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I'm trying to wrap my head around a pretty basic concept. But I'm not 100% sure my reasoning is right.

I have an event that happens 10 times, at the same time, for thousands of occurrences. All of these 10 events are independent.

For each of these events, there is a 5% chance I will have a bad outcome, and 95% chance I will have a good outcome.

So first occurrence:

  1. 95% Good Outcome
  2. 95% Good Outcome

Etc... until 10 times.

Would it be fair to say that over a few thousand occurrences, that the bad event will happen at least once 50% of the time in each occurrence, twice 25%, thrice 12.5% etc... since there is a 5% chance of it happening for every event, and there are 10 events each occurrence?

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The number of failures in each group of ten simultaneous, independent events is something called a binomial random variable. The probability distributions of variables like this are very well known. To exactly describe such a distribution, you quickly get into notations that look like $\binom nk$, called (not so coincidentally) binomial coefficients. But there are some things that you can say about the distribution without getting into such details. (You can get the details if you want by looking up "binomial distribution".)

One thing we can very easily do is to figure out the probability that there will be no "bad" events (that is, there will be ten "good" events) in a set of ten simultaneous independent and identically-distributed events. The probability of ten good events goes like this:

  • The probability the first event is good is $0.95$.
  • The probability the first two events are good is $0.95$ (first event good) times $0.95$ (second event good). That's $0.95 \times 0.95 = 0.95^2 = 0.9025$.
  • The probability the first three events are good is $0.95^2$ (first two events good) times $0.95$ (third event good). That's $0.95^2 \times 0.95 = 0.95^3 \approx 0.8574$.

The probabilities of the first four events being good, the first five events, and so forth follow this same pattern: $0.95^4$, $0.95^5$, etc. The probability of the first ten events (that is all ten events) being good is $0.95^{10} \approx 0.5987.$ That's just about $60\%$, leaving barely more than $40\%$ of the time that you will (on average) observe one or more bad events. So the statement that you will have at least one bad event $50\%$ of the time is inaccurate; $60\%$ would be much more accurate.

Further, a general rule for random variables is that if they have finite expectations, the expectation of a sum of variables is the sum of their expectations. (This is true even if the variables are not independent, although that's not a concern here.)

So consider each of the ten events that occur simultaneously. Call these events $X_1, X_2, \ldots, X_{10}.$ If we want to count bad outcomes, we can say $X_1=1$ (a bad outcome) $5\%$ of the time, and $X_1=0$ (a good outcome) $95\%$ of the time, and the same for $X_2, \ldots, X_{10}.$ Use $X$ to denote the number of bad outcomes in this set of ten events; then $X = X_1 + X_2 + \cdots + X_{10}.$ The expected value of $X_1$ is $$E(X_1) = 0.05 \cdot 1 + 0.95 \cdot 0 = 0.05,$$ and the same for $X_2, \ldots, X_{10}.$ So the expected value of $X$ (the expected total number of bad outcomes in the set of ten events) is $$\begin{align} E(X) & = E(X_1 + X_2 + \cdots + X_{10}) \\ & = E(X_1) + E(X_2) + \cdots + E(X_{10}) \\ & = 10\cdot0.05 \\ & = 0.5. \end{align}$$

Another way to compute expected value is to multiply the value of each possible outcome by the probability of its occurrence, and add all the resulting products together. So the expected number of bad outcomes should be $$\begin{align}E(X_1 + X_2 & + \cdots + X_{10}) = \\ & 1 \cdot P(X = 1) \\ + \; & 2 \cdot P(X = 2) \\ + \; & 3 \cdot P(X = 3) \\ + \; & \cdots \\ + \; & 10 \cdot P(X = 10). \end{align}$$

We have already found that $1 \cdot P(X=1) \approx 0.40$, so that leaves about $0.10$ to be equal to the total sum of $2 \cdot P(X = 2) + \cdots + 10 \cdot P(X = 10).$ Right away we can see that $P(X=2) + \cdots + P(X=10)$ cannot possibly be much more than $0.05$, which is much, much less than a $25\%$ chance of two or more bad events.

Compare the expected number of bad events predicted by the $50\%$, $25\%$, $12.5\%$ (etc.) model. It says the probability of exactly one bad event is $0.5-0.25 = 0.25$, of exactly two bad events is $0.25-0.125=0.125$, etc., but the probability of exactly $10$ bad events is the same as the probability of at least $10$ (because there cannot be more), so $$\begin{align} E(X) &\;{\stackrel ?=}\; 1 \cdot 0.25 + 2 \cdot 0.125 + 3 \cdot 0.0625 + \cdots + 9 \cdot (0.5)^{10} + 10 \cdot (0.5)^{10} \\ &\approx 0.99902. \end{align}$$

So altogether that's about twice as many bad events per set of ten events, on average, as there actually would likely be. In fact, among the proposed probabilities $50\%$, $25\%$, $12.5\%$, etc. for at least one bad event, at least two, etc., the only one that is even remotely close to true is $50\%$ (compared to the actual probability, about $40\%$).

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One issue is you don't really explain why you'd expect this to hold. You just suggest "maybe it works like this", and that's not the way to convince someone else something is true. You aren't even convinced yourself.

Start by checking small cases if possible (which it is in this case) to see if it could possibly be true. So for example, what if there was a 25% chance the outcome is B, and just 2 events happening at the same time. Then try to generalize based on that.

In this case, AA has a 9/16 chance of happening, AB and BA together have a 6/16 chance of happening, and BB has a 1/16 chance of happening. Thus the pattern you were hoping for doesn't seem to hold.

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  • $\begingroup$ I'm pretty sure the 50% figure holds up, the 25% does not. I'm more interested in the first figure. $\endgroup$ – Sefam May 5 '15 at 6:16
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Just to be clear about the subdivisions within the experiment, let us talk about individual trials, groups of 10 trials, and a run of 1000 groups.

We are looking for: the expected number of groups in the run that have a certain number of trials with bad outcomes.

The probability of a bad outcome on a trial is: $p=0.05$ as given.

The number of bad outcomes in a group will have a binomial distribution. $$X \sim \mathcal{Bin}(10,p)$$ That means the probability of exactly $x$ bad outcomes in a group is: $$\begin{align} \mathsf P(X=x) & =\binom{10}{x} p^x(1-p)^{10-x} & ,\text{ for }x\in\{0, 1, .., 10\} \\ & = \binom{10}{x}\frac{19^{10-x}}{20^{10}} \end{align}$$

Let $N_x$ be the number of groups with exactly $x$ bad outcomes in a run. This will also have a binomial distribution, with parameters $1000$ and $\mathsf P(X=x)$.

$$N_x \sim\mathcal{Bin}(1000, \mathsf P(X=x))$$

The expected value of a random variable with a binomial distribution $\mathcal{Bin}(n,p)$ is $np$.   So the expected number of groups with exactly $x$ bad outcomes in a run will be:

$$\begin{align} \mathsf E[N_x] & = 1000\mathsf P(X=x) \\ & =\binom{10}{x}\frac{10^3 \cdot 19^{10-x}}{20^{10}} \end{align}$$

For instance $\mathsf E[N_0] \approx 598.7, \mathsf E[N_1] \approx 315.1, \mathsf E[N_2] = 74.6,$ et cetera.


Well, that's the expected number of groups with exactly $x$ bad outcomes.

Can you now do the expected number of groups with at least $x$ bad outcomes?

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  • $\begingroup$ That somewhat answers my question. I wasn't looking for something this complex. This was more out of a question I had concerning a certain statistic that was pushed at a presentation at the Games Developer Conference. I was more looking towards how this holds up versus a run that goes up to infinity, but I haven't done distributions in a very long time. $\endgroup$ – Sefam May 5 '15 at 6:19

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