In Professor Stewart’s Cabinet of Mathematical Curiosities the following is asked:

You have $1000$ songs on your MP3 player. If it plays songs ‘at random’, how long would you expect to wait before the same song is repeated?

The author says that

If each song is chosen at random then you have to play a mere $38$ songs to make the probability of a repeat bigger $1/2$.

according to this approximation $$1.1774\sqrt n$$

where $n$ is the number of songs, which in this case is $1000$.

So here is the question in my mind. Lottery numbers have 5 digits, so there are $100000$ different lottery tickets. Imagine that there is a one guy who is obsessioned to win the lottery. He is buying every week a lottery ticket. He always plays the winning number of the current week for the next week. He starts to play at the beginning of 2015. How many weeks does he have to play to make the probability of winning the lottery bigger than $0.5$? Can we apply the same formula to answer this question? If not why?

Second pquestion: Is it the same if he played with the same number from the beginning?


Your model is different. If he buys a ticket with the winning number from the winning lottery, probability that it will appear next time, is $10^{-5}$. Now you have easy to solve inequality $$ \left(1-10^{-5}\right)^n\leq 0.5. $$

(And the answer to your second question is: yes. If choosing the winning ticket is really random, it is meaningless, which one is bought. In particular: always the same and this from the last lottery must lead to the same result).

  • $\begingroup$ Can you explain why it is different? It seems that we can apply your formula to songs as well. $\endgroup$ – newzad Apr 17 '15 at 21:35
  • 1
    $\begingroup$ @nikamed In the MP3 player's model one finds probability of repeating one of already played songs, not the last one. $\endgroup$ – Przemysław Scherwentke Apr 17 '15 at 21:37
  • $\begingroup$ yes I understand, now I will change the question. Thanks. $\endgroup$ – newzad Apr 17 '15 at 21:39

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.