Expected value of a randomly thrown dart

Given the following measurements of a regulation dartboard, what do you get for the expected value of a randomly thrown dart? Anything outside the double ring is worth 0, and assume all darts must hit the board. A standard dartboard is split into 20 congruent sectors, the sectors all have a unique integer point value from 1-20. Darts thrown in the double and treble rings are worth double and triple that sectors point value respectively. Bull's eye is worth 50 points regardless of the sector it is in, and the 25 ring is worth 25 points.

I get 7.22.


"Double' and 'Treble' rings inside width = 8.0 mm.
'Bull' inside diameter = 12.7 mm.
'25' ring inside diameter = 31.8 mm.
Outside edge of 'Double' ring to center = 170.0 mm.
Outside edge of 'Treble' ring to center = 107.0 mm.
Overall dartboard diameter = 453.0 mm.

I got the same. 7.224241

Are you calculating it where the entire area of the board has an equal chance of beign hit? Or are you weighting it towards the center? I sense you're giving it an equal chance.

Personally, I would calculate the area of each target area as you've done, then using a bell curve, calculate the volume based upon the height of the bell at that target area, etc.

Then the question becomes: What's the shape of the bell?

If your player throws darts in random direction, he is unlikely to even hit the board. So you need some assumption of your player to get reasonable results.
The simplest approach would be his accuracy (variance) and his aim (expected hit).
If he aims at the bulls eye, you get a gaussian weight of the target area. If he aims at triple-20, this gaussian weight is shifted.

So, to answer your question, you need a model of your player throwing the darts.

Quote: DJTeddyBear

Are you calculating it where the entire area of the board has an equal chance of beign hit? Or are you weighting it towards the center? I sense you're giving it an equal chance.

Personally, I would calculate the area of each target area as you've done, then using a bell curve, calculate the volume based upon the height of the bell at that target area, etc.

Then the question becomes: What's the shape of the bell?

differently skilled.players are going to have taller or flatter bells. That's before we even get into right and left handedness.

If I'm the player you need to consider a good chance that I miss the dart board all together and hit the wall.

It says "randomly" AND every dart hits the board.

Quote: nmacgre

It says "randomly" AND every dart hits the board.

I think that the real-world counter to your theoretical "random + no-misses" is that anyone skilled enough to hit the board every time will not have a random distribution.

It's a theoretical problem. I could have just as easily asked the value of a point randomly chosen on a dartboard.

Quote: rdw4potus

I think that the real-world counter to your theoretical "random + no-misses" is that anyone skilled enough to hit the board every time will not have a random distribution.

a random distribution of throwing a pair of dice will yield more 7s than 12s.

Quote: s2dbaker

a random distribution of throwing a pair of dice will yield more 7s than 12s.

Sure, and some no-rolls when I miss the table:-)

Okay, I think I have a solution to the problem of defining the problem. Let's say that the dartboard has a radius of 10 inches. Every two and a half inches away from the bulls eye's center will be one standard deviation. Now calculate!

Edit- and every throw is aimed directly at the bulls eye

This reminds me of a question I like ... Suppose you have a circle in the plane centered at the origin with a 1 unit diameter (a dart board) ... suppose you throw darts, and you are guaranteed they will always land in the rectangular region (-1,-1) to (+1,+1). All you know is that the darts are thrown randomly and will land in that region ... what is the probability that you'll hit the dart board?

It's pretty easy, I know, but I like the answer. You can program a computer using this model and a RNG to compute the value of pi -- it's the only way I know to get pi using a computer simulation.

--Ms. D.

Quote: DorothyGale

It's pretty easy, I know, but I like the answer. You can program a computer using this model and a RNG to compute the value of pi -- it's the only way I know to get pi using a computer simulation.

Nothing special, this is Monte Carlo. The problem with this method is, first convergence is pretty bad.
Second problem is, you need a large supply of random numbers, which are far from being cheap. Of course you can use a pseudo-RNG, but those algorithms are biased - you will end up with something different than pi.

Usually you use Monte Carlo as the last resort, when analytical methods fail. A good example are estimating path integrals which you need in calculating quantum field theories. Contemporary calculus has only a very poor understanding of those integrals.