Sunday, January 15, 2012

how many tosses to get 3 heads in a row

A common probability quiz -- Keep tossing a fair coin, how many tosses on average until you see 4 heads in a row? There's a Markov chain solution, but here's my own solution.

I will show a proposed solution using math induction. Suppose the answer to this question is A4. We will first work out A2 and A3.

Q1: how many tosses to see a head?

You can get h, th, tth, ttth, or tttth ...  Summing up 1/2 + 2/4 + 3/8 + 4/16 + 5/32 == 2.0. This is A1.

Q2: how many tosses to see 2 heads in a row?

t: Total tosses in this senario = 1+A2. Probabily = 50%
hh: 2. P = 25%
ht: 2 + A2. P = 25%

(1+A2:50%) + (2:25%) + (2+A2:25%) = A2. A2 works out to be 6.0

Q3: how many tosses to see 3 heads in a row?
After we have tossed x times and got hh, we have 2 scenarios only
.....hhh -- total tosses = x + 1. Probability = 50%
.....hht: x + 1 + A3 : 50%

(x+1: 50%) + (x+1+A3 : 50%) = x+1 + 0.5x A3= A3, So A3 = 2*(1+x), where x can be 2 or 3 or 4 ....

In English, this says

"If in one trial it took 5 tosses to encounter 2 heads in row, then this trial has expected score of 2*(1+5) = 12 tosses."
"If in one trial it took 9 tosses to encounter 2 heads in row, then this trial has expected score of 2*(1+9) = 20 tosses."

Putting on my mathematics hat, we know x has some probability distribution with an average of 6 because A2 = 6. We can substitute 6 for x, so

A3 = 2x(1+A2) = 14.0. This is my answer to Q3.

Now we realize A2 and A1 are related the same A2 == 2x(1+A1)

I believe A4 would be 2x(1+A3)==30.0. Further arithmetic shows A[n] = 2(A[n-1]+1) = 2^n - 2

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