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The original question is:

Suppose a man got two matchboxes in his pocket, each matchbox contains n matchsticks. Each time this guy would randomly pick up one matchbox, then consume a matchstick. One by one, he will eventually pick up a box and find it empty. Ask: what's the probability of having m matchsticks in the other matchbox?

The answer to this question is easy to follow:

  1. Since this guy picked up his matchboxes (2n+1-m) times, and every time he has two choices; overall there are $2^{2n+m-1}$ different ways

  2. Consider two different circumstances. It is possible that he picks up matchbox A the last time, so in the previous 2n-m rounds, he has picked matchbox A n times, and this yields $\binom{2n-m}n$ ways to pick up matchboxes. Similarly, he might pick up matchbox B and realize it's empty, both circumstances are possible and the results are symmetric. Therefore, the overall probability would be $$2\binom{2n-m}n/2^{2n+1-m}=\binom{2n-m}n/2^{2n-m}$$

Strangely, the final equation seems that one can simply use the binomial distribution to solve the problem.

However, I think the coincidence happens because two matchboxes have equal probabilities to be picked up. If I alter the question to: if this guy has probability of 1/3 to pick up box A and 2/3 to pick up box B, when he picks up an empty box (A or B is not specified), what's the probability of having m matchsticks in the other box? How will the answer be different?

ADDED:

I had thought about a binomial solution before, but then I felt something wasn't right. If I calculate the probability of picking up A as empty box and had B still have m matchsticks, and the probability of picking up B as empty box, then add them up; then in the initial question, why shoudn't the final answer be:

$$2*\binom{2n-m}n/2^{2n-m}$$ insteand?

Because $\binom{2n-m}n/2^{2n-m}$ seems to me only indicate probability of hitting either box A or box B n times in (2n-m) trails. However, in the original answer, one seems had already incorporated both circumstances into the consideration.

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If he finds out that $A$ is empty, then from beginning he would have picked up box $A$ $n+1$ times and box $B$ $n-m$ times, with picking up box $A$ at $2n-m+1$th pick. Thus if we consider all sequence of $A$s and $B$s which shows the order that he picked (except for the last one which is fixed as $A$), there would be total $\displaystyle{2n-m \choose n}$ such sequences containing $n$ $A$s and $n-m$ $B$s. From here we obtain that the probability is $\displaystyle{2n-m \choose n} (\frac{1}{3})^{n+1}(\frac{2}{3})^{n-m}$ (note that all sequences indicate distinct cases). Similarly the probability that he will find out $B$ being empty is $\displaystyle{2n-m \choose n}(\frac{1}{3})^{n-m}(\frac{2}{3})^{n+1}$. Thus the total probability is $\displaystyle{2n-m \choose n}\displaystyle\frac{2^{n-m}+2^{n+1}}{3^{2n-m+1}}$.

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  • $\begingroup$ Thank you for your answer. This is the answer I have thought about before, but if in this case, if you can simply add two probability up, in the first case, why the final answer is not 2*(2n-m)C(n)/2^(2n-m) instead? $\endgroup$ – Chloe Zhou Sep 10 '17 at 22:34
  • $\begingroup$ @ChloeZhou I think you thought about it too much and missed one little detail. It is the $+1$ in the denominator's exponent. The chance of hitting one box is WITHOUT the $2\cdot$ in the numerator and WITH the $+1$ in the denominator's exponent. While the overall probability is either WITH BOTH $2\cdot$ and $+1$, or WITHOUT BOTH, because they cancel each other out. Your intuition is right, but the devil is in the detail. $\endgroup$ – TStancek Sep 11 '17 at 8:12
  • $\begingroup$ I got what you mean. Thank you! By the way, is this question related to the concept of negative binomial distribution? I just started learning probability materials myself...Excuse me if I made silly mistakes... $\endgroup$ – Chloe Zhou Sep 11 '17 at 12:04
  • $\begingroup$ Yes, because negative binomial distribution counts the number of observations we make until certain number of 'failures' is achieved in consecutive Bernoulli trials. Thus, for example if we want to calculate the case in which box $A$ gets emptied first, we can define picking up box $A$ as failure and otherwise as success. Then, we are doing observations until we pick up box $A$ $n+1$ times, that is, $n+1$ failures are made, and calculating the probability that we have made $2n−m+1$ observations. $\endgroup$ – shrimpabcdefg Sep 12 '17 at 3:48

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