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Cardano Again20 November 2011
In November and December of 1999, I published a couple of articles entitled Independence Day and Cardano Revisited. The point of these two articles was to show how to solve a problem posed by the Italian mathematician Cardano centuries ago using the notion of probabilistic independence. I was reminded of these articles recently when I received a letter from one of my readers, Lew Wedgewood, who said he had what he felt was a simpler approach to Cardano’s problem than the one I had presented. He was right. My original solution, however, was to point out what Cardano had done wrong and how it could be corrected. Both of my 1999 articles are in the archives and you might want to go back and read them before reading this article. Just click on my name from the roster on the left and then click on the phrase “More than 150 articles by Don Catlin.”
So, if you read my earlier articles you now know that to address Cardano’s question you need to calculate the probability of rolling at least one snake eyes in n rolls of a pair of dice. The sample space here consists of n-tuples (i1 , i2 , … , in) where each entry is an integer from 2 to 12 (the possible dice totals). Lew’s solution to the problem was to define n events E1, E2, …, En where the event Ei consists of n-tuples where the first i-1 entries are any non 2s, the ith entry is 2, and the rest are anything at all. These events are pairwise disjoint sets. For if Ei and Ek have a point in common then i = k. Otherwise, if k < i and there is a 2 in the kth position, this contradicts the definition of Ei.
If at least one 2 occurs during our n rolls, then at least one of E1 or E2 or … or En must occur. Hence we want to calculate the probability: p(E1 or E2 or … or En). Since the events here are pairwise disjoint, this is easy to do:
p(E1 or E2 or … or En) = p(E1) + p(E2) + … + p(En) (1)
Because the sequence of rolls are independent, the probability of Ek occurring is just (35/36)k-1(1/36) since there is a 35/36 chance of rolling a non 2 and a 1/36 chance of rolling a 2. Hence
p(E1 or E2 or … or En) = 1/36[ 1 + 35/36 + (35/36)2 + … + (35/36)n-1] (2)
S = 1 + 35/36 + (35/36)2 + … + (35/36)n-1 (3)
and multiplying (3) through by 35/36 we have
(35/36)S = 35/36 + (35/36)2 + … + (35/36)n-1 + (35/36)n (4)
Subtracting (4) from (3), most of the terms on the right side cancel and we end up with
(1/36)S = 1 – (35/36)n (5)
S = 36[1 – (35/36)n] (6)
Substituting (6) into (2) we end up with
p(E1 or E2 or … or En) = 1 – (35/36)n (7)
Expression (7) is the probability we seek.
What if we ask for the probability that in n rolls of the dice there will be exactly one 2 rolled? Easy! We just define a series of events as above, but this time we specify that event Ek consists of k-1 non 2s, a 2 in the kth position, and non 2s thereafter. These events all have the same probability and are pairwise disjoint. I’ll bet you can calculate this probability just as Lew and I did. See you next month.
Don Catlin can be reached at firstname.lastname@example.org
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