Suppose I have a black box with a switch and a red light. Once you start up the box by flipping the switch, some sort of internal randomizer decides, moment to moment, whether or not to switch on the red light. However, you have no direct information about the workings of the box.
You turn the box on, and for the next year, the light doesn’t light up. What’s the probability that the light will light up on the 366th day?
-Ben
The probability based on purely empirical data would be 0, I believe. (But I would check to make sure the box was plugged in)
I don’t think there is enough information to give a numerical probability. So the randomiser is turning out a sequence of random digits, say 1 to n inclusive, of which individual digits in some subset, size m, trigger a “turn on.” This looks like a case of uniform probability so the probability of being lit in any time span in which the number is generated is p = m/n. That is, if the number is chosen once per day, then the probability is m/n in a day. If chosen once per hour then the probability in any hour is m/n and the probability for the day is 1 - (1-m/n)[sup]24[/sup] and so forth.
I’m not an expert on probability so I hope I didn’t screw this up too much. If I did, someone will correct it and you will have two examples, one bad and one good. You can’t lose.
Assuming the random switch can be in two states (e.g. on or off) and it is designed to choose one state and random, then the probability that it will turn on on the 356th day is 1/2. The outcome of the past 355 days random chances do not affect the next day. (Assuming the box is behaving as you said it would.)
As an anology, consider flipping a coin 355 times. Every time it comes up heads. The probability that it will come up tales on the 366th try is still 1/2.
But just becuase it’s random dosen’t mean it has a 50/50 probability. He said that he dosen’t know what the box does inside. Say it picks a random number between 0 and 1 but only turns the light on if the number is greater than 0.98. Although assuming that the box can turn the light on, the probablity is very small (say, perhaps it only turns on if the number is greater than .99999999) in which case I would venture to say that the empirical data collected is inconclusive. But anyways if I had that problem on a probability test and it was given to me exactly as stated above and I had to give a numerical answer, I would say the probability is 0.
Maybe I didn’t phrase the question clearly. Let me try again:
Suppose that each day the randomizer chooses a number from 1 to n, and if it’s 1, then the light goes on- but you don’t know the value of n which has been built into the machine.
If the light went on on the first day, you’d assume- in a very non-rigorous way- that the value of n was probably pretty low. Sure, it could be the case that n=10^100, and it just happened to choose 1 on the first day, but that’s unlikely.
If the light hasn’t gone on after a million tries, then yes, it could, theoretically, be the case that n=2 and it just so happened that “the coin landed tails” one million times in a row, but that’s unlikely. It’s more likely that you’re dealing with a large value of n.
If you know the value of n, things are different. If you flip a coin, then you know n=2. Even if it lands tails a million times, the probability of the next toss being heads is 50%.
But if you don’t know the value of n, and after a million tries the light never goes on, is any sane person going to bet even money on the next trial?
The question then is, if the only information we have is the fact that after x trials, the light never went on, then what can we determine about the probability that it will go on on the next trial? It seems to me that as more trials pass, it gets more and more likely that the value of n is large, and so you’re less and less able to expect a success on the next trial. But how does one rigorously quantify that commonsense expectation? What, precisely, is the probability of a success after x failures?
-Ben
Then I would say the prob is 1/(number of days +1). So if it’s already been 365 days, I would say the prob for tomarrow is 1/366. But this prob would change each day.
If you know the frequency of the randomizer, I think it is possible to give a probability of the light coming on on Day 356.
Is the following maths correct?
If the randomizer works daily , then you can say that IF the chance of the randomizer going for on is More than 1/1000, then the chance of getting 355 consecutive offs is Less than (999/1000 to the power 355)
And if it is, can you somehow combine the two probabilities to come up with a unfied probability for day 366?
The probability that it will light up on any given day is 1/n.
The probability that it will not light up on any given day is (n-1)/n.
The probability that it will not light up for x days running is [(n-1)/n][sup]x[/sup]
The probability that it will not light up for 365 days running is [(n-1)/n][sup]365[/sup]
I think that the most likely value of n wouls be the one that sets the above probability to 0.5:
[(n-1)/n][sup]365[/sup] = 0.5
[(n-1)/n] = 0.5[sup]1/365[/sup]
[1-(1/n)] = 0.5[sup]1/365[/sup]
(1/n) = 1 - 0.5[sup]1/365[/sup]
n = 1/{1 - 0.5[sup]1/365[/sup]}
n = 527 (and change)
So the probability that it will light on the 366th day is 1/527.
Sound right?
I think the answer you’re looking for can be found on pages 21 and 22 of this pdf document. It discusses the “rule of threes” which states that if after N chances, there have been zero occurences of an event, then there is approximately a 95% likelihood that the probability of the event is no greater than N/3. In your example, then, we have 365/3 = 121.67. So, you can be 95% certain that the chance of the light coming on is less than 1 in 121.67
Nice link, KG. And if anyone out there wants to get more in depth with this question after skimming over the first 21 pages of that .pdf, a good keyword to look for in a web/book search is ‘Bayesian analysis’. But you should walk this path only if you have a deep, abiding fascination with complex theoretical statistics, and your life ambition is to be an actuary. 
Calculating the chance of a complex event based on the known probabilities of the simple events that make it up is conceptually very simple. Calculating the probabilities of the simple events based on observed frequency of the complex event is a couple orders of magnitude tougher.
If you can instantly see why you can only estimate an upper bound on the ‘on’ probability (rather than an actual probability value), you may find it worth your while to look into it, though.
I don’t understand all the guesses that the probability asked for is zero.
A probability experiment consisting of generating a random number would do what is asked. If the number has a certain value the light turns on, otherwise not. The question is asked: If the light hasn’t turned on for 365 days, what is the probability that it will turn on the 366th day?
The fact that it hasn’t turned on for 365 days is irrelevant. The only thing needed is the probability that it will turn on any given day. If p is the probability that the random number picked will turn it on at any one trial of the experiment, then the probability that it will not turn on during any day is NOT = (1 - p)[sup]n[/sup], where n is the number of trials in a day. That makes the probability that it will turn on at least once equal to 1 - NOT.
If p = 0.3 and n = 24 (i.e. a trial every hour) then NOT = 1 - 2.82*10[sup]-13[/sup] and the probability that it will turn on on the 366th day, or any other day, is 1 - NOT = 0.999999999999718
If the random experiment is tried “from moment to moment” as the OP says, then I would bet quite a sum that the light will turn on at least once during the first 24 minutes.
The assumption that the light hasn’t turned on is not realistic since it is clear that almost certainly the light will turn on at least once during the first day. In fact, using my assumptions, the light will most probably turn on 3 times during any day. However, anything that can happen will, if you wait long enough and a run of 365 days without a light is possible.
I suspect that what the poster is angling for is an answer that depends on the fact that the probability of a light on is increased as the number of days that no light turns on increases.
Great balls of fire, my credibility just took a big hit (or maybe I didn’t have any to start with)! Somehow (carelessness comes to mind, but I don’t like that answer) I used p instead of 1 - p in the above computations. Wouldn’t you know?
Anyway, if the correct numbers are used:
p = 0.3
1 - p = 0.7, NOT = (1 - p)[sup]24[/sup] = 7.98*10[sup]-2[/sup]
and the probability that the light will come on at least once during is 1 - 7.98*10[sup]-2[/sup] or 0.9202.
If the trial is made “moment to moment” then NOT = 1.29*10[sup]-66[/sup] which is virtually zero, so the probability that it will come on at least once is nearly equal to 1.
Somebody give me a clout upside the head every now and then - please.
I think Karl has expressed it most succinctly.
<< But anyways if I had that problem on a probability test and it was given to me exactly as stated above and I had to give a numerical answer, I would say the probability is 0. >>
No, not zero, but “very close to zero.”
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Yes, but the entire point of the problem is that you don’t know the probability, and have to infer it from the fact that the light hasn’t turned on.
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But it’s only unrealistic because you’re addressing a completely different problem, in which you have chosen a value of p that makes it unrealistic.
Why would the probability of a light on increase as the number of days with no light increases?
(Emphasis mine.)
Thanks, KarlGauss, this is exactly what I was looking for. I’ll read the pdf when I get the chance- it sounds interesting.
-Ben
Is that the principle of three insufficient reasons?
It wouldn’t. That was my point. But I seem to have misunderstood what the OP was asking for so it is moot anyway.