The game starts with two empty suitcases. The showmaster then tosses a coin. If the outcome is HEAD, then he will add another empty suitcase and continue tossing the coin, else (TAIL) he will stop. Afterwards he will put 10.000$ into one of the suitcases. You know that one of the suitcases contains the money, but you don't know which one. You can pick one suitcase and keep whatever is inside.

  • Let $N$ be a random variable describing the number of suitcases after the showmaster hides the money. What's the distribution of $N$?
  • What's the probability of picking the suitcase with the money? $\text{(Hint: } \forall p \in (0, \infty)\forall k \in \mathbb{N}: \frac{p^{k+1}}{k+1} = \int_0^p x^k dx \text{)}$
  • What's the expected value of the amount of money you get?

Trying to get my head around this exercise. I might have solved the first task.

The distribution of $N$ should be $$\mathbb{P}(N=k) = \frac{1}{2^{k-1}} = 2 \left(\frac{1}{2} \right)^k$$

The distribution function of $N$ should therefore be $$F_N(k) := \mathbb{P}(N \leq k) = \sum_{i=2}^{k} \mathbb{P}(N=k) = \sum_{i=2}^{k} \frac{1}{2^{k-1}}$$

Regarding the two other tasks. I don't understand the "Hint". If there are $k$ suitcases, then the probability of selecting the suitcase with the money inside has to be $\mathbb{P}(N=k)$, but this seems a bit to simple.. I'm pretty sure i need to introduce another random variable to answer the last question, but i can't get my head around how to properly model it..

Question: Is the above correct? How do i need to model this experiment to complete the task?

  • $\begingroup$ If there are $k$ suitcases then the probability of selecting the suitcase with the money in it is $\frac1{k}$. $\endgroup$
    – drhab
    Jul 9, 2017 at 9:33

1 Answer 1


Let $E$ denote the event of picking the suitcase with the money in it. Then:$$P(E)=\sum_{k=2}^{\infty}P(E\mid N=k)P(N=k)=\sum_{k=2}^{\infty}\frac1{k}2^{1-k}$$

We can write this as $2f(0.5)$ where: $$f(r)=\sum_{k=2}^{\infty}\frac{r^k}{k}\tag1$$

Now we deduce with termwise differentiating: $$f'(r)=\sum_{k=2}^{\infty}r^{k-1}=\frac{r}{1-r}=\frac{1}{1-r}-1$$

Combining this with $f(0)=0$ we arrive at: $$f(r)=-\ln(1-r)-r\text{ and consequently }P(E)=2\ln 2-1\thickapprox0.386294361$$

Then the money you can expect to win and expressed in dollars is:$$10000\times(2\ln 2-1)+0\times(1-2\ln2+1)=10000\times(2\ln 2-1)\thickapprox3862.94361$$

Another way of finding the function in $(1)$ is working out: $$f(x)=\sum_{k=1}^{\infty}\frac{r^{k+1}}{k+1}=\sum_{k=1}^{\infty}\int^r_0x^kdx=\int_0^r\frac{x}{1-x}dx$$ There the hint is used.

  • $\begingroup$ Thanks a lot! But i do still have some questions, could you please tell me why the first equation holds? ($P(E) = \sum_{k=2}^{\infty} P(E \mid N=k)P(N=k)$) Also i can't seem to reproduce $P(E)$ by $0.5f(0.5)$, i get $\sum_{k=2}^{\infty} \frac{1}{k} 2^{-1-k}$.. Also at the last line for $P(E) = \ln{2}-0.5$, i get $0.5(\ln{2}-0.5)$ (assuming $P(E) = 0.5f(0.5)$) $\endgroup$
    – nobody
    Jul 9, 2017 at 10:21
  • 1
    $\begingroup$ Denoting the probability space by $\Omega$ we have $\Omega=\bigcup_{k=2}^{\infty}\{N=k\}$. Then $E=\bigcup_{k=2}^{\infty}(E\cap\{N=k\})$. The sets on RHS are disjoint, so $P(E)=\sum_{k=2}^{\infty}P(E\cap\{N=k\})$ and $P(E\cap\{N=k\})=P(E\mid N=k)P(n=k)$. I might have made a mistake where it concerns the other things you mention. After a good look I will come back with a second comment, and - if necessary - some repairs in my question. $\endgroup$
    – drhab
    Jul 9, 2017 at 10:32
  • $\begingroup$ Discovered two (repaired) mistakes. $0.5$ must be $2$ and also I forgot to take this factor in account by calculating $P(E)$. $\endgroup$
    – drhab
    Jul 9, 2017 at 10:41

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