# Discrete distribution

1. Feb 13, 2008

### Somefantastik

I need help getting started on this.

Want to generate a random variable X, equally likely 0, 1, using biased coin (heads probability p).

1. Flip coin, result is labled $$0_{1}$$
2. Flip coin, result is labeled $$0_{2}$$
3. $$0_{1} = 0_{2}$$=> return to step 1
4. $$0_{2}$$= heads => X = 0, $$0_{2}$$ = tails => X = 1

Show X is equally likely to be 0 or 1.

A good place to start would be showing P{X = 0} = P{X = 1}. Can you show me how to find those two probabilities?

Last edited: Feb 13, 2008
2. Feb 13, 2008

### mathman

In order to get past step 3, you have to have had one head and one tail on the two flips. Since the probs of HT and TH are equal, step 4 will give you 2 equiprobable results.

3. Feb 18, 2008

### Somefantastik

Keeping in mind that this is a biased coin, can someone please show me how to explicitly find P{X = 0}? I understand that it is equal to 1/2 but I need to see how to get there.

4. Feb 18, 2008

### mathman

Prob(HT)=Prob(TH)=p(1-p). Prob(get to step 4) is 2p(1-p), therefore prob(X=0)=prob(X=1)=p(1-p)/(2p(1-p))=1/2.

5. Feb 19, 2008

### Somefantastik

Is there a simpler way to do this where you continuously flip a coin until the last 2 results are different, that sets X = 0 if the final flip is a head, X = 1 if final flip is a tail?

6. Feb 19, 2008

### mathman

No. A sequence of heads followed by one tail has a different probability than a sequence of tails followed by one head. You always need to start fresh as described in your original statement.