Friday Puzzler -- Three Geysers

Re: Friday Puzzler -- Three Geysers

Larry Barrett

Since you did not specify how long the eruptions last, I assume the question has to do with the odds of the sight of the beginning of an eruption for A, B, or C.

In a 12 hour period, A will erupt 6 times, B 3 times, and C 2 times. A total of 11 eruptions. So the odds of seeing A first is 6/11, B first is 4/11, and C first is 2/11.

Correct assumption, but not the answer

Alex Y

Yes, we are looking at an instantaneous event--the beginning of an eruption, not an event with duration.

And your answer of 6/11, 3/11 and 2/11 would be right if the eruptions occurred at random, ON AVERAGE every 2, 4, and 6 hours, but surprisingly, the answer is different with the fixed spacing. Without giving away too much, geyser A is somewhat more likely, and both the others less likely to be seen first with the fixed spacing.

Hint

Alex Y

You are guaranteed to see at least one eruption in the first two hours.

Hint 2

Alex Y

Cases and conditional probabilities

Hint 3

Alex Y

Doesn't look like this one is going anywhere, so I will give one of he four cases when observing for two hours.

Case AC -- Both A and C will erupt in the next two hours, but B will not.

Probability of case AC, p(AC), is

Probability that A will erupt, p(A), = 1

x Probability that B will not erupt, p(~B), = 1/2

x Probability that C will erupt, p(C), = 1/3

=1/6

In Case AC, there is an equal chance that A or C will erupt first.

So Case AC's contribution to the answer is 1/6* 1/2 that A will erupt first and 1/6 * 1/2 that C will erupt first, and 0 that B will erupt first.

Re: Hint 3

Larry Barrett

The complete set of cases (using bold to indicate the event and italics to indicate the 'not x' event*) is ABC, ABC, BAC, CAB, ABC, ACB, BCA, ABC.

When observing for 2 hours, A is certain to erupt so the probability of all events which include A is 0, which eliminates all but four: ABC, ABC, ACB, ABC.

Alex has already shown us the probability for the ACB event occurring in a 2 hour period: 1*1/2*1/3= 1/6, and the probability that A occurs first is 1/2*1/6.

The probability of ABC occurring is 1*1/2*2/3 = 2/6, and the probability that A occurs first is 1*2/6.

The probability of ABC occurring is 1*1/2*2/3 = 2/6. and the probability that A occurs first is 1/2*2/6.

The probability of ABC occurring is 1*1/2*1/3 = 1/6, and the probability that A occurs first is 1/3*1/6.

So the probability that A occurs first in the 2 hour observation period is

1/2*1/6 + 1*2/6 + 1/2*2/6 + 1/3*1/6 = 3/36 + 12/36 + 6/36 + 2/36 = 23/36.

Similar calculations for the prob of B occurring first = 8/36 and for C occurring first = 5/36. The sum of all three probabilities is 23/36+8/36+5/36= 36/36= 1.

*My keyboard is configured so that the key which contains the 'not' symbol is used to toggle between the english and thai alphabets, so I needed another way to indicate the 'not x' event.

Correct!

Alex Y

This one gave me problems, even after seeing the answer. I know it is too easy to get tricked by a flawed calculation, so like to apply a gut check. WHY is the answer different when the eruptions occur on average every 2, 4, and 6 hours versus occurring exactly every 2, 4, and 6 hours?

The best that I can come up with is that with even spacing, you are always relatively close to a time for A to erupt, so the likelihood of B or C erupting first is relatively low. With average spacing, there will be some instances when A's eruptions are close together, say 1/2 hour apart, and probability of seeing A first is greater if you get to the park between these two eruptions. However, you are much more likely to get to the park between two eruptions of A that are, e.g., 3 1/2 hours apart, when it is less likely that you will see another A before B or C.

But this is pretty soft reasoning -- if you can come up with a better explanation, please share!

Re: Correct!

Larry Barrett

I do not only have a better explanation, I have some more questions about the results.

What bothers me is the possible dependence on the observation period of 2 hours. It seems to me that the result should not vary if you select a different observation period, but I am not sure of that. In any case, I repeated the calculations using an observation period of one hour. Now the probability of A occurring is 1/2, where it was 1 for the 2 hour period, and the probability of not A is also 1/2, so all eight cases are now in play.

I get the following result for someone watching for 1 hour:

Probability of seeing A first is 116/288

Probability of seeing B first is 50/288

Probability of seeing C first is 32/288

Probability of not seeing any is 90/288.

The sum of all four is 288/288, so it is likely that I did the calculations correctly, (but there could be compensation errors).

For the 2 hour observation, the results were 23/36, 8/36, 5/36, (and 0 for not seeing any eruption, since in that case A is certain to erupt in the 2 hour period).

Clearly the probabilities for the 1 hours observation period are lower for seeing A first, etc. but I think the non-zero 'no eruption' case should be eliminated. Doing this leaves the following:

Probability of seeing A first is 116/198 (=.59) which is less than 23/36 (=.64). Why are the two not equal?

Re: Correct!

Alex Y

I think what you are calculating is the conditional probability of seeing each first, given that you see an eruption in the first hour. But to determine the probability of each being seen first, you need to determine which is seen first if none is seen in the first hour. I calculated those probabilities as 34/45, 7/45, and 4/45. multiply that by the case probability for no eruptions in the first hour, add to the probabilities you have, and you will get the same totals. The probability of seeing A first, given that you have not seen any in the first hour is 76%. And this makes sense if you think of the extreme--if you have been watching for 1 hour and 59 minutes with no eruption, it is certain you will see A in the next minute, but pretty unlikely B or C will erupt in that narrow time window.

Odd probabilities, aren't they?