# Probability of a fair sequence of tosses ending on two successive tails given the first toss was a head?

Suppose a coin is tossed repeatedly until either two successive heads appear or two successive tails appear. Then, assume that the first coin toss results in a head. I would like to find the probability of this tossing ending on two successive tails. One way I know of doing this is by creating a Markov chain and then doing first step analysis. I was wondering if there was perhaps an easier way to do this type of problem using some other principles. thank you.

• @David How do you know it is fair?
– BCLC
Feb 9 '16 at 0:39
• I assume it is fair, as did @lulu, but I suppose it isn't specified and hence should remain arbitrary. Feb 9 '16 at 0:41
• Sorry, forgot to mention that the coin is fair, thanks! Feb 9 '16 at 0:45

Let $p$ be the probability you want.

Look at the possible sequences of three tosses (including the initial $H$). We have $\{HH*,HTT,HTH\}$ Conditional on the first being $H$, the probability of $HH*$ is $\frac 12$, and the probability of the other two are each $\frac 14$. The first is a win for $H$, the second is a win for $T$, and the third restarts the game. Thus $$p=\frac 12\times 0+\frac 14 \times 1 +\frac 14\times p\implies p=\frac 13$$

Note: if you want to allow for a weighted coin, the same calculation works but we need to correct for those initial probabilities. Thus assume that the coin comes up $H$ with probability $\phi$. Then $HH*$ has probability $\phi$, $HTT$ has probability $(1-\phi)^2$, and $HTH$ has probability $(1-\phi)\phi$. So the recursion is now $$p=\phi \times 0+ (1-\phi)^2 \times 1 + (1-\phi)\phi \times p\implies p=\frac {(1-\phi)^2}{1-(1-\phi)\phi}$$ Comparison shows that this matches the result obtained by looking at infinite sums, as in the posted solution of @BCLC (with different notation, unfortunately).

• (i'll read later so assuming this is right) how elegant!
– BCLC
Feb 9 '16 at 1:27

Let $p$ be the probability of heads.

It seems to me that we have the following possible scenarios with the following probabilities:

HH --> $p^2$

H T HH --> $p(1-p)p^2$

H T HT HH --> $p(1-p)[p(1-p)]p^2$

H T HT HT HH --> $p(1-p)[p(1-p)]^2p^2$

H T HT HT ... HT HH --> $p(1-p)[p(1-p)]^np^2$

H TT --> $p(1-p)^2$

H TH TT --> $p[(1-p)(p)](1-p)^2$

H TH TH TT --> $p[(1-p)(p)]^2(1-p)^2$

H TH TH ... TH TT --> $p[(1-p)(p)]^n(1-p)^2$

So we have:

$$P(H...TT) = \sum_{n=0}^{\infty} p[(1-p)(p)]^n(1-p)^2$$

$$= p(1-p)^2 \sum_{n=0}^{\infty} [(1-p)(p)]^n$$

$$= p(1-p)^2 \frac{1}{1-(1-p)(p)}$$

If we have $p=1/2$, then $P(H...TT) = 1/6$.

Similarly $P(H...HH or HH) = 1/3$

Of course $P(T) = 1/2$.

Check if it adds up: $P(T) + P(H...HH or HH) + P(H...TT) = 1/6 + 1/3 + 1/2 = 1$.

Just realised (before lulu pointed out) that I didn't condition on $H...$ so just divide all (or most? eg $P(T)$?) above by p or something. Anyway:

$P(T|H...) = 0$

$P(H...HH or HH|H...) = \frac{1/3}{1/2} = 2/3$

$P(H...TT|H...) = \frac{1/6}{1/2} = 1/3$.

Check if it adds up: $P(T|H...) + P(H...HH or HH|H...) + P(H...TT|H...) = 0 + 2/3 + 1/3 = 1$.

• Note: you should not be factoring in the probability that the first one comes up $H$, that is a given of the problem. Thus the factor of $p$ in your final answer shouldn't be there, and if $p=\frac 12$ the answer is $\frac 13$, not $\frac 16$.
– lulu
Feb 9 '16 at 0:54
• @lulu Oh yeah, noticed that. Edited. Thanks
– BCLC
Feb 10 '16 at 11:08