Finding the probabiliy that event THTH occurs before HTHH

  • Context: Undergrad 
  • Thread starter Thread starter SeekerofMath
  • Start date Start date
Click For Summary

Discussion Overview

The discussion centers on calculating the probability that the sequence THTH occurs before the sequence HTHH in a series of coin tosses. Participants explore different methods and reasoning related to this probability problem, considering both finite and infinite sequences of tosses.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants reference a source that claims the probability is 9/14, questioning how this result is derived.
  • One participant suggests that while THTH and HTHH might seem equally likely in a finite scenario, the infinite nature of the tosses introduces asymmetry, making THTH more likely to appear first.
  • Another participant proposes a method involving a transition matrix with states representing the outcomes of the last three tosses, suggesting that this could lead to determining the long-term probabilities of THTH and HTHH occurring.
  • Another approach is presented using a system of equations based on the last three tosses, leading to a calculation that supports the 9/14 probability result.

Areas of Agreement / Disagreement

Participants express differing views on the methods to calculate the probability and the implications of infinite tosses, indicating that multiple competing views remain without a consensus on the best approach.

Contextual Notes

Some participants note the complexity of the problem, including the dependence on the definitions of states and the assumptions made in the calculations. There are unresolved mathematical steps in the proposed methods.

SeekerofMath
Messages
2
Reaction score
0
I was refreshing myself of the basics of probability on wolfram for coin toss.
mathworld.wolfram.com/CoinTossing.html

At 1) , how did they get 9/14 for the probability that THTH occurs before HTHH ?
 
Physics news on Phys.org
Welcome to PF!

Hi SeekerofMath! Welcome to PF! :smile:
SeekerofMath said:
I was refreshing myself of the basics of probability on wolfram for coin toss.
mathworld.wolfram.com/CoinTossing.html

At 1) , how did they get 9/14 for the probability that THTH occurs before HTHH ?

By working it out! :rolleyes:

Have a go … show us what you get. :smile:
 
I'd like some direction...
 
SeekerofMath said:
I'd like some direction...

if there were only 4 tossing THTH and HTHH would be equally likely but if you consider infinite tossings then things changes:

Now through this infinite serie you will find situations like:

THTHTH (two appearances of THTH in 6 tosses)

Which is not possible for HTHH in just 6 tosses and the best you can do is

HTHHTHH (two appereances of HTHH in 7 tosses)

This asymmetry makes THTH more likely to appear before HTHH, anyway, the actual calculations should be in the reference given by the article you posted, that is (Gardner 1988, p. 64)

Good Luck :)
 
No idea if there's a better way, but here's a method which I believe would give the right answer.
Consider 10 states, X0 to X7, Y, Z.
X0 to X7 represent the outcomes of the last three tosses, TTT, TTH, .. , HHH.
Y, Z represent THTH, HTHH having occurred, first, at some point, respectively.
We take as initial states X0 to X7 equally likely (i.e. result of first 3 tosses).
Thereafter, each toss gives a transition to a new state. If we ever enter Y or Z we stay there. This gives a 10x10 transition matrix.
Extracting the eigenvalues allows us to find the long term odds of being in state Y or Z.
 
haruspex said:
No idea if there's a better way,
Indeed there is...:blushing:

Let p0 be the prob of THTH before HTHH if the last 3 were TTT.
Let p1 be the prob of THTH before HTHH if the last 3 were TTH.
etc. to
Let p7 be the prob of THTH before HTHH if the last 3 were HHH.
By considering the next toss in each case, we get:
p0 = p0/2 + p1/2
p1 = p2/2 + p3/2
p2 = 1/2 + p4/2
p3 = p6/2 + p7/2
p4 = p0/2 + p1/2
p5 = p2/2
p6 = p4/2 + p5/2
p7 = p6/2 + p7/2
This simplifies quite easily to give
p0 = p1 = p4 = 5/7
p2 = 6/7
p3 = p6 = p7 = 4/7
p5 = 3/7
The overall probability of THTH before HTHH is the average of these, 9/14.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 45 ·
2
Replies
45
Views
6K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 41 ·
2
Replies
41
Views
9K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K