1. ## Probability

A game is played by tossing two unbiased coins repeatedly until two heads are obtained in the same throw. The random variable X denotes the number of throws required. Find an expression for P(X=r)
Before playing the game, the player has to guess the value that X will take. If the player guesses correctly, he wins $5 . For an incorrect guess , the player loses$1. Suppose the player guesses X=t, express in terms of t, the expected amount won in a game.

I am not sure how to get the expression for part (1). If only one unbiased coin is tossed repeatedly, the probability of getting a head is

(0.5)+(0.5)^2+(0.5)^3+...

It's a infinite geometric series.

But how if two coins are tossed, does tree diagram work here?

2. Originally Posted by hooke
A game is played by tossing two unbiased coins repeatedly until two heads are obtained in the same throw. The random variable X denotes the number of throws required. Find an expression for P(X=r)
Before playing the game, the player has to guess the value that X will take. If the player guesses correctly, he wins $5 . For an incorrect guess , the player loses$1. Suppose the player guesses X=t, express in terms of t, the expected amount won in a game.

I am not sure how to get the expression for part (1). If only one unbiased coin is tossed repeatedly, the probability of getting a head is

(0.5)+(0.5)^2+(0.5)^3+...

It's a infinite geometric series.

But how if two coins are tossed, does tree diagram work here?
Geometric distribution - Wikipedia, the free encyclopedia

3. Thanks Mr Fantastic but i am still unsure how to set up the probability equation. Maybe a little hints to give me a push.

4. If I understand to game correctly, the key concept here is the tossing of two coins and getting two heads.
The probability of that is $\displaystyle \frac{1}{4}$.
So fhe probability of failure is $\displaystyle \frac{3}{4}$.
Now set up the geometric distribution.

5. Originally Posted by Plato
If I understand to game correctly, the key concept here is the tossing of two coins and getting two heads.
The probability of that is $\displaystyle \frac{1}{4}$.
So fhe probability of failure is $\displaystyle \frac{3}{4}$.
Now set up the geometric distribution.
Thanks Plato, but how many trials are there? It seems to be infinity?

X~B(? , 1/4)

but you seem to be suggesting that the probability is as such,

1/4+(1/4)(3/4)+(3/4)(3/4)(1/4)+...

=1/4(1+3/4+(3/4)^2+...)

but this is not an expression. Sorry if my answers and thoughts are totally off track and not of your expectation.

6. Originally Posted by hooke
Thanks Plato, but how many trials are there? It seems to be infinity?

X~B(? , 1/4)

but you seem to be suggesting that the probability is as such,

1/4+(1/4)(3/4)+(3/4)(3/4)(1/4)+...

=1/4(1+3/4+(3/4)^2+...)

but this is not an expression. Sorry if my answers and thoughts are totally off track and not of your expectation.
You need to read the link I gave you more carefully. It is directly relevant to your question and Plato has told you what to do. What part of it don't you understand?

7. Originally Posted by mr fantastic
You need to read the link I gave you more carefully. It is directly relevant to your question and Plato has told you what to do. What part of it don't you understand?
ok so the expression is P(X=r)=(1/3)(3/4)^r ?

Is there a proof for that formula? Or is it derived from the binomial distribution?

8. Originally Posted by hooke
ok so the expression is P(X=r)=(1/3)(3/4)^r ?

Is there a proof for that formula? Or is it derived from the binomial distribution?
Many textbooks on mathematical statistics will have a proof of the pmf for the Geometric distribution and I have no doubt that proofs can be found on the internet.

9. I would like to do a final checking to my work.

The expression, P(X=r)=(1/3)(3/4)^r

As for the next part of the question,

E(x)= 5(1/3)(3/4)^t + (-1)(1-(1/3)(3/4)^t )

= 2(3/4)^t-1

10. Originally Posted by hooke
I would like to do a final checking to my work.

The expression, P(X=r)=(1/3)(3/4)^r

[snip]
I don't understand how you got this from the formula in the link I gave you ....

11. Originally Posted by mr fantastic
I don't understand how you got this from the formula in the link I gave you ....
If the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is
for k = 1, 2, 3, ....

So in my case, as Plato has stated, p=1/4 , q=3/4 , k=r.

P(X=r)=(3/4)^(r-1)(1/4)

= (3/4)^r x (4/3) x (1/4)

= (1/3)(3/4)^r

12. Originally Posted by hooke
If the probability of success on each trial is p, then the probability that the kth trial (out of k trials) is the first success is
for k = 1, 2, 3, ....

So in my case, as Plato has stated, p=1/4 , q=3/4 , k=r.

P(X=r)=(3/4)^(r-1)(1/4)

= (3/4)^r x (4/3) x (1/4)

= (1/3)(3/4)^r
OK, it all looks fine.

(Except with E(X), the expression has no x in it ..... I'd just call it 'expected amount won')