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{{FMM
|title=A Pair of Dice Games
|problem=

Consider two dice games:

In the first game, you toss a single die N1 times. The question is, what should N1 be in order to make the probability of seeing at least one 6 greater than 1/2?

(In gambling terms, what should N1 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 N2 times. The question is, what should N2 be to make the probability of seeing at least one double 6 greater than 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

P1 = probability of first seeing, with probability > 1/2, at least one 6 in N1 tosses of a fair die

P2 = probability of first seeing, with probability > 1/2, at least one double 6 in N2 tosses of a fair die

Pascal observed that for P1, if a die is tossed N1 times, there are 6^N1 ways it can land, and 5^N1 ways it can land without showing a 6.
Since "no 6 in N1 tosses" is a complementary event to "at least one 6 in N1 tosses", they are mutually exclusive, and inclusive (the only possible outcomes).

The probability of getting a single 6 on a single toss is 1/6, so probability of not getting a 6 is 5/6. The probability of not getting a 6 in N1 rolls is (5/6)^N1.

Thus, the probability of not not getting a 6 in N1 rolls is 1 - (5/6)^N1, in other words,

P1 = 1 - (5/6)^N1

via tabulation, N1 = 4

Similarly, we know the probability of throwing a double 6 on a single toss of two dice is 1/36, so the probability of not throwing a double six is 35/36. Similarly, the probability of not throwing a double six in N2 straight throws of a pair of dice is (35/36)^N2.

Thus, the probability of not not throwing a double six in N2 straight throws of a pair of dice is 1 - (35/36)^N2, in other words,

P2 = 1 - (35/36)^N2

via tabulation, N2 = 25 (which is one off from the 24 that Gombaud expected)

Gombaud's mistake:

He believed in strict proportionality, so he believed

N1/6 = N2/36

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.

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Unknown user: Created page with "{{FMM |title=A Pair of Dice Games |problem= Consider two dice games: In the first game, you toss a single die N1 times. The question is, what should N1 be in order to make t..."