Probability of heads and tails are $\frac{1}{17}$ and $\frac{16}{17}$ . Find expected number of flips so that at least one head and one tail turns up?

Question

QUESTION: An unfair coin lands heads with probability $\frac{1}{17}$ and tails with probability $\frac{16}{17}$ . Matt flips the coin repeatedly until he flips at least one head and at least one tail. What is the expected number of times that Matt flips the coin?

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MY ANSWER: Assume that $E_h$ is the expected number of moves left after a head turns up and $E_t$ is the expected number of moves left after a tail turns up.. If $E$ denotes the total number of turns Matt has to toss, we can set up relations like - $$$$ $E=$ $\big($a head turns up$\big)$and$\big($number of moves left until a tail turns up$\big)$ OR $\big($a tail turns up$\big)$and$\big($number of moves left until a head turns up$\big)$ $$\implies E=\frac{1}{17}E_t+\frac{16}{17}E_h$$ Similarly, $E_t=$ if head turns up then sequence is over (since according to our assumption the previous flip was a tail) OR if tail turns up then we start over again.. $$\implies E_t=\frac{1}{17}+\frac{16}{17}E_t$$ Just like this $E_h=$ if a tail turns up then the sequence is over (since the last flip was a head) OR if a head turns up, we start over again.. $$\implies E_h=\frac{16}{17}+\frac{1}{17}E_h$$ Solving the equations we get $E_h$ and $E_t$ both to be $1$.. Which yields $E$ to be $1$.. But this is clearly wrong! $$$$

I do not understand where is my reasoning wrong? I have presented my logic in clear words before you.. Can someone please correct me..

Also, is/are there different way/ways to solve this sum? If so, it will be very much helpful if you introduce me to it/them.. Thank You so much..

Answer 1

You're on the right track, but the precise meaning of your definitions of your quantities $\ E_h\ $ and $\ E_t\ $ isn't very clear, and you've forgotten to count the toss needed to get to your first head or tail. Your $\ E_h\ $ needs to be the expected number of tosses needed to get a tail given that the first toss is a head, and your $\ E_t\ $ the expected number of tosses needed to get a head given that the first toss is a tail. That these quantities are somewhat confusingly labelled presumably explains why you got the labels reversed in your equation for $\ E\ $, which should be $$ E=1+\frac{1}{17}E_h+\frac{16}{17}E_t\ . $$ Similarly, your equations for $\ E_h\ $ and $\ E_t\ $ should be $$ E_h=\frac{16}{17}+\frac{1}{17}\left(E_h +1\right)\\ E_t=\frac{1}{17}+\frac{16}{17}\left(E_t +1\right)\ . $$ These give you $\ E_h= \frac{17}{16}\ $, $\ E_t=17\ $, and therefore $\ E=17\frac{1}{16}\ $, as Henno Brandsma obtained in his answer.

Answer 2

If $X$ is the number of tosses we still need to throw, we can condition on the first toss $T_1$:

$E(X) = 1 + E(\text{nr of tosses after } T_1 = t)P(T_1 = t) + E(\text{nr of tosses after } T_1 =h)P(T_1 = h)$, where the $1+$ is to account for the first throw.

$E (\text{nr of tosses after } T_1 = t)$ is just the expectation of a geometric distribution with success chance $P(h)=\frac{1}{17}$ so this is $17$.

Similarly $E (\text{nr of tosses after } T_1 = h)$ is just $\frac{17}{16}$.

So in total we get $$1+ 17\cdot \frac{16}{17} + \frac{17}{16} \cdot \frac{1}{17} = 17\frac{1}{16}$$

Answer 3

The idea of your method is fine as well, but you forget to count the throws you have done. With $E, E_h, E_T$ as you have defined them you namely get \begin{align*} E &= \frac{1}{17}(1+E_t) + \frac{16}{17}(1 + E_h), \end{align*} as you have to count in the fact that in both cases (heads and tails) you already have thrown once, this should also be counted. Then with your definition of $E_t$ you would e.g. get \begin{align*} E_t = \frac{16}{17} + \frac{1}{17}(1+E_t) \end{align*} and solving this gives $E_t = \frac{17}{16}$. The same can be done for $E_h$ giving $E_h = 17$ and thus resulting in \begin{align*} E = \frac{1}{17}(1 + \frac{17}{16}) + \frac{16}{17}(1 + 17) = 17\frac{1}{16}. \end{align*}