or H, TH already occurs before HH, right?
Just the case if you obtain all H and no T at the first 2 tosses will give you HH before TH..

I am thinking in this way

Yes.

 

[ichabodgrant]

so the probability of getting HH is p²?
Then the required probability is 1- p²?

 

[haruspex]

so the probability of getting HH is p²?
Then the required probability is 1- p²?

Yes, but can't you assume it's a fair coin?

 

[ichabodgrant]

then 1 - 0.5² = 3/4

 

[ichabodgrant]

thank you for your kindness