Mathematics Report - Probability - Darts

Hello,

I need some help with my maths assignment. The mission is to investigate the difficulty of achieving an outshot of 141 in a game of 501. I need help with displaying the sample space and calculating the probability of each segment. Here are the measurements given for the dartboard layout;

Bullseye = 12.7mm

Double ring and triple ring = 8mm wide

Outer bull = 31.8mm

Dartboard = 451mm

Centre bull to outside of triple ring = 107mm

Centre bull to outside of double ring = 170mm

^ Please help me! Also, I need some help as soon as possible. (Headbang)

Re: Mathematics Report - Probability - Darts

Hey alexa01.

You will need to explain what an outshot is and what kind of model you are using and the goal of the game (do you have to get to zero or past zero first? Do you have to get exactly zero?)

Are you assuming that the probability of hitting any point is the same as the other? If not, is it something like where hitting the outer edges is more probable than the inside?

Re: Mathematics Report - Probability - Darts

I'm reflecting on the perfect game, which is a nine-dart finish. Scoring 360 in the first two visits is the most common way of achieving a nine-dart finish. This leaves 141 to score in the last visit; this is the outshot. The goal is to get to zero in a nine-dart finish. I believe I'm assuming that hitting the outer edges is more probable than the inside. I'm really confused with calculating the probability of each segment! As a hint for my homework, it says to calculate area.

Re: Mathematics Report - Probability - Darts

Well to give you an example of a uniform distribution over the whole board, you calculate the area and the probability across the whole board (at each point that it is: aka a continuous distribution) will be 1/Area (since integrating this over the whole circle in R^2 will give 1).

If you based it on area alone then getting in the region of the bullseye relative to the rest of the board will be very small even with the uniform assumption.