# What does "twice as likely" mean?

Once in a while I hear people say something like

X is twice as likely as Y.

What they usually mean is:

$$p(X) = 2 \cdot p(Y)$$

and - in the context they refer to - they usually have $p(Y) < \frac{1}{2}$. But what do you do if $p(Y) > \frac{1}{2}$? Can there be an event $X$ that is twice as likely as $Y$? It also feels wrong to me to say that $p(X) = 100 \%$ is twice as likely as $p(Y) = 50\%$.

Is there a good definition what twice as likely means?

Let's call this "twice as likely" a function

$$d: D \rightarrow [0, 1]$$

I would expect $d$ to have the following properties:

• $D = [0, m]\subseteq [0,1]$
• $d(0) = 0$
• $d$ is monotonous
• Perhaps they mean $E(X) = 2E(Y)$ Apr 20, 2014 at 9:18
• There is hardly ever a good definition for what you "hear people say". But it seems that working with odds $\frac{p(X)}{1-p(X)}$ somhow comprises what people feel. This would make $99\%$ twice as likely as $98\%$, insofar as the event fails only half as often ... Apr 20, 2014 at 9:19
• @HagenvonEitzen I think there might be a typo in your comment. When you have $p(X) = 99\%$, you get $\frac{0.99}{1-0.99} = \frac{0.99}{0.01} = 99$. Did I misunderstand you? Apr 20, 2014 at 9:24
• It might help define your function if you can say what probability represents something "twice as likely" as a certainty. Apr 20, 2014 at 9:35
• @moose But this is exactly what Hagen means, since for $p(X) = 98\%$ you get $49$, which is roughly half of $99$. Apr 20, 2014 at 10:18

Personally I don't think "double as likely as $p$" (or "twice as likely", which sounds more English?) sounds like it ought to be meaningful for $p\ge\frac12$.

However, if we want it to have a meaning, then it seems reasonable to require that "$q$ is double as likely as $p$" should mean $q=2p$ for small $p$ and $(1-q)=\frac12(1-p)$ for $p$ close to $1$. The problem is, of course, that these two expressions don't agree for $p$s in the mid-range, so we have to bridge them somehow.

Simply switching formulas at $p=\frac13$ -- which is the point where they give the same $q$ -- is insufferably simplistic, of course. It creates ugly discontinuities.

The most mathematically principled way to bridge the two ends would seem to be to say that $q$ should be the number whose logit is $\log2$ more than the logit of $q$. This gives us the equation $$\log\frac{q}{1-q} = \log\frac{p}{1-p}+\log 2$$ which simplifies to $$2p-pq-q = 0$$ or $$q = \frac{2p}{p+1}$$ which is the same as user2345215 reached.

(This is, by the way, the rational function of lowest degree that has the required values and slopes at $p=0,1$).

Unfortunately, its results in the mid range are not quite intuitive. For example, it claims that "double as likely as" 33% ought to be 50% rather than the 66% one would probably expect.

• Is there any benefit to saying that the logit increases by log 2 instead of just saying that the odds ratio is doubled? Is there some additional intuition attached to the logit that I'm missing here? Jul 30, 2014 at 10:12
• "twice as likely" sounds more natural than "double as likely". Jul 30, 2014 at 10:22
• If the question is asking what ought "twice as likely" mean, then I like your second part - doubling the odds ratio. If this question were posed on English Stack Exchange and were asking what is typically meant when a non-mathematician says "twice as likely", then perhaps your first part is closer (which you noted with the $p=33\%$ example). Jul 30, 2014 at 10:40
• On second thought, I think KennyLJ's answer is the correct one, but I can't undo my upvote. Aug 1, 2014 at 22:39

There's one interpretation which I think makes the most sense: $$p\longrightarrow p:1-p\longrightarrow 2p:1-p\longrightarrow \frac{2p}{1+p}$$ One could also interpret it as picking the best result out of $2$, in which case you would get $$p+(1-p)p=2p-p^2$$ Both are ${\sim}2p$ for low values of $p$.

I think the first is right, because then half as likely matches twice as unlikely. But if you generalize the second to positive reals, you won't get that picking the worst result out of $2$ is the same as being half as likely in the generalization.

• I think the odds interpretation is what I was taught in elementary school. Also, consider efficiency. "Twice as efficient" basically means "half as wasteful". Apr 27, 2014 at 0:14

The 'people' you refer to who say "twice as likely" are not usually professional (or even recreational) mathematicians. Their use of this expression is therefore limited to instances where $P(Y) \leq 0.5$.

What 'people' mean when they say most things mathematical is usually the simplest thing it sounds like and which you've already stated in your question: To say that $X$ is twice as likely as $Y$ is to say that $P(X) = 2P(Y)$.

If $P(Y) > 0.5$, then nobody would ever say that some event $X$ is twice as likely as $Y$.

Of course, you are welcome to invent a "twice as likely" function with all the bells and whistles you like, but that will never be what 'people' mean.

Edit to elaborate: You add also that

It also feels wrong to me to say that p(X)=100% is twice as likely as p(Y)=50%.

I think it only 'feels wrong' because if $P(X)=1$ and $P(Y)=0.5$, then 'people' would typically say something (much) more emphatic like "$X$ will happen for sure, whereas $Y$ happens only half the time." The point of saying that $X$ is "twice as likely" as $Y$ is to emphasize that $X$ is much more likely to happen. But if $P(X)$ is going to happen for sure, you might as well simply say that it is going to happen for sure.

• It seems you are right and that in the statistics literature, this is what it means. For example, the Wikipedia page on the likelihood ratio say: "the likelihood ratio, which expresses how many times more likely the data are under one model than the other...", by which they mean $\frac{P(\text{model 1})}{P(\text{model 2})}$. Aug 1, 2014 at 22:37

This leads to the same answer as @user2345215's response, but hopefully motivates it more. In Pokemon (at least in Pokemon Go) you have regular balls with which to catch pokemon, as well as "enhanced" balls which have a higher chance to catch a given pokemon, called great balls. I may not have the details exactly right, but the way they handle this "higher chance" is nice, and could impart meaning to "twice as likely".

Suppose the chance to catch charizard with a regular ball is 10%. We want the chance of catching it with a great ball to be "doubled" in some sense that still holds were we to apply this discussion to pikachu, whose chances are 60%. What Pokemon does is to give great balls a $$2 \times$$ "multiplier": the chance of catching charizard with a great ball equals the chance of catching it given two regular balls, which equals the chance of catching it on the first throw + the chance of catching it on the second: $$p_{great} = p_{reg} + (1-p_{reg})*p_{reg}$$

Thankfully this never exceeds 1 (which can be checked) and gives a nice meaning to twice as likely. In the example above, a great ball has a 19% chance of catching a charizard. In fact, "twice as likely" only ever equals $$2p_{reg}$$ when the probability is zero; otherwise it's always strictly less than 2.

I'm trying to think of real life examples where "twice" or "$$5$$ times" as might be used. These facts tend to be quoted for things with already very low chances, like smokers dying of lung cancer, in which case the author simply means $$2p$$, or $$5p$$. So let me invent an example, which MIGHT be true, where the distinction matters:

A randomly chosen male is twice as likely to have experienced a car accident as a female.
70% of females have experienced a car accident.


Here the "pokemon" are car accidents and pokeballs are sequential draws of males or females, respectively. The chance of seeing accident history by drawing one random male, equals the chance of seeing accident history with two randomly drawn female lifetimes; 91%.

I think separate "definitions" of "twice as likely" for $p < 1/2$ and $p > 1/2$ are both meaningful.

For $p < 1/2$, if you think of probability as successful outcomes over total outcomes, then a statement "twice as likely" means that the ratio of successful outcomes over total outcomes is doubled. Thus, it does not make sense to define the expression "twice as likely" if a probability is already more than $\frac{1}{2}$.

For $p > 1/2$, then interpret the statement as "an unsuccessful outcome is half as likely." Thus, we get that $99\%$ is "twice as likely" as $98\%$. Note that this is the same definition of the first except we are focusing on the other side of the coin, so to speak.