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.
Best Answer
There are four possibilities:
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.