TITLE: dartboard probability equivalence?
QUESTION [1 upvotes]: Let's say we have a dartboard and a dart will be thrown at it at random. You can buy an area of the dartbord of say 1 cm$^2$ and if it hits your area you win some money. Consider the following two situations.
You buy a 1 cm$^2$ portion and a dart will be thrown and do this three times.
You buy a 3 cm$^2$ portion one time and a dart will be thrown.
The question: will the probability of winning some money be the same in these two situations.
Intuitively it doesn't make a difference but if we consider the case that you buy the whole area of the board then the probability of winning is one and that is isn't the case if you buy the same amount of 1 cm$^2$ chunks.

REPLY [1 votes]: It's clear from context that the probability of hitting a certain portion is proportional to its area.
Suppose that the probability of hitting a 1 cm2 portion is $p$. The proability of not hitting it $1-p$. The probability of none of the three darts hitting it is $(1-p)^3$. Hence the probability of winning some money is $1-(1-p)^3 = 3p-3p^2+p^3$.
In contrast, the probability of hitting a 3 cm2 portion is $3p$.
Since $3p - 3p^2 + p^3 = 3p - p^2(3-p) < 3p$ (since $p > 0$), it follows that the probability in the second case is better.
Why is this the case? Here are two answers.
First answer. We can think of the second case as an experiment the following three step experiment:

Throw the dart at a 1 cm2 portion. If it hits the portion, we win.
Throw the dart at a disjoint 1 cm2 portion, under the condition that it doesn't hit the first portion. If it hits the portion, we win.
Throw the dart at a 1 cm2 disjoint from the other two, under the condition that it doesn't hit the first two portions. If it hits the portion, we win, otherwise we lose.

The probability that we hit the portion in the second and third steps is strictly larger than $p$: it is $\frac{p}{1-p}$ in the second step, and $\frac{p}{1-2p}$ in the third step. Compare this to the first scenario, in which the hitting probabilities are just $p,p,p$. It follows that the probability that the dart hits none of the three portions is smaller in the second case.
Second answer. In both cases, the expected number of darts hitting the portion is $3p$. In the first case, there is positive probability that more than one dart hits the portion, whereas in the second case the probability is zero. Hence the probability of no darts hitting the portion is larger in the first case.
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