Since Jesus was very much a fellow

*Homo sapiens*, we can presume that he had a colon and intestinal flora pretty much like ours, and that the second person of the Trinity, the Savior of mankind also moved his bowels and farted as the rest of us do. Now what do you think is the probability that in the time it took you to read to this point you've already breathed in at least a molecule of the last couple of farts of the Son of God (perhaps those expelled soon after the last supper or during his crucifixion)?

Let's assume that the expelled gas molecules, given the twenty centuries, have become uniformly distributed all over the world. The estimated number of air molecules in the atmosphere is around 1 x 10

^{44}. While at rest a single breath contains about 1 x 10

^{22}molecules of air. For ease let's assume that the volume of Jesus' last couple of farts was equal to one breath.

Let

F = number of gas molecules in Jesus' flatus = one breath's worth = 1 x 10

^{22}

B = number of gas molecules in a specified number of breaths = (1 x 10

^{22}) x number of breaths

N = total gas molecules in the atmosphere = 1 x 10

^{44}

P(f) = probability of breathing in at least one molecule of F

F/N therefore is the relative frequency of the Jesus' flatus molecules in the atmosphere. It's also the probability of breathing in one molecule of F we were to breathe just one molecule of air. Its complement is (1 - F/N), the probability of not breathing in a molecule of F.

Since we don't just breathe in one molecule but 1 x 10

^{22 }of them, the probability of

*not*inhaling a single molecule of F is (1 - F/N)

^{B}. This is the probability that the first, second, third, ..... , and Bth molecule are all non-Fs. Its complement is therefore the probability of breathing in at least a molecule of F. And that's what we're looking for:

1 - (1 - F/N)

^{B}

If we use various values for B--one breath, two breaths, etc. (B, 2B, 3B,....)--we obtain the following results (see Note 1 for computational details):

Number of breaths | P(f) = probability of inhaling at least one molecule of a specified historical person's flatus |

1 | 63.2% |

2 | 86.5% |

3 | 95.0% |

4 | 98.2% |

5 | 99.3% |

6 | 99.8% |

7 | 99.9% |

In just half a dozen breaths you're practically certain to have inhaled at least a molecule of the Savior's last farts.

What this also means is that every hour, everyday we are inhaling rather odious molecules of practically everyone who's ever lived in the past. And remember we were computing the probability of breathing in at least a molecule of the flatus of one particular person. Consider how many humans have lived over the hundreds of thousands of years and how much gas they've expelled. The probability that we are taking in at least a molecule is 100% for every

*one trillionth*of a breath we take--that's much less than a whiff (see Note 2).

And you wonder why halitosis exists.

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NOTES:

1. If you plug in the values of F, B, and N into a calculator or even a spreadsheet program you'll exceed their limits and you won't get a result anywhere close to accurate. F/N for instance is an infinitesimal figure, while raising a number to the power of B is almost unimaginable. Even resorting to logarithms won't help much. One way around this is to use smaller values of F, B, and N. For example instead of 1 x 10

^{22}use F = 1 x 10

^{6}and N = 1 x 10

^{12}. Note that the exponent of N must be twice that of F since it is so in the original values. For one breath B = 1 x 10

^{6}. Try using other values for F such as 1 x 10

^{4}for instance (and corresponding appropriate values for B and N). You should get practically the same P(f) values.

However, a more graceful way of doing it is to use a different equation altogether. If you examine the above problem you'll see that it is basically a binomial probability distribution and can be solved using the general equation

_{}

_{n}C

_{r}p

^{r}(1 - p)

^{n-r}

where

_{n}C

_{r}is combinatorial notation for n taken r at a time. For our problem above n = B, r = 0 (since we're computing for zero flatus molecules), and p = F/N. Because r = 0 ,

_{B}C

_{r}= B! / [r! (B -r)!] = 1 and p

^{r}= 1. Therefore, the equation reduces to (1 - p)

^{n}. Substituting, we have: (1 - F/N)

^{B}, the very equation we have above.

Because p is very small and n very large the conditions are satisfied for using the Poisson distribution as an accurate approximate for the binomial probability. The Poisson distribution is given by the equation

(np)

^{r}e

^{-np}/ r!

where e = 2.71828. Since r = 0 the equation reduces to e

^{-np}. Substituting, we have e

^{-BF/N}. This can be easily handled by a scientific calculator. Because this equation gives us the probability of not having a single flatus molecule in B, we are interested in its complement: 1 - e

^{-BF/N}.

2. To estimate the amount of flatus molecules ejected by humans throughout the ages, I assumed 1 x 10

^{22}gas molecules expelled per person three times a week, with each person having an average lifespan of only 30 years, and a total (deceased) human population of 10 billion. F then is = 4.68 x 10

^{35}. For 1 x 10

^{10}molecules inhaled the P(f

_{total}) = 99.9%. 1 x 10

^{10}is one trillionth of 1 x 10

^{22}.

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This math problem is a variation of the Caesar's-dying-breath example by mathematician John Allen Paulos (

*Innumeracy: Mathematical Illiteracy and Its Consequences*. 1988. New York: Vintage. p. 32.). However, Paulos does not offer the reader the use of the Poisson distribution. The idea of substituting flatus for breath comes from The New England Skeptical Society's podcast (2006).

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