A TIP OF THE HAT TO PROBABILTY THEORY!

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Probability theory is a branch of mathematics concerned with the analysis of random phenomena. The outcome of a random event cannot be determined before it occurs, but it may be any one of several possible outcomes. The actual outcome is considered to be determined by chance.

It is increasingly making its way into all branches of mathematics including number theory where at times mathematicians are looking for answers in a statistical sense.

Here is an innocuous-sounding probability question that is not as easy as it sounds.

What is the expected number of tosses of an unfair coin needed to get two heads in a row (assume probability p of a head)? Same question for three heads in a row and j heads in a row. 

Start with a simple case First: What if you only need the expected number of tosses required to get one head? Let N be the number of coin tosses, then we want to find E(N|1H) (expected number of tosses given that you seek only one head). Toss the coin once. Either you get a head (with probability p) or you get a tail. If you get a tail, then you are recursively back where you started. That is, there is probability p that N=1, and probability 1-p that you still have E(N|1H) tosses to go after the one you already tossed.

In other words

E(N|1H)=(p.1)+(1-p).[1+E(N|1H)]

Solving for E(N|1H) gives E(N|1H)=1/p. Check the higher is p, the lower is E(N|1H), and when p=0.50 (a fair coin), E(N|1H)=2, all of which seems reasonable.

Now consider the case of two heads in a row. Well, to get two heads in a row, you first need one head “in a row,” which requires the expected 1/p tosses just calculated. If you have this one head already, then there is probability p that your next toss will be a head, and probability (1-p) that you are back where you started having performed E(N|1H) plus one tosses already.

In other words

E(N|2H) =p.[E(N|1H)+1]+(1-p).[E(N|1H)+1+E(N|2H)]

                =E(N|1H)+1+(1-p).E(N|2H)

This last line implies that

In this case of a fair coin, this gives E(N|2H) =6. Exactly the same reasoning in the case of three heads in a row lead us to


In the case of a fair coin, this gives E(N|3H)=14. It should be clear that there is a pattern: in the case of J heads in a row.

This can be proved formally using numerical induction if you wish.

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