Answer: Tuesday Teaser #12
It’s easy enough to solve this puzzle by working through all the combinations, but there’s a more elegant and adaptable solution using conditional probability. Let’s define the events:
A: I throw three numbers in strictly increasing order;
B: I throw three different numbers;
Then we can easily see the probability of B and the probability of A given B.
P(B) = 1 x 5/6 x 4/6 = 5/9
P(A|B) = 1/3 x 1/2 x 1 = 1/6
Applying conditional probability [ namely that P(A|B) = P(A∩B) / P(B) ], we see that P(A∩B) = 5/54. But then the probability of A and B is just the probability of A since A can’t happen without B happening.
Therefore the probability of throwing three numbers in strictly increasing order is 5/54.
As a further example, imagine I split a pack of cards into four piles, one for each of the suits. If I draw one card from each in turn, what is the probability that I pick cards in a strictly increasing order (let’s say that Ace is high)? Using the above method, it’s easy. Define the events:
A: I draw four cards in strictly increasing order;
B: I draw four different numbered cards;
Then:
P(B) = 1 x 12/13 x 11/13 x 10/13 = 1320/2197
P(A|B) = 1/4 x 1/3 x 1/2 x 1 = 1/24
So the probability is 1/24 x 1320/2197 = 55/2197 (or just better than 1 in 40).
