Continuing the discussion from Comments on Gpuccio: Functional Information Methodology:

Pardon me, but my pet peeve has been perked! @AllenWitmerMiller, bring me my *soapbox!*

I am quoting @Gpuccio here, but only because this is the latest example of an error I see regularly, and darnit, I’m getting tired of correcting some basic probability calculations. The error is, **probabilities do not multiply this way.**

If **X** is some event and the probability of **X** occurring in a single trial is * 0.1*, then the probability of

**X**occurring in two trials generates the following probability distribution:

The probability **X** does not occur in either of two trials: (1.0-0.1)\times(1.0 - 0.1) = 0.9 \times 0.9 = 0.81

The probability **X** occurs in the first trial but not the second: (0.1)\times(1.0 - 0.1) = 0.1 \times 0.9 = 0.09

The probability **X** does not occur in the first trial but does in the second: (1.0-0.1)\times(0.1) = 0.9 \times 0.1 = 0.09

The probability **X** occurs in both the first and second trials: (0.1)\times( 0.1) = 0.1 \times 0.1 = 0.01

If we add up all these possibilities they sum to **1.0**.

0.81 + 0.09 + 0.09 + 0.01 = 1.0

AND finally the probability that **X** occurs at least once in two trials is the sum of the last three terms:

0.09 + 0.09 + 0.01 = 0.19

OR we can take the complement; the compliment of a probability § is one minus that probability (1-p), so the probability that **X** does not occur in two trials:

1-0.81 = 0.19

More on this complementary trick, but first a word from our sponsor, the Binomial distribution, which is where we end up if we generalize this example.

Now (back to our program and) athe compliment trick; with the same **X** and five **N** trials instead of two.

The probability **X** does not occur in any of **N** trials: (1.0-0.1)^N = 0.9^N

and the complimentary probability that **X** occurs at least once:

(1 - 0.9^N)

and if **N**=2:

(1 - 0.9^2 = 1 - 0.81 = 0.19

and finally (FINALLY!) generalizing to some probability *p*; if the probability of **X** is *P*, then the probability of **X** occurring at least once in **N** trials is:

(1 - (1-p)^N)

which will always top out at 1 even for an infinite number of trials.

So to my point, quoting Gpuccio with no malice intended,

Of course, it is 1 billion times more likely to find the target in 1 billion steps then in 1 step.

I’m not using Gpuccio’s stated probability because it does not matter - the statement is incorrect. If the original probability is greater than 10^{-9} = 0.000000001, then you end up with a probability greater than one, which is NOT a probability by definition. This simple multiplication is not a valid probability calculation. It’s actually the number of times we expect **X** to have occurred in **N** trials.

So it is not *“a billion times more likely to find a target”*, in this example. That might be correct to within an order of magnitude for *p* much less than 1\over N, but not in general.

Remember this compliment trick, because the same calculation comes up regularly in discussions of ID.

OK, rant over.