# Game of probability

In a game, played between $2$ players there is a circular field and one of the players is blindfolded, who stands in the center of the field. The other player stands at a fixed point on the circumference of the circular field. On the word GO, the blind-folded player starts running towards the edge of the field while the second player's aim is to run in and catch him before he moves out of the circle.

If the blind-folded player runs in a random direction with a constant speed $v$ while the second player runs towards the first player with a constant speed $m$ times $v$, what should be the value of $m$ such that the probability that the second player wins is $0.50$?

• I have no idea. I haven't solved many continuous probability problem of this types. I don't know how to approach it. – guest123456 Jul 5 '15 at 5:31
• With player on circumference at 12 on a clock dial, consider blindfolded one's movement from 0 through π radians. – true blue anil Jul 5 '15 at 6:32

Let at time t, the first player has moved a distance R1 while the second player who was situated at a point on the circumference at the start has caught the first player at the same time t. Let the distance R2 be between his initial point on the circumference to the point where he caught the first player. Let R be the radius of the circular field. Let the angle $\alpha$ be subtended by the second player to the radius. As the player 1 moves towards the circle edge, R1,R2 and $\alpha$ all change. At any instantaneous point, we have a relation amongst R1,R2,R and $\alpha$ by cosine rule.

From cosine rule,

we get $R_1^2 = R_2^2 + R^2 - 2R_2Rcos\alpha$

Taking the derivative, we get

$2R_1\frac{dR_1}{dt} = 2R_2\frac{dR_2}{dt} + 2R\frac{dR_2}{dt}cos\alpha -2RR_2sin\alpha \frac{d\alpha}{dt}$

$\frac{d\alpha}{dt} = \frac{mv}{R_2}$

$R_1v = R_2mv + Rmv(cos\alpha - sin\alpha)$

$R_1 = mR_2 +Rm(cos\alpha - sin\alpha)$

$R_1 = m^2dR_1+ mR(cos\alpha - sin\alpha)$

$R_1(1-m^2) = mR(cos\alpha - sin\alpha)$

$R_1 = \frac{mR}{1-m^2} (cos\alpha - sin\alpha)$

Area of the Apollonius circle $= \int_{0}^{\pi}R_1dR_1d\alpha$

$=\int_{0}^{\pi}\int_{0}^{R}\frac{m^2R}{1-m^2)^2}(cos\alpha - sin\alpha)^2dRd\alpha$

$=\frac{m^2R^2}{2(1-m^2)^2} \int_{0}^{\pi}(cos^2\alpha+sin^2\alpha - 2sin\alpha cos\alpha)d\alpha$

$= \frac{m^2R^2}{2(1-m^2)^2}(\int_{0}^{\pi}(1-sin2\alpha)d\alpha$

$=\frac{m^2R^2}{2(1-m^2)}(\pi)$

Probability = Area of Apollonius Circle/ Area of the circular field ($\pi R^2$)

Required probability $= \frac{m^2}{2(1-m^2)^2} = \frac{1}{2}$

$\frac{m^2}{(1-m^2)^2} = 1$

Rearranging the terms you get $m^4-2m^2 +1 = m^2$

$m^4-3m^2+1 = 0$

Solving this you get $m = \dfrac{1+\sqrt{5}}{2}$

Thanks

Satish

• In the actual question, value of $R$ is unknown. $m$ should be independent of $R$. – guest123456 Jul 5 '15 at 17:57
• Is the answer what I have indicated? – Satish Ramanathan Jul 6 '15 at 5:09
• No, :( $\sqrt \frac{1}{2}$ is not the correct answer. – guest123456 Jul 6 '15 at 5:33
• Seems like you are aware of an answer being wrong but at the same time you do not know the answer. If you know the answer, let me know. Do you understand the approach. If you can give me the answer, I can further tweak the solution to fill the gap of where I am missing. – Satish Ramanathan Jul 6 '15 at 6:50
• No no. I don't know the answer. But I check if an answer is correct by submitting at erdos.sdslabs.co/problems/15 . I mentioned the source at the end of my question. You can submit yourself and check. If you get the answer please tell me the approach. – guest123456 Jul 6 '15 at 7:34

The locus of a catch positions is a circle of Apollonius defined by the players initial positions and the ratio of their velocities. The probability of the second player win is a relative part of the 'game circle' not contained in the Apollonius' circle.

0.5 probability of player 2 winning means that if player 1 runs perpendicular to the line joining the initial positions of the players, then player 2 will catch him on the boundary of the circle.

In that case the path of player 2 will be the hypotenuse of an isosceles right with side length $r$, the radius of the circle. the length of this path is $l=\sqrt 2 r$

since player 2 must cover $\sqrt 2 r$ in the same time as player 1 takes to cover the distance $r$, they must be moving $\sqrt 2$ times as quickly.

So $m=\sqrt 2$

• Unfortunately the answer is not correct. :( – guest123456 Jul 5 '15 at 6:18
• The second player is not very bright. It always runs towards the first, not towards the place where the first will exit the field. – André Nicolas Jul 5 '15 at 6:24
• Then what approach should I take. Can you please explain? – guest123456 Jul 5 '15 at 6:42
• @guest123456, Could you tell me the answer. I tried solving it and have an answer. I just want to check the answer and type it out. – Satish Ramanathan Jul 5 '15 at 7:53
• I don't know the answer yet. $\sqrt 2$ is not the correct answer. – guest123456 Jul 5 '15 at 8:20