Gushing Geysers

It’s Friday!! Time for “Riddle” 2, which isn’t much of a riddle, but that’s OK.  It is titled, “Which Geyser Gushes First?”  The riddle is quoted below.

You arrive at the beautiful Three Geysers National Park. You read a placard explaining that the three eponymous geysers — creatively named A, B and C — erupt at intervals of precisely two hours, four hours and six hours, respectively. However, you just got there, so you have no idea how the three eruptions are staggered. Assuming they each started erupting at some independently random point in history, what are the probabilities that A, B and C, respectively, will be the first to erupt after your arrival?

Here is a quick explanation of how independence works.  If there is a 50% chance of rain today, and a 50% chance of rain tomorrow, and the events of rain today and/or tomorrow are independent, then the chance that it rains both today and tomorrow is (0.5)(0.5)=0.25, or a 25% chance.  In other words, independence allows us to multiply probabilities together.

So, to our problem.  We know for a fact that we will definitely see one of the three geysers erupt within 2 hours of our arrival.  So, let’s focus on that 2 hour interval.

Case 1: All three geysers erupt within this 2 hour interval.  There is a 100% chance that geyser A will erupt in this interval, a 50% chance geyser B will, and a 33.3% chance that geyser C will (or 1/3 probability).  Thus, the probability that all three will erupt within 2 hours of arrival is 1(1/2)(1/3)=1/6.
Now, if all three are within that interval, they can erupt in the following six orders: ABC, ACB, BAC, BCA, CAB, CBA.  Each of these are equally probable (given the condition), so the probability that A is the first geyser you see is 1/3, and likewise for B and C.  So, the probability that all three geysers erupt in the first two hours AND any one of them go first is (1/6)(1/3)=1/18.

Case 2: Geysers A and B erupt in the first two hours, and C erupts after 2 hours.  There is a 100% chance A will erupt within the first two hours, a 50% chance B will, and a 66.7% chance (or 2/3 probability) that C will not.  Thus, the probability of of this case is 1(1/2)(2/3)=1/3.

Now, if geysers A and B erupt in the first two hours, they can erupt in order AB or BA with the equal conditional probability of 1/2.  So, the probability of Case 2 occurring AND A or B going first is (1/3)(1/2)=1/6.

Case 3: Geysers A and C erupt in the first two hours, and B erupts after 2 hours. There is a 100% chance A will erupt within the first two hours, a 33.3% chance (or 1/3 probability) that geyser C will, and a 50% chance that geyser B will erupt after 2 hours.  Thus, the probability of this case is 1(1/3)(1/2)=1/6.

Analogous to Case 2, we have a 1/2 conditional probability that either A or C goes first. So, the probability of Case 3 occurring AND A or C going first is (1/6)(1/2)=1/12.

Case 4: Geyser A is the only geyser to erupt in the first two hours.  There is a 100% chance that A will erupt within the first two hours, a 50% chance that B will erupt after 2 hours, and a 66.7% (or 2/3 probability) that geyser C will erupt after 2 hours.  Thus, the probability of this is 1(1/2)(2/3)=1/3.

Now, summing the cases that A goes first, we get (1/18)+(1/6)+(1/12)+(1/3)=23/36.  Summing the cases that B goes first, we get (1/18)+(1/6)=8/36.  Summing the cases that C goes first, we get (1/18)+(1/12)=5/36.

And you thought you understood probability.

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