I was wondering if the reciprocal of P(X = 1) represents anything in particular?

  • 3
    $\begingroup$ Maybe something related to odds $\endgroup$
    – BCLC
    Oct 30, 2016 at 4:44
  • 1
    $\begingroup$ why X=1 in this case? can X be anything? $\endgroup$
    – mandata
    Nov 1, 2016 at 17:57

6 Answers 6


Yes, it provides a 1-in-$n$ scale for probabilities. For example, the reciprocal of .01 is 100, so an event with probability .01 has a 1 in 100 chance of happening. This is a useful way to represent small probabilities, such as .0023, which is about 1 in 435.

  • 9
    $\begingroup$ +1 This is a form of "rarity" measure sometimes used in talking about rare events (akin to "a one-in-a-hundred-year flood"). When dealing with various aspects of insurance of unusual events such measures are of interest. In the case of P(X=1) it mightn't be quite as relevant though. $\endgroup$
    – Glen_b
    Oct 29, 2016 at 0:47
  • 16
    $\begingroup$ Somewhat related is the number needed to treat (NNT). $\endgroup$ Oct 29, 2016 at 16:03
  • 1
    $\begingroup$ So basically, the reciprocal of a probability is the rarity of something. Probability = .0023, rarity = (1 in) 435 $\endgroup$
    – Cullub
    Oct 31, 2016 at 17:09

$\frac 1p$ does not mean anything in general (but for a particular meaning for a specific random variable see the answer by Alex R.). However, the logarithm of $\frac 1p$ to base 2, viz., $\log_2 \frac 1p = -\log_2 p$ is the amount of information (measured in bits) that you receive when you are told that the event (of probability $p$) has occurred. If the event has probability $\frac 12$, then you receive one bit of information when you are told that it has occurred. In a different answer, Kodiologist has suggested that if $N$ is chosen as $\left\lfloor\frac 1p\right\rfloor$ or $\left\lceil\frac 1p\right\rceil$, then one can say that

$$\textrm{an event of probability } p ~\textrm{has approximately } 1 ~ \textrm{chance in } N ~ \textrm{of occurring}$$

So, since $2^{20} \approx 10^{6}$, the occurrence of an event that has $1$ chance in a million of occurring conveys only 20 bits or so of information to you, far less than is needed to transmit "Cubs win!" in ASCII ! :-)

  • 4
    $\begingroup$ It's worth pointing out that $\log$ is monotonic, so for probabilities $p$ and $q$ we can state $p > q \implies \frac{1}{p} \leq \frac{1}{q} \implies \log \frac{1}{p} \leq \log \frac{1}{q}$ $\endgroup$ Oct 29, 2016 at 20:18

In the case of a geometric distribution, the reciprocal $1/p$ represents the expected number of throws you need to make to see one success. For example if a coin has probability $0.2$ of landing on heads, then you'd need to throw it around 5 times to see one head.

  • $\begingroup$ Isn't that proba P(get a head in 5 runs) = 1 - P(not getting a head in 5 runs) = 1 - (0.8)^5 = 0.67... This way you can see that 4 runs are enough to get more than 50% chance of seeing a head. $\endgroup$ Feb 7, 2017 at 10:52
  • $\begingroup$ @David天宇Wong: No. Let $\tau$ be the waiting time, in throws, until the first coin. We are saying that $E[\tau]=1/p$. On the other hand, $P(\tau=1)=p$, $P(\tau=2)=2p(1-p)$. $\endgroup$
    – Alex R.
    Feb 7, 2017 at 23:16
  • 1
    $\begingroup$ I figured it out, this is the expectation of the random variable X := number of tries until a head is observed. E(X) = 1 * P(X = 1) + 2 * P(X = 2) + ... = 5 $\endgroup$ Feb 8, 2017 at 17:30

What are sometimes called European odds or decimal odds if fair are the reciprocal of the probability of winning, which might be a Bernoulli random variable $P(X=1)$.

For example if the quoted odds are "1.25" and you bet $8$ then you get $8 \times 1.25=10$ back if you win (including your original stake, so a gain of $2$) and nothing back if you lose. This would be a fair bet if the probability of winning was $\dfrac{8}{10}=0.8$, which has a reciprocal of $\dfrac{1}{0.8}=1.25$.

Similarly if the quoted odds are "5.00" and you bet $8$ then you get $8 \times 5=40$ back if you win (including your original stake, so a gain of $32$) and nothing back if you lose. This would be a fair bet if the probability of winning was $\dfrac{8}{40}=0.2$, which has a reciprocal of $\dfrac{1}{0.2}=5.00$.


In the context of survey design, the inverse of the probability of being included in the sample is called sampling weight.

For example, in a representative sample of some population, a respondent with the weight of 100 has 1/100 chance to be included in the sample, in other words, this respondent represents 100 similar people in the population.


In statistical mechanics, a system has a large number of microstates, and it is a fundamental principle that these are all assumed to be equally likely. The reciprocal of the probability of a particular microstate is therefore the number of possible microstates, and this has a name in physics; it is (confusingly) called the thermodynamic probability.

The log of the thermodynamic probability is the entropy of the system, up to a constant.


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