In a comedy TV programme, four men are sitting at the bar.
The barman tells them: "Did you know that, statistically, one out of four men is having an affair?".
The first man replies "Not me, I love my wife."; the second man "Me neither, I would never do this to her!"; and the third man "Same here, I've only been married for two weeks...".
Before the fourth man can open his mouth, his wife, who was within earshot and had listened to the whole conversation, angrily walks up to him, slaps him in the face and shouts "You swine! I'm leaving you!".

Silly joke, of course, but it made me think about the implications of revealing information on the probability of an event.

Obviously the wife's conclusion that her husband was certainly cheating on her, based on the fact that 25% of men cheat on their wife, and in a group of 4 men, 3 (75%) of them stated that they were not cheating, was incorrect.
Even assuming that the 3 men who spoke were telling the truth, how did their statements change the probability that the 4th man was unfaithful?

Initially I thought: it does not change anything, because the event 'a randomly selected man is cheating on his wife' still had a probability of 25%.

However, on further reflection, it occurred to me that if the whole population of men had been those 4, the wife would have been right!

In a more general case, where the population (let's say of $N$ subjects) is indeed larger than the sample one is observing, how would one correct the probability of an event based on the disclosure of some information about the sample?

Suppose $N = 12$.
When we know that $n_f = 3$ men in the sample of $n = 4$ men were not cheating, we still need to have $N \cdot p = 12 \cdot 0.25 = 3$ cheating men in the remaining $N - n_f = 12 - 3 = 9$ part of the population.
So the probability that the 4th men was cheating increased to:

$$p \cdot \frac {N}{N - n_f} = 0.25 \cdot \frac {12}{12 - 3} = 0.25 \cdot \frac {12}{9} = \frac {1}{3}$$

In situations where $N >> n_f$, $p$ can probably be used for any single event.
I suppose that is where the binomial distribution can be used with good approximation.

When $N$ and $n_f$ are closer in value, do you think my method above is correct?

More formally, I would think that the situation I am describing may be handled by the hypergeometric distribution (sampling without replacement from a finite population), but then I am not exactly sure how.

If I ask: what is the probability that only 1 man out of a sample of 4 men, taken from a population of 12 men where 3 of them in total are cheating, is cheating?
The formula gives:

$$P(X=1) = \frac {\binom {K}{k} \binom {N-K}{n-k}}{\binom {N}{n} } = \frac {\binom {3}{1} \binom {12-3}{4-1}}{\binom {12}{4} } = \frac {\binom {3}{1} \binom {9}{3}}{\binom {12}{4} } = \frac {28}{55}$$

which makes sense:

$$\frac {3}{12} \frac {9}{11} \frac {8}{10} \frac {7}{9} \binom {4}{1} = \frac {28}{55}$$

But if my previous approach is correct, then this result is not what I am after.

Perhaps should I use conditional probability, so I can account for the first 3 drawn samples that turned out to be negative?

$P(\text{4th man sampled cheating} | \text{first 3 men sampled not cheating})$

Would this work?

  • 2
    $\begingroup$ This seems like gambler’s fallacy. Anyone else see it that way? $\endgroup$
    – Dave
    Apr 1, 2022 at 20:03
  • 2
    $\begingroup$ Re " if the whole population of men had been those 4, the wife would have been right": this confuses probability with proportion. Although one is often used to measure the other, they are not the same things. $\endgroup$
    – whuber
    Apr 1, 2022 at 20:09
  • 1
    $\begingroup$ Thanks @whuber, I think I see what you mean. Even if the whole population of men were $N=4$, the information that $p=1/4$ does not constitute certainty that one of them is cheating, if $p$ is information about generic male populations. Perhaps my confusion came from seeing this in a hypergeometric setup, where you have an urn with $N$ balls, $0<K<N$ of which are black, the rest white, and you sample $n\le N$ balls. If $n=N$, you are sure that $K$ balls in your sample are black, so if you have looked at the first $N-K$ and found them to be white, you are sure the rest are black, I suppose. $\endgroup$ Apr 2, 2022 at 8:13

2 Answers 2


I see this as gambler's fallacy.

The coin (husband) has come up heads (faithful) a bunch of times in a row. I'm due for a tails (cheater)!

In the gambler's fallacy, the gambler mistakenly believes information to be informative of events that are independent of the observed events, such as previous flips of a coin or previous inqueries about marital fidelity. However, the events are independent of one another. The first three instances had a $1/4$ chance of coming up a certain way. When you get to the fourth one, it's still just a $1/4$ chance.

Let's simulate this in R.

N <- 10000
X <- matrix(rbinom(N*4, 1, 1/4), N, 4)
husband4 <- c()
for (i in 1:N){
  if (X[i, 1] == 0 & X[i, 2] == 0 & X[i, 3] == 0){
    husband4 <- c(husband4, X[i, 4])

mean(husband4) # 0.2466667

(There are better ways to write that, yes.)

When we repeat the comedy skit many times and limit our investigation to the fourth husband after the first three were faithful, it turns out that the fourth husband was a cheater only about a quarter of the time, completely consistent with the $1/4$ probability of cheating.

In order for your argument to make sense, you would have to tweak the situation. Instead of saying that a quarter of men cheat on their wives, you could limit it to saying that exactly one of those four men has cheated on his wife. Let's give our characters names.

Husband 1 is John, and his wife is Jen.

Husband 2 is Ed, and his wife is Eleanor.

Husband 3 is Peter, and his wife is Patricia.

Husband 4 is Mike, and his wife is Maggie.

When it's John's turn, Maggie is feeling good. There's only a $1/4$ chance that Mike has cheated. When John is shown to have been faithful, now Maggie has new information and can update her beliefs. One of the remaining three husbands has cheated. By the time Ed is shown not to have cheated, Maggie is nervous. Either Peter has cheated on Patricia, or Mike has cheated on her. When Peter is shown not to have cheated, Maggie is certain that Mike cheated.

  • $\begingroup$ Thank you Dave, this is very clear. Your example where there is certainty that exactly one out of the 4 men has cheated was what I thought was the case with $N = 4$. But actually I see now that it is not, as per @whuber's comment. Your initial simulation is, I believe, valid if we consider $p =1/4$ to be a piece of information about populations of men in general, not the proportion of cheating men in the sample. $\endgroup$ Apr 2, 2022 at 8:02
  • $\begingroup$ Exactly! If we know that one of the four men cheated, then we know it was Mike if it wasn’t any of the first four. If we know that Mike’s probability of cheating if $1/4$ and independent of the first three husbands, then the probability of Mike having cheated by the time the bartender asks him is still $1/4$. $\endgroup$
    – Dave
    Apr 2, 2022 at 12:15

One way to treat such a situation is to use a Bayesian argument.

Suppose you have an opinion, perhaps based on personal experience, perhaps based on some data, or perhaps based on a combination of personal experience and data that about $1/4$ of men cheat on their wives.

You might say that the proportion $p$ of cheaters is roughly expressed by a beta prior distribution $p \sim \mathsf{Beta}(4,12),$ which has mean $1/4,$ probability 95% of being in the interval $(0.078, 0.481),$ and density $f(p) \propto p^3(1-p)^{11},$ where the symbol $\propto$ indicates that the 'kernel' of the density function is shown (without the constant that makes it integrate to unity).

qbeta(c(.025,.975), 4, 12)
[1] 0.07787155 0.48089113

curve(dbeta(x, 4, 12), 0, 1, lwd=2, ylab="Density", 
  xlab="p", main="BETA(4,12)")
abline(v=0:1, col="green2")
abline(h=0, col="green2")
abline(v=c(0.078, 0.481), col="orange", lty="dotted")

enter image description here

Now, suppose you are able to get reliable data on $n = 50$ married men from some population of interest, finding $x = 10$ cheaters among the fifty.

Then the binomial likelihood function of these data is $$g(x|p) = p^x(1-p)^{50-x} \propto p^{10}(1-p)^{40}.$$

Using Bayes' Theorem, we can get the posterior distribution $h(p|x)$ by multiplying the beta prior density by the binomial likelihood function:

$$h(p|x) = f(p)\times g(x|p) \propto p^{13}(1-p)^{51},$$

where we recognize the result as the kernel of the distribution $\mathsf{Beta}(14, 52).$

Thus, the posterior mean is $14/66 = 0.212$ and a 95% posterior probability interval for $p$ based on the prior and the data is $(0.123, 0.318).$

qbeta(c(.025,.975), 14, 52)  
[1] 0.1230531 0.3176880

enter image description here

If you get additional data later, you can use the posterior distribution $\mathsf{Beta}(14, 52)$ as the new prior, and the additional data to get a new posterior and probability interval.


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