The following question is posed:

A coin is tossed 3 times, and among the 3 coin tosses, X heads show. Then the same coin is tossed X additional times, and with the X coin tosses, Y heads show. It should be noted that the coin is balanced.

Find the distribution for Y

So I think I have most of this correct but I do feel this is not 100% I said in order to find the distribution we would need to plug into this:


The denominator is the total number of tosses choose the total number of heads. Meanwhile, the first tosses are represented by 3 tosses choose $3-X$, total number of heads and the second set of tosses are represented by $X$ tosses choose $X-Y$ total number of heads.

I know the next step would be plug in $p(0)$, $p(1)$, etc. but why I don't think I did this correct is that we have two variables in here, $X$ and $Y$ so how would I proceed with that?

  • $\begingroup$ I guess you can marginalize X out. $\sum_{X=0}^3 Bin(X;3,0.5) Bin(Y;X,0.5)$ $\endgroup$ – Eric Feb 16 '20 at 17:29
  • $\begingroup$ Is that a disturbution for Y or X though? The question is asking for a distribution of Y $\endgroup$ – user748468 Feb 16 '20 at 17:32
  • $\begingroup$ That gives you the marginal distribution of Y. $\endgroup$ – Eric Feb 16 '20 at 17:35
  • $\begingroup$ I am most likely looking for a distribution not marginalized but thanks for your input $\endgroup$ – user748468 Feb 16 '20 at 17:40
  • $\begingroup$ I am confused why am I getting a downvote for this question. I made sure to share my logic to the question when posting what I am plugging into. I know how to solve it, the setup is confusing me that is what I am asking $\endgroup$ – user748468 Feb 16 '20 at 18:06

The analysis for the case with a biased coin is much the same as for the unbiased coin. Let $0<p<1$. Define $B_1,B_2,B_3\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Ber}(p)$ and $X= \sum_{i=1}^3 B_i$. Then $$ \mathbb P(X = j) = \binom 3j p^j(1-p)^{3-j},\ \ j=0,1,2,3. $$ We compute the distribution of $Y$ by conditioning on $X$: \begin{align} \mathbb P(Y=k) &= \sum_{i=0}^3 \mathbb P\left(\sum_{j=0}^i C_j = k\mid X=i \right)\mathbb P(X=i)\\ &= \sum_{i=0}^3\binom ikp^k(1-p)^{i-k}\binom 3ip^i(1-p)^(3-i)\\ &= (1-p)^{3-k} p^k \left(p^3 \binom{3}{k}+3 p \left(p \binom{2}{k}+\binom{1}{k}\right)+\binom{0}{k}\right), \ \ k=0,1,2,3,4,5,6. \end{align}

In particular the mean of $Y$ is given by \begin{align} \mathbb E[Y] &= \sum_{k=0}^6 k\cdot\mathbb P(Y=k)\\ &= \sum_{k=0}^6 k(3 p^6 + 2 (1 - p) p^2 (3 p^2 + 3 p^3) + (1 - p)^2 p (3 p^3 + 3 p (1 + 2 p)))\\ &=3p^2 \end{align}


For the distribution of $Y$ you want to know, for each possible value $y$ that $Y$ can take, what is $P(Y=y).$

Let's start with $y=3.$ So the question is, what is $P(Y=3)$?

Notice that in order to have $Y=3,$ we must toss the coin at least three more times after the first three tosses, otherwise it is not possible to have three heads when counting heads for $Y$. So we must toss three heads in the first three tosses, $X=3,$ and then in the three tosses after that we must toss three heads again. In short we must toss six heads in a row, and that is the only way we can have $Y=3.$

If $p$ is the probability of heads on one toss, the probability of six heads in a row is $p^6.$ If $p = \frac12$ (a fair coin) then $$P (Y=3) = p^6 = \left(\frac12\right)^6 = \frac1{64}.$$

For $P(Y=2)$ the calculation is already more complicated, because we only need two heads after the first three tosses, so $X=2$ is enough. But the probability that $Y=2$ is greater when $X=3$ than it is when $X=2.$ That means at least two cases to consider.

I will not take this further, because you already have an answer that lays out exactly what you need to compute. I merely wanted to give some motivation for why such a calculation was necessary.

Trying to work out the probability using a formula in the form of a single ratio, as you attempted, is (I think) a highly impractical way to solve the problem. Remember that by the end of the exercise you might have tossed the coin $6$ times (and will, if the first three tosses are all heads) but you might have tossed the coin only $3$ times altogether (if the first three tosses are all tails). Alternatively you might have $4$ tosses or $5.$

Computing a probability in the simple form $\frac MN$ requires that your probability space have $N$ equally likely outcomes, but the probability of each string of $6$ tosses is much less than the probability of any of the $5$-toss strings. While it might be possible to "pad" the shorter strings so that you end up with $2^6$ equally-likely outcomes, I would expect this to add much more complexity to the counting than you save by making the outcomes equally likely.


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