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A friend of mine posed this problem and we have had a disagreement on the answer.

The problem:

  • There is a 90% chance that some event will happen in the next year.
  • There is a 95% chance that the event will happen eventually.
  • What is the probability that it happens after this year, if it does not happen in the next year?

Hover over here for a description of our answers and arguments:

I told my friend that it is 95%. My reasoning is that there is a 95% chance that it happens from now to infinity. The probability that it happens from now to infinity minus one year is still 95%. In one year, when it hasn't happened, events that didn't happen don't affect your "eventual" odds.

My friend on the other hand, believes it to be 50%. He had two arguments. First, he said if you flip a coin 4 times, the chance that you get heads the first time is 50%, but the chance that you get heads eventually is 93.75% (15/16). If you don't get it the first time, what is the chance that you get it after the first time? 87.5% (7/8), in other words, it has decreased.

To this I responded that the "eventually" is based on a finite set of events, each with their own probability and after the first event has passed, you have fewer chances. In his problem, on the other hand, it is a summary of the probability of unknown events broken in two time periods.

He continued to insist that it is the same because you have two time periods with known odds. The 95% is the combination of two probabilities, 90% and 50%. Eliminating the first 90%, the probability of the remaining period is 50%.

I agreed that's an accurate way to look at it from now, but I don't think that's accurate once you know it did not happen in that time period.

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Let's use Bayes' Theorem: \begin{equation} P(A|B)=\frac{P(B|A) P(A)}{P(B)}.\end{equation} In our case, $B$ is the event whatnot does not occur in year 1 (so with probability $P(B)=.1$, and $A$ is the event whatnot occurs at some point (so with probability $P(A)=0.95$). Of course, $P(B|A)$ is the only thing left to determine, but notice that $P(A|not(B))=1$ since if whatnot occurs in year 1, then it occurs at some point. Therefore, again by Bayes' Theorem \begin{equation} P(not(B)|A)=\frac{P(A|not(B)) P(not(B))}{P(A)}=\frac{90}{95}.\end{equation} Since $P(not(B)|A)+P(B|A)=1$, we have that $P(B|A)=\frac{5}{95}$. Hence plugging this all in, we find \begin{equation}P(A|B)=\frac{\frac{5}{95} \frac{95}{100}}{\frac{10}{100}}=\frac{1}{2}. \end{equation} So it looks like your friend was right!

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Calculate it this way. Given that the probability that it occurs during or after year 2 is $x$%. And the probability it occurs in year 1 is 90%, what is the probability it occurs?

Now, consider $x=0$, then the probability the event occurs is 90%.

If $x=50$, then it is 95%. (because it is $0.9 \times 1 + 0.1 \times 0.5 \times 1= 0.95$)

This informal argument should convince you.

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After talking about it some more, we came up with this scenario which illustrates why he is correct, and I am wrong.

  1. Say I want to give him a 90% chance that he will get $100 this year, 95% chance total, and 5% chance that he gets nothing.

  2. So, I take 100 sheets of paper and write dates in the next year on 90 of them, dates after that on 5, and "Bad luck" on the other 5.

  3. I draw one at random and keep it a secret.

  4. A year passes and my friend got nothing. Now he knows that all 90 dates this year are still in the pile. He now knows I must hold one of the other 10 sheets. That makes a 5/10 chance that I have a sheet with a date on it, and 5/10 chance that the sheet I hold has "Bad luck" on it. 50% chance he will get $100.

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Nice. Beats Bayes. – André Nicolas Jan 18 '13 at 21:15

Let $A$ represent the cases in which the event happens eventually. Let $B$ represent the cases in which the event happens next year. You've stipulated $P(A)=0.95$ and $P(B)=0.90$. Because $B$ is a subset of $A$, $P(A\cap B)=P(B)=0.90$, and $$P(A\cap \neg B)=P(A)-P(A\cap B)=0.95-0.90=0.05.$$ By the definition of conditional probability, then, $$ P(A|\neg B)=\frac{P(A\cap \neg B)}{P(\neg B)}=\frac{0.05}{0.10}=\frac{1}{2}. $$ So if I propose a deal right now, under the terms of which (a) you'll put in $\$1000$ if the event hasn't happened after one year and (b) you'll get back $X$ if the event happens after next year, then you should take the deal if $X>\$2000$. In this sense, your friend is right.

Your point, though, is different. You're wondering what deal you should take a year from now if the event hasn't happened by then. The answer is that it depends on probabilities not stipulated in the problem. In particular, you may learn something from the fact that it didn't happen this year, which may raise or lower these odds.

Suppose the event is the earth being hit by a meteor, that this depends on the solar system's meteor density, and that we're building a meteor shield that will be operational in just over a year. Based on what we know now, the meteor density might be high or low, with equal likelihood; if it's high, then the earth will be hit in the next year with probability $1$ (no shield yet!); if it's low, then the earth will be hit next year with probability $0.8$ and eventually with probability $0.9$. A year from now, if the earth hasn't been hit yet, you will know that the meteor density is low, and that the remaining probability that we'll ever be hit is $0.1$. In this case, the odds were lowered.

On the other hand, the choice may be between cases where the event will happen in either eighteen months with certainty (so next year with probability $0$ and eventually with probability $1$), or at some indeterminate time (next year with probability $0.9/(1-\varepsilon)$ and eventually with probability $(0.95-\varepsilon)/(1-\varepsilon)$), based on some unknown variable that might point to the first case (with probability $\varepsilon$) but almost surely points to the second (with probability $1-\varepsilon$). Here, a year from now you'll be much more likely to believe that the unknown variable is pointing to the first case, which will increase the odds that the event is still forthcoming.

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There could be slight issue with the phrasing of the problem. Your interpretation is that the statement is true no matter what time period you are in. I.e. in 5 years time, there is still a 90% chance that some event will happen in that year. However, you're forgetting to account for the fact that the event didn't happen in the first year.

Your friend's interpretation, is that your statement is true for this year. He has ignored the probability space in which the event occurred this year, and considered the remaining probability space. This now has measure $1-0.9=0.1$, and we know that the event has $0.95 - 0.9 = 0.05$ measure to occur, so it has $\frac {0.05}{0.1} = 50\%$ chance to happen.

For example, consider the event where outcomes are based on drawing a uniform random variable on $[0,1]$. Let $C$ be your favorite number from 0 to 0.9.

If we draw a number from 0 to 0.9, the event happens this year.
If we draw a number from $X$ to 0.95, the event happens next year.
If we draw a number from 0.95 to 1, the event never happens.

A possible scenario describing your interpretation, could be

If we draw a number from 0 to 0.9, the event happens every even year.
If we draw a number from 0.05 to 0.95, the event happens every odd year.
If we draw a number from 0.95 to 1, the event never happens.

Then, it will always be true, that there is a 90% chance of it happening this year, and a 95% chance of it happening eventually. However, not that if we know it doesn't happen in this year (doesn't matter whether it's even or odd), then there's only a 50% chance that it will ever happen again.

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Thanks for the alternate scenario, that's interesting. This is the kind of thing that made me feel like it's hard to guarantee an answer in the first place. – NickC Jan 18 '13 at 21:25
@NickC The hard part is considering completely what it means for an event to not occur. You should look at this recent post, which had a lot of different answers initially. – Calvin Lin Jan 18 '13 at 21:29

In case the mathematical proof already mentioned above isn't enough to convince you, pretend you're a statistician collecting your own results. You have 100 people infected with a disease that has a 95% death rate and a 90% death rate in the first year. Year one is finished, 90 dead, as predicted. 10 people left. We know 95% will be dead at some point. How many more are likely to die if the odds hold true? 5/10. Chances once the first year has passed is 50%. You don't need to be a statistician to see that. It has nothing to do with a finite number of events or an infinite time frame. That part has already been calculated for us in the given 95%.

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The analysis for this appears to be relatively straightforward. Assume the probability of event in year 2+ is $x$. Next, the probability that the event does not occur in the first year is $0.1$ and that it does not occur any time thereafter is $(1-x)$. The product of the two is the probability that the event never happens, which we know to be $ 0.05$. So,

$0.1 \times (1-x) = 0.05$

solving for x yields $x=\frac{1}{2}$

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