Wikipedia:Reference desk/Archives/Mathematics/2013 November 11

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November 11

Question about probability

The following statement is asserted: "When a person goes to sleep with his contact lenses in, he increases his chances of getting an eye infection by 10%." If that statement is correct, does it also – by laws of math and probability – follow that this statement is correct: "If a person goes to sleep with his contact lenses in for 10 days, he increases his chances of getting an eye infection by 100%." ... ? Obviously, 10 times 10% equals 100%. So, is this math correct or is there some flaw in the logic and reasoning? The conclusion can also be re-stated as follows: "If a person goes to sleep with his contact lenses in for 10 days, it is absolutely certain (i.e., 100% chance) that he will get an eye infection." The re-stated conclusion seems mathematically correct (I think?); but it does not seem to hold up by common sense. Also, a related question, does the probability in any way change if the person sleeps 10 days in a row consecutively with the contact lenses in versus if he sleeps just 10 days total, randomly, here and there (not necessarily consecutively, ten days in a row)? Does that make any difference in the mathematical conclusion? Thanks. Joseph A. Spadaro (talk) 16:47, 11 November 2013 (UTC)

What goes wrong with 10 times 10% = 100% is that you are adding probabilities which seems correct, but it's only correct when the probabilities refer to mutually exclusive events. So, if the probability of drawing a red ball is 1/3 and the probability of drawing a blue ball is 1/3 then you can add up these probabilties to conclude that the probability of drawing a red or blue ball is 2/3. If the two events are not mutually exclusive, you will have overcounted the probability that both occur. In case of the chance of getting an eye infection, you are overcounting the cases where you get an eye infection more than once in the ten day period. The easiest way to deal with this problem is to consider the probability of not getting an eye infection in the ten day period and then subtract that from 100%. Count Iblis (talk) 18:10, 11 November 2013 (UTC)

So, you are saying that each (individual) night that he goes to bed is not mutually exclusive of each of the other nights? Joseph A. Spadaro (talk) 18:32, 11 November 2013 (UTC)

There is a 10% chance that an event happens on a certain night that leads to an infection. The next night you again have that 10% probability for that event happing on that night. The probability that you'll get an infection after two days is equal to the probability that the infection will happen on night one but not on day two +llus the probability that on night one nothing happens while it happnes on night two plus the probability that it happens on both nights. This is the correct sum over all the mutually exclusive possibilities that lead to the the infection being there after 2 days. This is thus 0.1*0.9 + 0.9*0.1 + 0.1*0.1. Now, if you compoute it as 0.1 + 0.1, the first 0.1 can be written as 0.1*1 = 0.1* (0.1 + 0.9), which is the probability of getting the infection on night one and either getting an infection or not getting an infection on night two (the latter two possibilities are obviously mutually exclusive). Similarly the second 0.1 refers to the probability of getting an infection on night 2 while on night one anything can happen. You then have added up correctly the probabilties for precisely one infection happening on one of the two days, but the probability for the infection happening on both days has been counted twice, it is included in both the first 0.1 and the second 0.1. Count Iblis (talk) 19:07, 11 November 2013 (UTC)

There's also the issue of percent vs percentile. Suppose that going to sleep without contact lenses resulted in a 20% chance of eye infection that night. Suppose also that going to sleep with contact lenses increases the chance by 10%, as stated in the original question. That means the chance of an eye infection when you go to sleep with your contact lenses is 22%, not 30%. "Increase by X%" means "multiply by (100+X)/100". If you started with Y% and wanted to get to (Y+X)%, you would say "increase by X percentile". However, many people get this wrong, so it's usually best to seek clarification.

There's also also the issue of how long the event takes. The statement is being interpreted as "a person who sleeps a night with their contacts in increases their nightly chance of an eye infection by 10%", but it could just as easily mean "a person who sleeps a year with their contacts in increases their yearly chance of an eye infection by 10%". Or any other amount of time.--80.109.80.78 (talk) 18:29, 11 November 2013 (UTC)

That's a good point about the distinction between percent and percentile. One does not add a "flat" (fixed) number of 10 each time, but rather only 10 percent of the original base (whatever that might be). Thanks. Joseph A. Spadaro (talk) 18:36, 11 November 2013 (UTC)

The problem here isn't adding probabilities. The probability of getting an eye infection is small, or we'd all be getting those, contacts or not. When you have two independent events A and B with small probabilities P(A)=P(B)=ε of occurring, you can add them with small error, because the probability of both occurring is so small: P(A or B) = P(A)+P(B)-P(A and B) = ε+ε-ε² ≈ ε+ε. The problem is in not understanding what "increases by 10%" means. It means that if your initial probability of having an eye infection was ε, it is now increased by 10%, i.e. gets 0.1ε added to it, so it becomes 1.1ε. Still assuming independence, your probability of getting eye infection in ten days is now about 11ε instead of about 10ε. Zilch increased by 10% is still zilch. It's like saying that the probability of you being struck by lightning increases by 100% (i.e. doubles; I pulled this out of my hat) when you go outside during a thunderstorm, and yet you probably never get struck by one even if you always go outside to watch. -- 212.149.196.26 (talk) 20:29, 12 November 2013 (UTC)

Let's say the probability of getting an infection in one night (or year, or whatever) is p without the lens in but is q = 1.1×p with the lens in. So the probability of not getting an infection is 1-p or 1-q. Then assuming every night is independent from every other night, the probability of not getting an infection after say 10 nights is (1-p)¹⁰ or (1-q)¹⁰, and the probability of getting an infection in one or more of those nights is one minus that. So that's what you need to compare.

In answer to your other question, as to whether it matters whether the nights with the lens in are consecutive, yes it matters if the lens is never taken out for cleaning in the consecutive-nights case, because then the probability presumably would be higher in the second night, higher still in the third night, etc., whereas your given information treats it as a fixed probability for any night. Duoduoduo (talk) 18:50, 11 November 2013 (UTC)

The previous poster and I hand an edit conflict. We are probably saying the same thing, Nonetheless, here's my post: The question is about probability, and in this type of questions it is easy to go wrong. You (rightly) suspect that there is something murky somewhere, because it all doesn't add up to common sense. Often, the easiest approach to this kind of problem is to ask an equivalent related question. What is the probability that the sleeper does not get an eye infection sleeping with his lenses for N days? It is p = 0.9^N. The probability that he does catch an eye infection is P = 1 - p. Now, this makes sense, right? The error is in one of your early assertions (first post). Math doesn't go wrong and your common sense doesn't go wrong either (but your math does). Best is if you find the error yourself. YohanN7 (talk) 18:58, 11 November 2013 (UTC)

Thanks, all. Much appreciated! Joseph A. Spadaro (talk) 19:09, 13 November 2013 (UTC)