### Probabilities

Categories: [ Science ]

Before Easter, the supermarket was selling Rölli suprise-eggs, announcing that every fifth egg contained a figurine related to Rölli's universe. This made me wonder: how many eggs you need to buy to ensure that you get at least one such figurine?

The following formula gives the probability (*p*) of getting at least one Rölli
figurine given that one has bought *n* eggs, and that every *k* egg contains
such a figurine:

p = 1 − (1 − 1/k)^{n}

The answer to the first question is not straightforward, though. To be absolutely sure to get at least one figurine, you need to buy 4/5 of the egg production plus one egg, because there is always an (admitedly slim) chance that the 4/5 of are made entirely of eggs containing something else than a Rölli figurine, and that the 1/5 that is left is made only of eggs containing Rölli figurines. The extra egg that you need to buy is therefore taken from this last 1/5, and is guaranteed to contain a Rölli figurine.

If you are not willing to spend so much time and money tracking and buying most
of the egg production, you can trade time and money for a tiny bit of
uncertainty. For example, if you can accept to have only 90% chance of getting a
Rölli figurine instead of 100%, it is enough to buy 11 eggs. If you want a
better chance yet and want to go for 95%, you need to buy 14 eggs. Finally, if
you want a 99% chance, you need to get 21 eggs. These values were computed from
the formula above, setting *k* = 5 (“every fifth egg”), *p* = 0.90 or
*p* = 0.95 or *p* = 0.99, solving the equation for *n* and rounding the result
to the nearest, larger integer.

It is also worth noticing that if you decide to buy 5 eggs (because every 5th egg contains a Rölli figurine), you have only about 2 in 3 chances of getting a Rölli figurine.

The table below gives the minimum values of *n* for a given value of *k* and
different probability thresholds. It also gives the ratio *n* over *k*, i.e.,
given a “one in *k*” probability, how many times *k* does one need to get to
have a probability greater than the thresold. The second column also indicates,
given a “one in *k*” probability, what are you chances of getting what you
want if you get *k* items. Notice that these values grow toward a given, finite
limit when k grows larger.

k | p(n=k) | p≥0.90 (n/k) | p≥0.95 (n/k) | p≥0.99 (n/k) |
---|---|---|---|---|

2 | 0.750 | 4 (2.000) | 5 (2.500) | 7 (3.500) |

3 | 0.704 | 6 (2.000) | 8 (2.667) | 12 (4.000) |

4 | 0.684 | 9 (2.250) | 11 (2.750) | 17 (4.250) |

5 | 0.672 | 11 (2.200) | 14 (2.800) | 21 (4.200) |

6 | 0.665 | 13 (2.167) | 17 (2.833) | 26 (4.333) |

7 | 0.660 | 15 (2.143) | 20 (2.857) | 30 (4.286) |

8 | 0.656 | 18 (2.250) | 23 (2.875) | 35 (4.375) |

9 | 0.654 | 20 (2.222) | 26 (2.889) | 40 (4.444) |

10 | 0.651 | 22 (2.200) | 29 (2.900) | 44 (4.400) |

20 | 0.642 | 45 (2.250) | 59 (2.950) | 90 (4.500) |

37 | 0.637 | 85 (2.297) | 110 (2.973) | 169 (4.568) |

50 | 0.636 | 114 (2.280) | 149 (2.980) | 228 (4.560) |

100 | 0.634 | 230 (2.300) | 299 (2.990) | 459 (4.590) |

500 | 0.632 | 1151 (2.302) | 1497 (2.994) | 2301 (4.602) |

1000 | 0.632 | 2302 (2.302) | 2995 (2.995) | 4603 (4.603) |

One can use this table to find out how many times one needs to play the roulette
in a casino to have 95% chance of winning at least once: a european roulette has
37 numbers (*k* = 37), and the limit of the n/k ratio is about 3; one
therefore needs to play about *n* ≅ 37 × 3 = 111 times (the row for k = 37
indicates the actual value is *n* = 110).