4

Two parents decide to have children until they have 3 children of the same gender one after another (3 in a row). If p(boy)=p(girl)=1/2, how many children are they expected to have?

I have tried to draw a tree diagram to analyse the problem, but i can't generate a useful expression for the probability that a 'success' will happen on the nth trial.

So far P(3)=1/4, P(4)=1/8, P(5)=1/8, P(6)=3/32, P(7)=3/64, P(8)=5/128 and P(9)=1/32.

I have read a similar article on this forum about finding E(x) and Var(x) for 2 consecutive successes, but generating an expression here is far more difficult. In the case of first 2 consecutive successes in k trials, every success must be followed by a failure until the (k-1)th trial. Here not every success needs to be followed by a failure, though 2 consecutive successes must be followed by a failure until the (k-2)th trial.

many thanks, Yun Fei

Did
  • 279,727
Y-dog
  • 627

2 Answers2

3

They have a kid, cost $1$. We will add this at the end.

After the first kid, the parents are in one of two states: State A if the two preceding children are of the same sex (so they are almost there); State B is there is a previous child, but they are not in State A.

Let $a$ be the expected number of additional children if they are in State A, and $b$ the expected number of additional children if they are in State B.

If they are in State A, then with probability $\frac{1}{2}$, they get their three in a row, and breeding is over. And with probability $\frac{1}{2}$ they go into State B, and their expectation becomes $b$. The cost in either case is $1$ more child. Thus $$a=\frac{1}{2}(1)+\frac{1}{2}(1+b).$$

If they are in State B, then with probability $\frac{1}{2}$ they get a child of the same sex as the previous one, and enter State A. With probability $\frac{1}{2}$ they get a child of the opposite sex, and stay in State B. Thus $$b=\frac{1}{2} (1+a)+\frac{1}{2}(1+b).$$

Solve. We get $b=6$. Add the initial $1$ mentioned in the first paragraph. The mean number of children is $7$.

Remarks: $1.$ We can also express the expectation as a series, by calculating the probability that breeding ends after $3$ trials, $4$, and so on. The seres can be summed explicitly. But the conditional expectation approach we used above is much smoother, and is of wide applicability.

$2.$ I would prefer a phrasing in terms of coin tossing, since in fact boy births and girl births are not equally likely. And the tacit independence assumption that we made is also not correct.

André Nicolas
  • 507,029
  • I like your approach because it really simplifies the problem. Would it be possible to generate a probability mass function for this situation and sum to infinity to confirm this result, as you wrote in remark 1? – Y-dog May 13 '13 at 07:54
  • 1
    @YunFeiOuYang: Please note I had by mistake written down the value of $a$, not $b$. Yes, in this case it would be possible. There are some situations, such as when one uses indicator functions, such that I wouldn't know how to do it. If you want to try it, I suggest starting with one birth, and then instead of recording sexes, record letters S and N, for shift and no shift. We want two N in a row. – André Nicolas May 13 '13 at 08:54
2

Let $t_i$ denote the expected number of children to be born for 3 of the same sex to be born consecutively, after $i$ children of the same sex were born. Thus, $t_3=0$ and one is looking for $t_0$. Conditioning on the sex of the next child, one gets $t_0=1+t_1$, $t_1=1+\frac12t_1+\frac12t_2$ and $t_2=1+\frac12t_1$. Hence $t_0=7$.

For example, assume the last children born are ...MFMFF. Then $i=2$. If the next child is F, one goes to $j=3$ and the mean time to wait after this step is $0$ since the last children born are ...MFMFFF. If the next child is M, one goes back to $j=1$ and the mean time to wait after this step is $t_1$ since the last children born are ...MFMFFM. Thus, $t_2=1+\frac12\cdot0+\frac12t_1$. Likewise for $i=0$ (always goes to $1$) and for $i=1$ (goes to $2$ or to $1$).

Did
  • 279,727