Calculate the probability that a point in a circle will land within an inner circle with half the diameter

In the video game Player Unknown's Battlegrounds, you start with a circular safezone of diameter $n$. After some time, the safezone instantly shrinks to a circle with a diameter $\frac{n}{2}$.

The catch is that the smaller safezone will be entirely contained within the bounds of the previous, larger circle. With that in mind, the smaller circle can be anywhere within the larger circle and its position is randomly chosen each time. Here's an illustration to clarify what I mean:

How would you calculate the probability that a randomly chosen point within the larger circle will be contained by the smaller circle? I believe it's a 1:4 probability, but I'd like confirmation as math is not my domain.

Secondly (and more importantly), what distance from the center of the larger circle has the highest probability of being within the smaller circle (and what is that probability)?

If the radius of the big circle is $R$ and you're a distance $\rho < R$ from the center, then you're safe if the center of the little circle lands within a distance $R/2$ from you. That is, the center of the little circle has to be within the lens-shaped intersection of two circles of radius $R/2$ whose centers are $\rho$ apart. The area of such an intersection is readily found to be $$\frac{1}{2}R^2\cos^{-1}\left(\frac{\rho}{R}\right)-\frac{1}{2}\rho\sqrt{R^2-\rho^2}$$ (e.g., see this MathWorld article). The probability that you're safe is obtained by dividing this by $\pi R^2/4$, the area of the little circle: $$P(\rho)=\frac{2}{\pi}\cos^{-1}\left(\frac{\rho}{R}\right)-\frac{2}{\pi}\frac{\rho}{R}\sqrt{1-\frac{\rho^2}{R^2}}.$$ Plotting this against $\rho/R$ shows that, not surprisingly, you're safest right in the center. However, that requires you to stand at a single location, which is probably dangerous in and of itself... if you assess the utility of being at radius $\rho$ with some benefit for having room to move around, you may find a nontrivial answer.

• That's exactly what I'm looking for! It's unsurprising that the center is the safest point, but I thought it was worth the sanity check. Thank you very much! – Kieran E Sep 13 '17 at 17:11

A person can land anywhere in the area of outer circle,

We want person to land in area bounded by inner circle ,

$$P=\frac{\text{area of inner circle}}{\text{area of outer circle}}$$ $$P=\frac{π(\frac{r}{2})^2}{πr^2}$$ $πr^2$ gets cancelled, $$P=\frac{1}{2^2}$$ $$P=\frac{1}{4}$$

As both areas are independent of distance of inner circle from centre, wherever the circle be, probability remains the same.

• Thank you! I thought that was the case. Do you have any advice for the second part of my question? – Kieran E Sep 13 '17 at 16:39
• Probability is independent of distance from centre of larger circle of smaller circle – neonpokharkar Sep 13 '17 at 16:41
• My answer is only true if the game's algorithm has no bias towards the inner circle being more in the centre. – neonpokharkar Sep 13 '17 at 16:42
• I don't understand. Shouldn't there be a ring around the center of the larger circle in which the probability that you'd land in the smaller circle would be higher? There is no bias for inner circle location - it's random. – Kieran E Sep 13 '17 at 16:43
• My math is awful, but bear with me. If the circle has a radius of 1 and it's centered around 0, 0, the point 1, 1 has a much smaller chance of being in the smaller circle than the point 0.5, 0.5. I'm trying to find the most optimal distance from the center of the larger circle. – Kieran E Sep 13 '17 at 16:46