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I drove my motorcycle to a fast food restaurant the other day. As I was waiting for my lunch, I noticed they still had their coffee condiments out. Not having any at home, I decided I'd grab a small handful and toss them into my motorcycle bag for later.

I always put two sweeteners in my coffee, but one day I pulled a sweetener out of my motorcycle bag and I didn't see a second.

I thought to myself, "I should keep looking; there's a 50/50 chance there's another one in here."

I'm a math novice so I asked some of my friends if they thought I was right. They weren't willing to assert either way, so I thought I'd ask here. Was I statistically sound in my conclusion?

EDIT: I haven't accepted an answer on this question yet, but I will be reviewing the answers soon. Part of the reason I haven't is because I feel like a lot of the answers are a smidgen pedantic (and I don't mean that in a negative way at all).

While I recognize that it's impossible to predict the distribution of a handful of sweeter packets because the universe is or is not random (whatever the case), I feel like a lot of people would fail a statistics course if they were to tell the professor a coin flip isn't 50/50 because, "nobody shuffles cards completely randomly, no die is unbiased, and you impart bias in a coin flip."

I was really hoping to get answers to this question based on the same simple model of the universe that allows statistics professors to teach and allows casinos to make fortunes based on the knowledge that a six sided die has, ceteris paribus, 6 equally likely outcomes.

I'm by no means saying that my original conclusion was correct, however, I'm not prepared to accept that it was wrong because of the "orientation of the hairs on my skin, the amount of blood distending my vessels, or the imperfections in my skin."

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A lot of people seem to be interpreting the question as "Is the probability that the OP grabbed an odd number of sweeteners 50%?" But if we take the framing story seriously and not just as code for that particular question, there are other factors to consider. For instance, how big is his bag and how easy is it to see what is in there?... –  Pete L. Clark Jun 28 '11 at 18:01
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As for me, my backpack has random amounts of change in it, coming for instance from when I go through airport security and dump the contents of my pockets into my backpack. If I want change and don't find enough in my pocket, I often look in my backpack; sometimes I find change there and sometimes I don't. But when I don't, I am aware that the probability that there really is change in there somewhere and I didn't look hard enough is positive. This probability depends a lot more on how many times I've ransacked my bag since my last trip than any parity considerations. –  Pete L. Clark Jun 28 '11 at 18:03
    
Casinos make money because their devices have very precise physical symmetries. Without those physical symmetries, all we can say about questions of probability is, "It depends." (See math.stackexchange.com/questions/48557/…) Your question lacks this element of precise physical symmetry, so it seems perfectly appropriate to answer with, "It depends." That said, though, I think it is a great question, and I think it is important and valuable to understand what the answer depends on, and how it depends on it. –  Jason Swanson Jul 5 '11 at 15:47

7 Answers 7

up vote 3 down vote accepted

Let \begin{align*} A &= \text{"I pulled a sweetener out of my motorcycle bag."}\\ B &= \text{"I didn't see a second sweetener."}\\ C &= \text{"There's another sweetener in there."} \end{align*} You want $P(C\mid A \cap B)$. Applying Bayes' theorem to the probability measure $P(\;\cdot\mid A)$ gives $$ P(C\mid A\cap B) = \frac{P(C\mid A)P(B\mid A\cap C)} {P(C\mid A)P(B\mid A\cap C) + P(C^c\mid A)P(B\mid A\cap C^c)}. $$ Now, $P(C^c\mid A)$ is the probability that your bag is empty after taking the first sweetener that day, computed without regard for the fact that you did not immediately see another one. It could, in principle, be anything. For example, if you have any rough idea about how many sweeteners you originally took (about 5? about 50?), and if you have a rough idea about how many days ago it was, then this could influence your estimate of $P(C^c\mid A)$.

But let us assume that you are happy to model this as $P(C^c\mid A)=1/2$ by considering no other information but the fact that you use two sweeteners per day, and by assuming that you were as likely to have originally grabbed an odd number of sweeteners as to have grabbed an even number. Then the above probability reduces to $$ P(C\mid A\cap B) = \frac{P(B\mid A\cap C)} {P(B\mid A\cap C) + P(B\mid A\cap C^c)}. $$ Now notice that if $C^c$ holds, then the bag is empty and you cannot see a second sweetener. Thus, $P(B\mid A\cap C^c)=1$. If we let $p=P(B\mid A\cap C)$, then we now have $$ P(C\mid A\cap B) = \frac p{p + 1}. $$ The only way for the above probability to be 1/2 is if $p=1$. Verbally, $p$ is the probability that you would fail to see another sweetener, despite the fact that there is one in there.

So even if you assume that an odd and even number of sweeteners were equally likely to have been originally taken, and even if you ignore whatever information you may have about how many you took and how long ago it was, the probability that you have another sweetener, given that you did not immediately see one, can still only be narrowed down to somewhere between 0 and 1/2. The messier and more cluttered your bag is, the closer it is to 1/2. The cleaner and more organized your bag is, the closer it is to 0.

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This depends on the distribution of the number of sweeteners you grab. It is not possible to have a "randomly chosen integer" in which all integers are equiprobable.

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As a programmer, I've heard the arguments that randomness doesn't exist and I agree that insofar as we're capable of comprehending, that's currently true, but is it possible that "nature" has a bias for odd numbers vs. even numbers or vice versa? I feel like that's what you're saying; am I wrong? –  D. Patrick Jun 28 '11 at 4:36
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I don't think ncmath is saying randomness doesn't exist. There is such a thing as a uniform distribution on the reals, but there is no such thing on the integers. It's not that nature has a bias; it's that the guy grabbing the sweeteners has an unknown bias. –  Gerry Myerson Jun 28 '11 at 7:26

Suppose the number of packets follows a Poisson distribution, that is, the probability of grabbing $k$ packets is $\lambda^ke^{-\lambda}/k!$, where $\lambda$ is the average number of packets grabbed. Then the probability of grabbing an even number of packets is $e^{-\lambda}\cosh\lambda$, and the probability of grabbing an odd number is $e^{-\lambda}\sinh\lambda$. So, no matter what $\lambda$ is, you're more likely to grab an even number, since $\cosh\lambda\gt\sinh\lambda$ for all $\lambda$. BUT

  1. The difference is very small: $e^{-\lambda}\cosh\lambda-e^{-\lambda}\sinh\lambda=e^{-2\lambda}$ is miniscule for, say, $\lambda=6$, and

  2. I haven't the faintest idea whether it's reasonable to assume a Poisson distribution.

So all I'm really doing here is pointing out that if you hypothesize a distribution, you can get an answer, and there is one well-known and highly useful distribution that has a (tiny) bias toward even numbers.

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Now I see. It's not 50-50; $$$$ it's 49.99993-50.00007 ! –  The Chaz 2.0 Jun 28 '11 at 12:37

Is there a 50/50 chance that you put an even/odd number of sweeteners in your bag?

Yes.
(That statement requires too many disclaimers at best, and is just wrong at worst!)

But as to the probability that, even after not seeing one during an initial glance, you still had one left, ... can't say!

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You've given no reason to think there's a 50-50 chance of an even number, and I don't think there is any reason to believe it. –  Gerry Myerson Jun 28 '11 at 4:23
    
@Gerry: Could you answer/comment in the affirmative? What would be a chance with reasoning to support it? Can you give a concrete example? Thanks, –  The Chaz 2.0 Jun 28 '11 at 4:47
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There's no reason to believe it's precisely 50-50, but I think 50-50 is a good estimate (say, plus or minus 5%). "Taking a handful of sugar packets" is a fine event to model probabilistically. We might do a thousand trials, and find a range of 5 to 20 packets, with most samples being around 12. If the percentage of even numbers were not between 45% and 55%, I would be very surprised. There's nothing wrong with this intuition, as long as 50-50 is recognized as an estimate. –  Jonas Kibelbek Jun 28 '11 at 5:04
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I think Chaz's second point is more relevant. You could easily have a sufficiently jumbled bag that it's likely at least one sugar is still hiding in there! –  Jonas Kibelbek Jun 28 '11 at 5:14
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@The, to get any kind of answer to the question, you have to hypothesize some distribution, some figure for the probability of grabbing exactly one packet, exactly two, exactly three, etc. Absent such a distribution, there's no sensible answer to the question. And I don't know a good way to hypothesize a distribution (although this is surely my failing, as questions of this nature are near enough to the surface that surely statisticians have put some effort into answering them). –  Gerry Myerson Jun 28 '11 at 7:32

I think you are wrong. You are more likely to have initially put an odd number of candy in your bag. This is because the positive natural numbers start with an odd number. For any number there can be more odd numbers then even numbers less then it but not the other way around.

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So, you're saying by virtue of the fact that I grabbed some, then I must not have grabbed 0. As a result, there is one more odd than even number in the infinite set of integers? That's interesting. So, if I were to grab two sets and put them together, then it should be slightly more likely to be even than odd, right? Who's to say I didn't grab two sets in one hand? –  D. Patrick Jun 28 '11 at 4:20
    
If you grabbed two sets in one hand, then you cant have grabbed 1. But you can cant you? –  outsider Jun 28 '11 at 4:25
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I can't accept the reasoning here. What if we know he always grabs at least $2$. Would that change the answer? Do we know that he never grabs just one? In fact, he says he grabs "a small handful." Sounds like more than $1$. –  Gerry Myerson Jun 28 '11 at 4:26
    
It sounds like more then two, id say exactly 7, so 0%. –  outsider Jun 28 '11 at 4:28
    
I do have pretty big hands; I could totally grab two handfuls in one hand. In any case, I feel like that should negate the "more odds than evens" argument, n'est-ce pas? (I'm not making fun, by the way . . . just kidding around, but I'm serious about the question). –  D. Patrick Jun 28 '11 at 4:31

I'll interpret the question as: is there a mathematical model of this situation for which the probabilities are not equal? The answer to that is yes and an example is a binomial distribution: if you grab on average six packets, then you will usually grab an even number, and if you grab on average seven packets, then the number will usually be odd. If you grab on average exactly six and a half packets then the probabilities will be closer to equal, as can be seen from the symmetry of the normal distribution which the binomial approaches. This simple model is not really of any use if you want to know the truth: to do that, you must calculate the distribution by performing repeated experiments. But the model does suggest that the answer depends to first order on whether the sample mean is closer to an even or an odd.

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Actually, wolframalpha bit.ly/kEzVmv seems to be saying that even numbers are always at least as likely as odds in the binomial distribution; I can't explain this yet and now I doubt my answer is correct. –  Dan Brumleve Jun 28 '11 at 5:49
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Suppose you flip a fair coin 12 times. On average you get 6 heads, and you're exactly as likely to get an even number as an odd. Even is exactly as likely as odd if you flip the fair coin 13 or 14 or any number of times. –  Gerry Myerson Jun 28 '11 at 7:29
    
The fact that even and odd numbers are equally likely in binomial distributions is easy to see with a simple, clever calculation. We look at $(a+b)^n$ and compare the sum of coefficients with $b$ to an even power and the sum of coefficients with $b$ to an odd power. I'll take the first sum minus the second sum to see the difference. But this is just $(1-1)^n=0$, so they are the same! –  Jonas Kibelbek Jun 29 '11 at 1:42

There is a famous problem with a related flavour called the Banach matchbox problem. The original problem goes like this. You have two matchboxes, each originally with $n$ matches, one in each pocket. Every time you smoke (Banach was a heavy smoker) you pick a pocket at random, and remove a match. What is the probability that when you first find a matchbox to be empty, the other box is empty?

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This is rather Socratic, isn't it?? –  The Chaz 2.0 Jul 5 '11 at 16:11
    
@The Chaz: Perhaps more unhelpful than Socratic. I was directing attention away from the question asked, for which I think no plausible probabilistic model exists. –  André Nicolas Jul 5 '11 at 16:34
    
I don't know if that's a good thing or not! But thanks for introducing me to the match problem. –  The Chaz 2.0 Jul 5 '11 at 16:53

protected by Qiaochu Yuan Jun 28 '11 at 22:13

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