How many times would you expect to tosss a balanced coin to obtain the first head?


Ok, I have p(y) = q exp (y-1) * p
Is q = 1-p? and I'm trying to find y?

Confused!!

The first assumption in each individual toss is that heads and tails are equally likely. Probability 1/2. Also, the probability with any pattern of length n is obtained by the multiplication rule: (1/2)(1/2)(1/2)......(1/2)=1/(2^n).

In the simplest case, the game ends as soon as a single head is tossed.
The outcomes are the positive integers 1,2,3,.....(the possible durations of the game) and they occur with probabilities 1/2, 1/4, 1/8,.....(the likelihood of the patterns H, TH, TTH,....). Thus, the expected number of tosses is given by:

E=SUM(n/2^n, n=1..infinity)

Since the nth partial sum of the series is equal to 2-(n+2)/2^n, it follows that E=2. Therefore, Two tosses on average are needed to produce a single head


This is an interesting problem. One could say, "on average, how many times must a fair coin be tossed unitl a run of n consecutive heads appears?".

This problem is more interesting for two heads or more. It's derivation involves the Fibonacci recursive formula. Skipping ahead, we have

2^(n+1)-2.

2 tosses on average for 1 head

6 tosses on average for 2 consecutive heads.

14 tosses on average for 3 consecutive heads.
.
.
.
.
.

And so forth.

Brilliant!! I'm not familiar with Fibonacci recursive formula though. Will look into it more.

So how does the probability density function (pdf) of a geometric p(y) = q^y-1 * p comes into play?


Thanks

Oh come on! If X is a geometric random variable with probability p then E(X)=1/p.
P(H)=1/2 so E(H)=1/(1/2)=2.

I thought the E(x) of a geometric random variable is 1-p/p. And the variance is 1-p/p^2.

No you have that wrong.
If X is a geometric random variable with probability p then:
E(X)=(1/p) and V(X)=(1-p)/p^2 = q/p^2.

Thanks Plato.

hello
there is two kind of geometric distribution .
the fisrt and second geometric distribution .
that both of you , say true.

x have the first geometric distribution :

if x is the number of failure till come to first success

pdf(x)=p*(q^x) x=0,1,2.......
E(X)=q/p
VAR(X)=q/(p^2)

x have the second geometric distribution :

if x is the number of experiment till come to first success .

pdf(x)=p*(q^(x-1)) x=1,2,3.......
E(X)=1/p
VAR(X)= q/(p^2)