FMM21

Friday Morning Math Problem

A Pair of Dice Games

Consider two dice games:

In the first game, you toss a single die $ N_1 $ times. The question is, what should $ N_1 $ be in order to make the probability of seeing at least one 6 greater than $ \tfrac{1}{2} $?

(In gambling terms, what should $ N_1 $ be to make an "even money" bet on seeing at least one 6 advantageous to the gambler instead of the house?)

In the second game, you toss a pair of dice $ N_2 $ times. The question is, what should $ N_2 $ be to make the probability of seeing at least one double 6 greater than $ \tfrac{1}{2} $?

Source: Will You Be Alive 10 Years From Now? by Paul Nahin

Original source: Antoine Gombaud and Blaise Pascal, correspondence, 1654

Solution

Let $ P_1 $ = probability of first seeing, with probability $ > \tfrac{1}{2} $, at least one 6 in $ N_1 $ tosses of a fair die $ P_2 $ = probability of first seeing, with probability $ > \tfrac{1}{2} $, at least one double 6 in $ N_2 $ tosses of a fair die Pascal observed that for $ P_1 $, if a die is tossed $ N_1 $ times, there are $ 6^{N_1} $ ways it can land, and $ 5^{N_1} $ ways it can land without showing a 6. Since "no 6 in $ N_1 $ tosses" is a complementary event to "at least one 6 in $ N_1 $ tosses", they are mutually exclusive, and inclusive (the only possible outcomes). The probability of getting a single 6 on a single toss is $ \tfrac{1}{6} $, so probability of not getting a 6 is $ \tfrac{5}{6} $. The probability of not getting a 6 in $ N_1 $ rolls is $ \left(\tfrac{5}{6}\right)^{N_1} $. Thus, the probability of not not getting a 6 in $ N_1 $ rolls is $ 1 - \left(\tfrac{5}{6}\right)^{N_1} $, in other words, $ P_1 = 1 - \left(\tfrac{5}{6}\right)^{N_1} $ via tabulation, $ N_1 = 4 $ Similarly, we know the probability of throwing a double 6 on a single toss of two dice is $ \tfrac{1}{36} $, so the probability of not throwing a double six is $ \tfrac{35}{36} $. Similarly, the probability of not throwing a double six in $ N_2 $ straight throws of a pair of dice is $ \left(\tfrac{35}{36}\right)^{N_2} $. Thus, the probability of not not throwing a double six in $ N_2 $ straight throws of a pair of dice is $ 1 - \left(\tfrac{35}{36}\right)^{N_2} $, in other words, $ P_2 = 1 - \left(\tfrac{35}{36}\right)^{N_2} $ via tabulation, $ N_2 = 25 $ (which is one off from the 24 that Gombaud expected) Gombaud's mistake: He believed in strict proportionality, so he believed $ \dfrac{N_1}{6} = \dfrac{N_2}{36} $ since there are 6 ways for a die to fall in the first game, and 36 ways for a die to fall in the second game. on very careful observation of a large number of games, Gombaud found that this proportionality did NOT hold.