Probability question on fair coin

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Homework Help Overview

The problem involves determining the probability of the pattern T,H occurring before the pattern H,H when flipping a fair coin. The discussion revolves around the interpretation of the question and the implications of various sequences of coin flips.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the nature of the problem by considering different sequences of flips and their implications on the probabilities of T,H versus H,H. Questions arise regarding the likelihood of different outcomes and the reasoning behind them.

Discussion Status

The discussion is active, with participants questioning the assumptions of the problem and exploring various interpretations. Some participants suggest using conditional probability and consider the outcomes based on the first few flips. There is an ongoing examination of the relationships between the probabilities of the different patterns.

Contextual Notes

Participants note the importance of defining the events clearly and consider the implications of a fair coin on the probabilities involved. There is a recognition that the question may have been misstated initially, which has led to further clarification and exploration.

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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) (pr)(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?
 
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Is there any reason why one of the two results is more likely than the other?
 
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?
 
ichabodgrant said:
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.
 
Same?
 
ichabodgrant said:
Same?
Yes.
 
So simply calculate the first case?
 
ichabodgrant said:
So simply calculate the first case?
What will the two probabilities add up to?
 
Which two? TH and HT?

0.5*0.5 + 0.5*0.5 = 0.5?
 
  • #10
ichabodgrant said:
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?
 
  • #11
1?
 
  • #12
but it seems 0.5*0.5*0.5*... , gets you 0...
 
  • #13
ichabodgrant said:
1?
Yes.
ichabodgrant said:
but it seems 0.5*0.5*0.5*... , gets you 0...
What exactly is the event for which that calculation applies?
 
  • #14
Oh Sorry...

I have a typo in the question...
The correct question is "T,H before HH"
 
  • #15
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?
 
  • #16
ichabodgrant said:
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.
 
  • #17
Actually TT, HT,TH all means TH must occur before HH, right?
 
  • #18
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?
 
  • #19
ichabodgrant said:
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?
 
  • #20
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
 
  • #21
ichabodgrant said:
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.
 
  • #22
so the probability of getting HH is p2?
Then the required probability is 1- p2?
 
  • #23
ichabodgrant said:
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?
 
  • #24
then 1 - 0.52 = 3/4
 
  • #25
thank you for your kindness
 

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