# Probability of Same face in Coin flips

1. Dec 19, 2006

### rdmltrs

1. The problem statement, all variables and given/known data
You flip a fair coin until you get the same face twice in a row.
What is the probability that you stop on an even number of tosses?

Example: Probability of taking less than 6 tosses to stop is 15/16
In general, the probability of stopping on the nth toss is 1/2^(n-1)

2. Relevant equations

Example: Probability of taking less than 6 tosses to stop is 15/16
In general, the probability of stopping on the nth toss is 1/2^(n-1)

The sum of probabilities as n (the number of tosses) grows is 1.

3. The attempt at a solution

Pr(2 tosses) = 1/2
Pr(3 tosses) = 1/4
Pr(4 tosses) = 1/8

In general there are only two sequences for n tosses that end on the nth toss.

Either it starts with tails and alternates until the last 2 tosses or it starts with
heads and alternates until the last 2 tosses.

Example
HTHH
THTT
Only 2 sequences that end on the 4th toss.

2. Dec 19, 2006

### D H

Staff Emeritus
You are making this too hard. Don't worry about the sequences that end on 2 tosses, 3 tosses, ..., as you will quickly end up in a combinatoric nightmare. Look at things conditionally. For example, given that you have tossed 1000 alternating heads and tails, what are the odds the sequence stops on the 1001 toss? It becomes a simple proposition to compute the joint probabilities and then to compute the probability the sequence ends with an even number of tosses.

The problem statement already gives the joint probabilities. Add them up. (Show this is a valid thing to do.)

Last edited: Dec 19, 2006
3. Dec 22, 2006

### chanvincent

This is indeed a very easy problem.. you knew the probability of stopping on the 3rd toss is half of stoppoing on the 2nd toss, and stopping on the 5th toss is, again, half of the 4th toss, therefore, we have a relationship:
$$P_3=\frac{1}{2}P_2$$
$$P_5=\frac{1}{2}P_4$$
...........
In general,
$$P_{2n+1}=\frac{1}{2}P_{2n}$$
From the above equation, it is not difficult to draw the conclusion that the total probability of stopping on odd is half of stopping on even... applying $$P_{odd}+P_{even}=1$$, you will have the answer...

Last edited: Dec 22, 2006