Thread: Probability (Is what I did correct?)

A fair coin is tossed 5 times. What is the probability of getting at least 3 heads on consecutive tosses?

I said HHHHH, HHHHT, HHHTH, HHHTT, THHHT, THHHH, TTHHH, HTHHH are all the possible ways this can happen.

Since there are 32 possible ways of doing five coin tosses:

8/32 = 1/4 = .25

Yes, that looks good. That's a brute force approach. Works OK for smaller trials, but what if it were the probability of 8 consecutive heads in 20 tosses?.

You could show off and use a Markov chain.

The entry in the upper right hand corner tells us the probability of getting 3 conscutive heads by the time we throw 5 times. See?. It matches yours.



I hope I set that up right.

could you kindly explain how did you set up the Markov chain for those of us not yet inducted into the Markov world?

Anyway, I present my brute force method using calculation:

We let Event "At Least 3 Consecutive Heads in 5 Throws" be R.
Let H= Number of Heads Thrown in 5 Throws

P(R) =
P(R|H=0).P(H=0) + P(R|H=1).P(H=1) + P(H|T=2).P(H=2) + P(R|H=3).P(H=3) + P(R|H=4).P(H=4) + P(R|H=5).P(H=5)

Let's get the trivial cases off first.

P(R|H=0), P(R|H=1). P(R|H=2) all equal 0.
P(R|H=5) = 1.

We only need to calculate P(R|H=3) and P(R|H=4).