Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose there is a biased coin that gives heads with probability $d$ such that $0<d<1$.

Every time, I toss the coin two times.

If both outcomes of the toss are the same, I ignore it.

Else, I would record down the result of the first toss. In other words, I would record Head if the first outcome is head then second outcome tail, and record Tail if the first outcome is tail then second outcome head.

I repeat this process until I managed to record down 100 Heads and Tails (ie, a sequence of heads and tails of length 100).

So, I want to find out the probability that I will record a Head at any point of time during the sequence of experiment.

Here's what I did:

Let Event A: $\{$ Coin flips Head $\}$ , Event B: $\{$ Outcome of Tail $\}$, Event W: $\{$ Able to record as Head $\}$

Given from the problem, $P(A)=d$, so $P(B)=1-d$.

So, $P(W)=P(A\cap B)$.

Since the events are independent, $P(W)=P(A)\cdot P(B)=d \times (1-d)=d-d^2$

But I am wrong because the only options available are: $50\%$, $d$, $1-d$ or 'depends on previous records in the sequence'.

What have I done wrong? I found what I have done logical and the available options are impossible to get.

share|improve this question
1  
The process you have chosen - flipping a pair and ignoring the result if they are the same, etc - amounts to flipping a fair coin. That's because, given that you have flipped a head and a tails in a toss of two coins, the probability that the first is heads is the same as the probability that the second is heads. –  Thomas Andrews Feb 7 '12 at 13:54
    
Thanks. I still don't quite get it. Yes, the probability of getting the first toss as head and the probability of getting the second toss as head are the same because they are independent. But since only the first toss head and second toss tail is recorded as Head, I still don't understand how they would amount to flipping a fair coin? –  xenon Feb 7 '12 at 14:04

1 Answer 1

up vote 2 down vote accepted

There are four possibilities:

  • HH. Heads followed by Heads: probability $d^2$
  • HT. Heads followed by Tails: probability $d(1-d)$
  • TH. Tails followed by Heads: probability $(1-d)d$
  • TT. Tails followed by Tails: probability $(1-d)^2$

So the conditional probability of Heads being recorded (i.e. HT happening) given something is recorded (i.e. HT or TH happening) is $$\frac{d(1-d)}{d(1-d)+(1-d)d} = \frac{1}{2}$$ equivalent to the probability of getting heads from a fair coin.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.