1
$\begingroup$

Intuition tells me the expected number of tosses of a fair coin needed to get 3 TOTAL heads is 6.

I'm trying to show this using purely (1) Bernoulli random variables and (2) purely Geometric random variables


For approach (1), Let $X_i = 1$ if the i-th toss gives heads and $X_i = 0$ if the i-th toss gives tells. Let $N$ be the number of tosses to get 3 heads. Then we have that $$ 3 = \sum_{i}^{E[N]} E[X_i] \\ 3 = E[N] \frac{1}{2} \\ E[N] = 6 $$ I'm now analyzing my own solution, and I feel that the first line $3 = \sum_{i}^E[N] E[X_i]$ seems odd, but it makes sense in my head. It seems odd because I haven't solved a problem before where the summation goes up to an expected value. Is this procedure sound? If so, is there a formal name for the formula that I applied in $3 = \sum_{i}^E[N] E[X_i]$?


Now for approach 2. I am having problems with formulating this problem in terms of geometric random variables. The way I've learned geometric random variables is that they're the count of the number of trials before of getting the FIRST success. In this case, the first success could simply be getting the first head. Would it still be a geometric random variable if I instead defined the first success as getting all 3 heads? If so, I am having trouble figuring out how to solve the problem from this definition of "first success."

Now if the first success was defined as just getting the first head, then the expected number of trials is certainly 2. It seems that I can define 3 geometric RVs, $X_1, X_2, X_3$ to represent the number of trials needed to get the first, second, and third head. In this case the answer is $E[X_1] + E[X_2] + E[X_3] = 6$. Is $X_2$ and $X_3$ independent of $X_1$? It seems that they are dependent because to get the second or third head, you'd need to get the first head first?

$\endgroup$
  • $\begingroup$ The number of tosses until the 3rd success are distributed according to the negative binomial distribution. $\endgroup$ – YJT Jul 1 at 6:42
1
$\begingroup$

Since expectations are additive, the second approach works as well and is used, for example, in the coupon-collector-problem :

If we have $n$ possible outcomes, every outcome equally likely, then the number of trials to get each outcome at least once has expectation $$\sum_{j=1}^n \frac{n}{j}$$ This can be seen as a chain for "first-success-problems" with probabilities $1,\frac{p-1}{p},\cdots,\frac{1}{p}$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.