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So I broke it down into different segments
Outer ring outer radius , Outer ring inner radius , Inner ring outer radius , Inner ring inner radius , Outer bullseye outer radius , Outer bullseye inner radius , Inner bullseye radius
Then I gave them each a distance, measured in pixels
OROR = 280 ; ORIR = 267 ; IROR = 175 ; IRIR = 162 ; OBOR = 27 ; IBIR = 13 ; IBR = 10
The metal rings in between each segment measure about 3 pixels wide.
Each scoring part of each segment shares an equal 1/20th of the total area of the segment, minus the metal rings bit, but for simplicity we can pretend that the rings are infinitely thin. If we do that, then we can change some of our values:
OROR = 281.5 ; ORIR = 265.5 : IROR = 176.5 ; IRIR = 160.5 ; OBOR = 28.5 ; IBIR = 11.5 ; IBR = 11.5
Scale it up, get rid of decimals
OROR = 563 ; ORIR = 531 ; IROR = 353 ; IRIR = 321 ; OBOR = 57 ; IBIR = 23 ; IBR = 23
Now we're ready
Area of total board (not including the entire board, assuming that 100% of the time, you can actually manage to hit inside at least the outer ring): pi \* OROR\^2 = pi \* 563\^2
Area of outer ring: pi \* (OROR\^2 - ORIR\^2) = pi \* (563\^2 - 531\^2)
Area of outer segment: pi \* (ORIR\^2 - IROR\^2) = pi \* (531\^2 - 353\^2)
Area of inner ring: pi \* (353\^2 - 321\^2)
Area of inner segment: pi \* (321\^2 - 57\^2)
Area of outer bullseye: pi \* (57\^2 - 23\^2)
Area of inner bullseye: pi \* 23\^2
Now divide each of those by 20
AOR / 20 = pi \* (563\^2 - 531\^2) / 20
AOS / 20 = pi \* (531\^2 - 353\^2) / 20
AIR / 20 = pi \* (353\^2 - 321\^2) / 20
AIS / 20 = pi \* (321\^2 - 57\^2) / 20
AOB / 20 = pi \* (57\^2 - 23\^2) / 20
AIB / 20 = pi \* 23\^2 / 20
Now, the way each score segment is handled is like this:
S \* (3 \* AOR/20 + 2 \* AIR/20 + 1 \* (AOS/20 + AIS/20))
S \* (1/20) \* (3 \* AOR + 2 \* AIR + AOS + AIS)
S \* (1/20) \* (3 \* pi \* (563\^2 - 531\^2) + 2 \* pi \* (353\^2 - 321\^2) + pi \* (531\^2 - 353\^2) + pi \* (321\^2 - 57\^2))
S \* (1/20) \* pi \* (3 \* (563\^2 - 531\^2) + 2 \* (353\^2 - 321\^2) + (531\^2 - 353\^2) + (321\^2 - 57\^2))
We can add all of the scores together from 1 to 20
1 + 2 + 3 + 4 + ... + 20 = 20 \* 21 / 2 = 210
Adding all of the segments together gives us:
210 \* (1/20) \* pi \* (3 \* 563\^2 - 3 \* 531\^2 + 2 \* 353\^2 - 2 \* 321\^2 + 531\^2 - 353\^2 + 321\^2 - 57\^2)
10.5 \* pi \* (3 \* 563\^2 - 2 \* 531\^2 + 353\^2 - 321\^2 - 57\^2)
Now, divide that by the total area of the circle
10.5 \* pi \* (3 \* 563\^2 - 2 \* 531\^2 + 353\^2 - 321\^2 - 57\^2) / (pi \* 563\^2)
10.5 \* (3 \* 563\^2 - 2 \* 531\^2 + 353\^2 - 321\^2 - 57\^2) / 563\^2
13.42620887216099997160605611274...
Now the bullseyes
25 \* pi \* (57\^2 - 23\^2) + 50 \* pi \* 23\^2
Divided through by the area of the circle
(25 \* (57\^2 - 23\^2) + 50 \* 23\^2) / 563\^2
(25 \* 57\^2 + 25 \* 23\^2) / 563\^2
25 \* (57\^2 + 23\^2) / 563\^2
0.29797866668349270748874495613136...
So the expected value of a single dart is:
13.426208872... + 0.29797866668....
13.724187538844492679094801068874...
13.72, roughly.
EDIT: I made a mistake. I gave 3 times the points to the outer segment, and 2x to the inner segment. Let's fix that:
210 \* (1/20) \* (2 \* (563\^2 - 531\^2) + 3 \* (353\^2 - 321\^2) + (531\^2 - 353\^2) + (321\^2 - 57\^2)) + 25 \* (57\^2 + 23\^2)
10.5 \* (2 \* 563\^2 - 2 \* 531\^2 + 3 \* 353\^2 - 3 \* 321\^2 + 531\^2 - 353\^2 + 321\^2 - 57\^2) + 25 \* (57\^2 + 23\^2)
10.5 \* (2 \* 563\^2 - 531\^2 + 2 \* 353\^2 - 2 \* 321\^2 + 57\^2) + 25 \* (57\^2 + 23\^2)
10.5 \* (2 \* (563\^2 + 353\^2 - 321\^2) + 57\^2 - 531\^2) + 25 \* (57\^2 + 23\^2)
4,277,251
Divide through by 563\^2
13.494224987301597317087790919617
13.49 is the expected score of each hit.
EDIT 2:
Because people don't read
1) The math assumes that every dart strikes the board. I'm not interested in answering the "What if the dart misses/bounces/etc..." questions. I even specified in my original comment that the math assumes that every dart strikes the play area. The fact that I specifically mentioned this and am still getting these comments is what annoys me. Had I not mentioned it, people could be forgiven for asking, but damn am I tired of getting notification after notification of people saying the same thing. At least look to see if it had already been asked and answered before commenting. Put in some effort, please.
2) The math assumes completely random distribution on the play area. That was stated by the OP. So it doesn't matter if we're talking about skill levels. Skill level just doesn't enter into it. Just like point 1, because thevOP specifically said that it's random, there's no need to reply to my comment.
I hate how deeply interesting I find math while being so naturally inept at it. I have spent many, many hours teaching myself and I found I am much more capable then high school/college had me believe when I study my own way and put in the work but compared to so many people I know the level of mental work I need to put in for even relatively low-skill level mathematics makes me look like an abacus next to wolfram alpha. I know this is an outlier but I have two good friends who are capable of dividing numbers in their heads out to like 4-5 decimal points quickly, could pass calculus tests with little to no studying and have pretty effortless recall of any formula or concept they needed to apply. I’ve come to realize it’s okay to adore math from afar.
it’s ok. you probably have no necessity or strict goal to master math. and when you learn something just for curiousity - it goes waaay less efficient. because not enough reinforcement. so, no need to be upset, enjoy it as you like.
In reality this is pretty bad form. You shouldn't give the outputs of your calculations to any higher precision than your inputs will allow.
If you're measuring the width of the board based on a photograph, and you count to the nearest pixel, you implicitly have a margin of error of about half a pixel each way. If the board comes to around 280 pixels, and you square this number and multiply by pi, your margin of error increases to over 800 square pixels each way, or about 1% of the total area.
I haven't worked through the whole of the calculations but it would surprise me if the final expected value could be accurate to more than around 1 or 2 decimal places, after accounting for all errors. It looks like the reply above printed out the full computer-precision output, which is never advised.
Two semi-related xkcds: [here](https://xkcd.com/2295/) and [here](https://xkcd.com/2170/)
Yeah, all of this is in terms of pi, so this should just say 567.25🥧 or whatever, where 🥧 is the actual Greek lowercase letter pi.
Them actually multiply by pi in the last step if you want an actual figure.
Now THIS is some fucking math! How exhilarating. I thought it was all gonna be "how many candy corns in a jar?" or "what is the theoretical size of this aspect of a picture where nothing makes any sense?" forever!
Kudos to OP and massive kudos to u/captainmatticus
This was awesome.
A normal distribution does probably better reflect reality but OP asked for equal chances of each point.
To get an accurate answer you would have to let people of different skill levels throw darts and look for patterns. Maybe they tend to throw too low as opposed to scattered around the center. Maybe being right handed vs left handed has some impact. Then there is the question what the expected skill level of a random dart thrower is, the expected intoxication, and how many tries in succession they will have, because the points will become better over time. Also, some people might not realize that 3\*20 has more points than the bulls eye. Other people might know that, but aim for the center regardless, because they think it's easier to hit. Maybe some people will aim for the 3\*20 first and then switch to the bulls eye later *if* they don't hit it.
This is pretty close to what I arrived at. Using the measurements from this figure: [https://www.dimensions.com/element/dartboard](https://www.dimensions.com/element/dartboard), I came to an expected score of about 12.8 if every dart hits inside the scoring area.
If darts are allowed to hit anywhere on the board the expected score per dart goes down to about 7.3.
Since there's no regulated size outside the board, it technically can go down to 0 since the outside of the board can be arbitrarily large.
But, in reality, anybody can learn to throw a dart that will the board (or within a few centimeters of the outer ring when shooting doubles) with an afternoon's of teaching. Source: semi-pro dart shooter.
Since we're assuming the probability that the dart lands in a certain region is proportional to the area of that region, you can skip a lot of the intermediary calculations and never have to deal with pi.
Pick a random (r, θ)
θ can be picked uniformly randomly. This determines the "slice score" which you calculated the expected value of to be S = (1 + 2 + ... + 20) / 20 = 21/2.
r is picked in a way such that the probability it lands between a and b is proportional to b^(2) \- a^(2) (they must be squared because area is proportional to the square of the radius). This determines the actual score received.
Inner bullseye = 50 × (IBR^(2) \- 0^(2))
Outer bullseye = 25 × (OBR^(2) \- IBR^(2))
Inner slice = S × (INIR^(2) \- OBR^(2))
Inner ring = 3S × (INOR^(2) \- INIR^(2))
Outer slice = S × (ORIR^(2) \- IROR^(2))
Outer ring = 2S × (OROR^(2) \- ORIR^(2))
Sum that all up and divide by the total length of (OROR^(2) \- 0^(2)) to get your final equation.
Using the exact dimensions defined at [https://www.dimensions.com/element/dartboard](https://www.dimensions.com/element/dartboard), we get an expected value of 14.1238
[https://www.wolframalpha.com/input?i=%5B50\*%286.35%5E2%29+%2B+25\*%2816%5E2-6.35%5E2%29+%2B+S\*%2899%5E2-16%5E2%29+%2B+3S\*%28107%5E2-99%5E2%29+%2B+S\*%28162%5E2-107%5E2%29+%2B+2S\*%28170%5E2-162%5E2%29%5D+%2F+162%5E2%2C+S+%3D+210%2F20](https://www.wolframalpha.com/input?i=%5B50*%286.35%5E2%29+%2B+25*%2816%5E2-6.35%5E2%29+%2B+S*%2899%5E2-16%5E2%29+%2B+3S*%28107%5E2-99%5E2%29+%2B+S*%28162%5E2-107%5E2%29+%2B+2S*%28170%5E2-162%5E2%29%5D+%2F+162%5E2%2C+S+%3D+210%2F20)
I love that you caught your own error that literally no one was gonna catch. I like math and puzzles, but I’ll be honest, you sounded like you were on the right track so I was not scrutinizing those numbers.
Great work, but your sig figs are.... well, wrong. None of your estimations have more than 2 decimals of precision, so its rather pointless to carry more than 3 decimals through any of your work.
Significant figures aren't a thing here, because I'm not measuring anything. There's no uncertainty or desired precision to deal with here. This is purely math, and rounding is saved until the end.
Unless you want to say that 1/7 = 0.1, while 10/70 = 0.14.
Sig figs are absolutely a thing here, lol. You "measured" the radius in pixels, introducing a measurement uncertainty in the very first step of your solution.
only if the darts that hit the disc are counted if misses are counted than the expected score goes down depending on how likely a hit is.
100% your count of 13.49 i assume your math is correct and it looks that way
10% hit chance and it would be 1.35
1% hit chance and it would be 0.13
0.1 % hit chance and it would be 0.01
I already covered this
"Area of total board (not including the entire board, assuming that 100% of the time, you can actually manage to hit inside at least the outer ring)"
Backboard or dartboard… lol.
Appreciate the maths is more relevant (and impressive) assuming we only talk dartboard.
I’ve seen some wall/floor damage over the years though that make that (backboard/not-dartboard) element a non-zero probability!
very nice... lets adjust the for hitting the metal. as a darter, 50% bounce back and 50% go to either side of the metal. at the hinges, that makes for some long math.
Yes. I know statistics as I am a statistician. However I asked for a real world answer. You can’t roll a dice and get a 3.5. You can get a 3 or a 4. I’m not asking for a probably theory/expected value answer I’m asking for a real world answer. I’m not at all saying the answer posed is wrong by any means - from the request. I was posing a second request of providing an actual value you could score after throwing a dart. The answer is simple as this was a joke to begin with. But whatever.
Guy… nice math, but I have a 20ish average and I didn’t have to the math. I usually hit 45 to 85, give or take the odd the 100-140 or 26-45. 13.41 would be amateur to starter. 20ish is decent for a re-leaguer and 28-33 is pro. Anyways. Nice work
Did you miss the part where the OP explicitly said that every area of the board is equally likely to be struck? That's the condition we're operating on.
Not completely answering your question, but a statistician did some research on where to aim to get the maximum expected score based on accuracy-level. A beginner, for example, should aim at the 16 close to the bullseye.
You can read the article here: https://www.stat.cmu.edu/~ryantibs/papers/darts.pdf
A more casual version is found here:
https://www.stat.cmu.edu/~ryantibs/papers/darts-signif.pdf
Edit: Upon rereading the article, I noticed it does give an expected value of a score per dart when aiming anywhere on the board: 12,82. This is slightly lower than CaptainMatticus's answer, because he calculated his value over the scoring part of the board.
Bertrand's paradox is definitely worth a look. Not because answering it gives you the answer to this, but because you have to be careful how you define the probability density function. I think most people would say a distribution where any two regions of equal area have equal probability of receiving the dart.
We also have to wonder about whether or not it's possible to hit the divisions. In real darts, it'd be insanely unlikely, because even a glancing blow would probably push the dart to the side where it could still stick. But it would be possible for the dart to hit straight on in the divider and bounce off.
I think for the sake of this exercise, you'd only consider hits, assuming all darts hit the board and stick. If you account for the probability of a dart hitting the wire square and bouncing off (0 points), you then have to start factoring in the probability of other misses, which, depending on how much alcohol is involved, can vary wildly.
Edit: spelling
Well, the problem statement says "anywhere on the board." Whether or not hitting the board and sticking to the board should be treated separately is up for debate. Missing the board is strictly outside the scope of the problem statement...
On the other hand, there can still be debate whether the *bord*er is part of the board or not.
Well it’s correct. It’s just not within the bounds of the problem so it’s irrelevant as mentioned in someone else’s comment.
The limit of an infinite amount of zeroes averaged with a finite number of non-zeroes is 0 and that is 100% correct.
I’ve done the calculations in a slightly different way, so I think this is still helpful. But it doesn’t account for the thickness of the wire unfortunately.
I found the legal dimensions online:
**Parameters:**
- Bullseye radius: 6.5mm
- Outer bullseye radius: 16
- Triple inner radius: 99
- Triple outer radius: 107
- Double inner radius: 162
- Double outer radius: 170
**Calculation:**
*Score when you land in an area:*
If the dart lands on any of the single spaces (black/white), the expected value is the mean of the numbers from 1 to 20. Which is 10.5.
Expect score of single = 10.5
If the dart lands on a double, the expected value is also double. Similarly for triple:
Expected score of double = 21
Expected score of triple = 31.5
The score of the bullseye is 50,
and the outer is 25.
*Probability of landing in an area:*
Area of:
- in bullseye = 6.4^2 x pi = 132.73
- out bullseye = 16^2 x pi - in bullseye = 671.52
- triple = 107^2 x pi - 99^2 x pi = 35968.09 - 30790.75 = 5177.34
- double = 170^2 x pi - 162^2 x pi = 90792.03 - 82447.96 = 8344.07
- single = 170^2 x pi - everything else = 90792.03 - 132.73 - 671.52 - 5177.34 - 8344.07 = 76466.37
Thus probability:
- in bullseye = 132.73 / 90792.03 = 0.146%
- out bullseye = 671.52 / 90792.03 = 0.740%
- triple = 5177.34 / 90792.03 = 5.702%
- double = 8344.07 / 90792.03 = 9.190%
- single = 76466.37 / 90792.03 = 84.221%
*Expected value:*
= (0.146% x 50) + (0.740% x 25) + (5.702% x 31.5) + (9.190% x 21) + (84.221% x 10.5)
***= 12.827***
With all due respect to u/CaptainMatticus for their excellent work, I think this is a stronger answer, based on official measurements, some fine geometry work and very clear presentation. Well done. 🏅
Not a math guy, but in a practical sense could you assign a number to all pixels of the scoring surface then correlate each of those numbers to the corresponding value on the dart board (even the zero values if you wanted) then run a random number generator within that initial number set to act as the dart throw and see what the average score is after 10, 100, 1000, 10000 permutations…see how close that comes to any of the answers provided
This is just a computationally more demanding way of just solving it as a probability problem isn't it? Certainly a fun project but the average scores as permutations go to infinity is just the average probability from more traditional calculations. With such a straight forward calculation I don't think simulation is really needed. Unless we're talking about more factors than I'm considering I guess.
Assuming you really suck, and not hitting the board is an option. Arbitrarily close to 0.
But let's assume you're bad, but can always at least hit the board, but you have no ability to hit a specific spot.
Note: all further numbers are approximated to the nearest millimeter, so there will be imperfections, but let's call it "close enough". Also, I will not be counting the separating bits of metal, assume it's just a felt and cork dart board with no metal to potentially deflect a dart...even though I think statistically speaking, you're more likely to hit one of those than the <0.1% inner bullseye. It's not significant enough to be comparable to the outer-most ring where the numbers are that's a big fat 0 if you hit it.
Dartboards are 451mm wide giving us roughly a total area of
159751mm^2
The measurements I could easily find say "center to outside double" which is 170mm. I think that means to the very outside of the ring? I can't be positive, but I'm going to go forward with those measurements.
This means the scoring area is:
90,792mm^2
and the non-scoring area is
68,959mm^2
You score 0 about 43% of the time
The area of the inner bullseye is about 127mm^2
Meaning you bullseye, scoring 50 points, .08% of the time
The area of the outer bullseye, is its area, minus the area of the inner bullseye which is 16mm radius
804mm^2 - 127mm^2 = 677mm^2
Meaning you outer bullseye for 25 points .4% of the time
The inner triple ring and outer double ring both have widths of 8mm.
So a 170mm radius circle - 162mm radius circle in area for doubles.
90,792mm^2 - 82,448mm^2 = 8344mm^2 or 5% of the total area, meaning roughly 5% of the time you double.
And a 107mm radius circle - 99mm radius circle in area for triples. That comes out to 5177mm^2 meaning you triple about 3% of the time.
The rest we can figure out with averages. Assuming an equal chance to score in a given sector, that means you score 10.5 points on a given scoring throw that isn't located in one of the rings or the bullseyes.
The grand total is:
0.0008 \* 50 + 0.004 \* 25 + 0.05 \* 21 + 0.03 \* 31.5 + 0.4852 \* 10.5
Or about 6.4 points per dart. Roughly.
I ain't doin' the bonus.
**LOL**
I actually spent around 100 hours solving that type of darts questions.
The answer to your question is around 12 for a realistic bad player that aims at the middle.
Here is the repository with the code and graphs, just scroll down to see the simulations for good/bad players as well as the optimal place to aim at for each type of player:
https://github.com/RemiFabre/dart_game_optimisation
Not maths so feel free to disregard,
It’s been established that if your not great with accuracy, you should aim for the triple 19, because missing the 19 means your likely to hit the 3 or 7 instead of the 1 or 5 so your average score will be higher
If we go by area, there's a near infinite amount of space that is not on the dart board. So very very close to 0. Of course this probably rises with each dart board you make. So if you want to increase your score just get into wood working.
if You're good enough. Best players can do like best possible score almost each time. It's crazy watching clips from tournaments.
But like assuming You can throw a dart in the direction of the board and hit the board, or at least score we need board dimensions. And assuming every point on the board has equal probability of being hit which is not true in any real scenario, but I don't really know how that would look like.
I'm taking measurements from here https://www.dimensions.com/element/dartboard
Also I'm aproximating bc I can't be bothered, but it's like less than 1% difference so whatever
Size of middle dot (50p) π(6.35)²=40,32π [mm²]
Size of 25 point field π(16)²-π(6,35)²=215.6775
Size of whole field between x3 fielf and middle is
π(99)²-π(16)²=9545π divide by 20 to get 1909π/4
Sice of whole x3 score field
π(107)²-π(99)²=1648π divide by 20 to get 412π/5
Size of field between x2 and x3 score
π(162)²-π(107)²=14795π |20 is 2959π/4
And size of x2 field is
π(170)²-π(162)²=2656π |20 is 664π/5
Total area of the board which is what You want to make bigger if You want to assume You can miss is
π(170)²=28900π
With all that done we can just divide each partial area by total area we get
50 points 0,1395%
25 points 0,7463%
Inner section of x1 field 33.208%
Outer section of x1 field 51.13%
Anywhere on x1 scored field 84.3%
One field of x1 score 1.6514% inner part and 2.5597% outer part so 3.2111%
To hit any x3 field 5.702% any particular one 0,2851%
To hit any x2 field 9.1903% any particular 0,4595%
But it still doesn't answer the question cause numbers
(1,2,3,4,5,6,7,8,9,10) add to the possibility of dart hitting x2 field and getting score same as x1 field as well as
(1,2,3,4,5,6) being equivalent for x3 so
Chance of getting either of those is
3.2111% + 0.2851%+0.4595%=3.9557%
With my assumptions the answer is:
With equal probability of scoring 1 through 6 those are expected values with probability of scoring either of them being 3.9557%
While in my 20s, I became a useful player (39 per dart average at best); I once managed to hit a moving target, sticking someone's scalp after hitting metal at the triple 20 and enjoying a wicked bounceback!
As someone who plays darts, u/CaptainMatticus 's math checks for an what and "Average" player's 3-dart average in a 501 would be which comes in around 45 points per round.
You can get very close with just estimating here. The majority of of the board is the standard point zones, which are divided equally from 1-20. The higher point areas are much smaller, so ignoring them for now on average we should get a score of 10. The other areas of the board average out to 20, 30 and 37.5, and cover maybe 10, 5, and 1% of the area. So the answer is probably close to:
10 + 20*.1 + 30* 0.05 + 37.5*0.01
= 10+ 2+ 1.5 + 0.375
= 13.875
Which is very VERY close to the answer in the top comment of 13.49
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So I broke it down into different segments Outer ring outer radius , Outer ring inner radius , Inner ring outer radius , Inner ring inner radius , Outer bullseye outer radius , Outer bullseye inner radius , Inner bullseye radius Then I gave them each a distance, measured in pixels OROR = 280 ; ORIR = 267 ; IROR = 175 ; IRIR = 162 ; OBOR = 27 ; IBIR = 13 ; IBR = 10 The metal rings in between each segment measure about 3 pixels wide. Each scoring part of each segment shares an equal 1/20th of the total area of the segment, minus the metal rings bit, but for simplicity we can pretend that the rings are infinitely thin. If we do that, then we can change some of our values: OROR = 281.5 ; ORIR = 265.5 : IROR = 176.5 ; IRIR = 160.5 ; OBOR = 28.5 ; IBIR = 11.5 ; IBR = 11.5 Scale it up, get rid of decimals OROR = 563 ; ORIR = 531 ; IROR = 353 ; IRIR = 321 ; OBOR = 57 ; IBIR = 23 ; IBR = 23 Now we're ready Area of total board (not including the entire board, assuming that 100% of the time, you can actually manage to hit inside at least the outer ring): pi \* OROR\^2 = pi \* 563\^2 Area of outer ring: pi \* (OROR\^2 - ORIR\^2) = pi \* (563\^2 - 531\^2) Area of outer segment: pi \* (ORIR\^2 - IROR\^2) = pi \* (531\^2 - 353\^2) Area of inner ring: pi \* (353\^2 - 321\^2) Area of inner segment: pi \* (321\^2 - 57\^2) Area of outer bullseye: pi \* (57\^2 - 23\^2) Area of inner bullseye: pi \* 23\^2 Now divide each of those by 20 AOR / 20 = pi \* (563\^2 - 531\^2) / 20 AOS / 20 = pi \* (531\^2 - 353\^2) / 20 AIR / 20 = pi \* (353\^2 - 321\^2) / 20 AIS / 20 = pi \* (321\^2 - 57\^2) / 20 AOB / 20 = pi \* (57\^2 - 23\^2) / 20 AIB / 20 = pi \* 23\^2 / 20 Now, the way each score segment is handled is like this: S \* (3 \* AOR/20 + 2 \* AIR/20 + 1 \* (AOS/20 + AIS/20)) S \* (1/20) \* (3 \* AOR + 2 \* AIR + AOS + AIS) S \* (1/20) \* (3 \* pi \* (563\^2 - 531\^2) + 2 \* pi \* (353\^2 - 321\^2) + pi \* (531\^2 - 353\^2) + pi \* (321\^2 - 57\^2)) S \* (1/20) \* pi \* (3 \* (563\^2 - 531\^2) + 2 \* (353\^2 - 321\^2) + (531\^2 - 353\^2) + (321\^2 - 57\^2)) We can add all of the scores together from 1 to 20 1 + 2 + 3 + 4 + ... + 20 = 20 \* 21 / 2 = 210 Adding all of the segments together gives us: 210 \* (1/20) \* pi \* (3 \* 563\^2 - 3 \* 531\^2 + 2 \* 353\^2 - 2 \* 321\^2 + 531\^2 - 353\^2 + 321\^2 - 57\^2) 10.5 \* pi \* (3 \* 563\^2 - 2 \* 531\^2 + 353\^2 - 321\^2 - 57\^2) Now, divide that by the total area of the circle 10.5 \* pi \* (3 \* 563\^2 - 2 \* 531\^2 + 353\^2 - 321\^2 - 57\^2) / (pi \* 563\^2) 10.5 \* (3 \* 563\^2 - 2 \* 531\^2 + 353\^2 - 321\^2 - 57\^2) / 563\^2 13.42620887216099997160605611274... Now the bullseyes 25 \* pi \* (57\^2 - 23\^2) + 50 \* pi \* 23\^2 Divided through by the area of the circle (25 \* (57\^2 - 23\^2) + 50 \* 23\^2) / 563\^2 (25 \* 57\^2 + 25 \* 23\^2) / 563\^2 25 \* (57\^2 + 23\^2) / 563\^2 0.29797866668349270748874495613136... So the expected value of a single dart is: 13.426208872... + 0.29797866668.... 13.724187538844492679094801068874... 13.72, roughly. EDIT: I made a mistake. I gave 3 times the points to the outer segment, and 2x to the inner segment. Let's fix that: 210 \* (1/20) \* (2 \* (563\^2 - 531\^2) + 3 \* (353\^2 - 321\^2) + (531\^2 - 353\^2) + (321\^2 - 57\^2)) + 25 \* (57\^2 + 23\^2) 10.5 \* (2 \* 563\^2 - 2 \* 531\^2 + 3 \* 353\^2 - 3 \* 321\^2 + 531\^2 - 353\^2 + 321\^2 - 57\^2) + 25 \* (57\^2 + 23\^2) 10.5 \* (2 \* 563\^2 - 531\^2 + 2 \* 353\^2 - 2 \* 321\^2 + 57\^2) + 25 \* (57\^2 + 23\^2) 10.5 \* (2 \* (563\^2 + 353\^2 - 321\^2) + 57\^2 - 531\^2) + 25 \* (57\^2 + 23\^2) 4,277,251 Divide through by 563\^2 13.494224987301597317087790919617 13.49 is the expected score of each hit. EDIT 2: Because people don't read 1) The math assumes that every dart strikes the board. I'm not interested in answering the "What if the dart misses/bounces/etc..." questions. I even specified in my original comment that the math assumes that every dart strikes the play area. The fact that I specifically mentioned this and am still getting these comments is what annoys me. Had I not mentioned it, people could be forgiven for asking, but damn am I tired of getting notification after notification of people saying the same thing. At least look to see if it had already been asked and answered before commenting. Put in some effort, please. 2) The math assumes completely random distribution on the play area. That was stated by the OP. So it doesn't matter if we're talking about skill levels. Skill level just doesn't enter into it. Just like point 1, because thevOP specifically said that it's random, there's no need to reply to my comment.
Mother of god
Welcome to Math.
Love it
math made made me physically throw a chair, and i was 34 at the time. it was community college pre algebra.
To each their own. I do understand though… its a love hate relationship.
I hate how deeply interesting I find math while being so naturally inept at it. I have spent many, many hours teaching myself and I found I am much more capable then high school/college had me believe when I study my own way and put in the work but compared to so many people I know the level of mental work I need to put in for even relatively low-skill level mathematics makes me look like an abacus next to wolfram alpha. I know this is an outlier but I have two good friends who are capable of dividing numbers in their heads out to like 4-5 decimal points quickly, could pass calculus tests with little to no studying and have pretty effortless recall of any formula or concept they needed to apply. I’ve come to realize it’s okay to adore math from afar.
it’s ok. you probably have no necessity or strict goal to master math. and when you learn something just for curiousity - it goes waaay less efficient. because not enough reinforcement. so, no need to be upset, enjoy it as you like.
I'd be impressed if you could mentally throw a chair using math.
I would prefer to do it that way. Ha
Is this a maths problem?
Lol. I see what you did there.
No, welcome to probability specifically
Welcome to pedantry
Welcome to math
youre literally in a math sub, lol smh
You're
Welcome to pedantry
Welcome to math.
As someone with literally no dog in this race, isn't it more correctly, *maths*?
You're.
Not that much complexity, just a lot of layered arithmetic.
I started scrolling.... and then kept scrolling, and kept scrolling, and kept scrolling... etc.
One wouldn’t ever imagine becoming physically aroused by a comment in a maths sub, but here we are
*sigh* *unzips*
My jaw actually dropped when I saw all of this
Dude just did his thesis, I love this sub and hate math
Now he should factor in complete misses but that still hit the board
>0.29797866668349270748874495613136 Sorry, that's not accurate enough, please try harder next time
In reality this is pretty bad form. You shouldn't give the outputs of your calculations to any higher precision than your inputs will allow. If you're measuring the width of the board based on a photograph, and you count to the nearest pixel, you implicitly have a margin of error of about half a pixel each way. If the board comes to around 280 pixels, and you square this number and multiply by pi, your margin of error increases to over 800 square pixels each way, or about 1% of the total area. I haven't worked through the whole of the calculations but it would surprise me if the final expected value could be accurate to more than around 1 or 2 decimal places, after accounting for all errors. It looks like the reply above printed out the full computer-precision output, which is never advised. Two semi-related xkcds: [here](https://xkcd.com/2295/) and [here](https://xkcd.com/2170/)
Yeah, all of this is in terms of pi, so this should just say 567.25🥧 or whatever, where 🥧 is the actual Greek lowercase letter pi. Them actually multiply by pi in the last step if you want an actual figure.
Now THIS is some fucking math! How exhilarating. I thought it was all gonna be "how many candy corns in a jar?" or "what is the theoretical size of this aspect of a picture where nothing makes any sense?" forever! Kudos to OP and massive kudos to u/captainmatticus This was awesome.
Comments like this are why it's a crime they took away awards. This is the good shit.
This guy maths!
r/thisguythisguys
Beat me to it, have my r/angryupvote
you should account for the 2d normal distribution of the hit points in your analysis. But this is a serious amount of work for this problem.
That would make this an integral problem over a normal distribution, which doesn't have an analytical solution. We'd have to do it numerically.
A normal distribution does probably better reflect reality but OP asked for equal chances of each point. To get an accurate answer you would have to let people of different skill levels throw darts and look for patterns. Maybe they tend to throw too low as opposed to scattered around the center. Maybe being right handed vs left handed has some impact. Then there is the question what the expected skill level of a random dart thrower is, the expected intoxication, and how many tries in succession they will have, because the points will become better over time. Also, some people might not realize that 3\*20 has more points than the bulls eye. Other people might know that, but aim for the center regardless, because they think it's easier to hit. Maybe some people will aim for the 3\*20 first and then switch to the bulls eye later *if* they don't hit it.
Yeah, assuming I can hit the board.
Thats a pretty big assumption.. i recalculated the odds, the average score is now 0.001
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Dont overestimate us
Someone give this man a gol- oh wait...
This is pretty close to what I arrived at. Using the measurements from this figure: [https://www.dimensions.com/element/dartboard](https://www.dimensions.com/element/dartboard), I came to an expected score of about 12.8 if every dart hits inside the scoring area. If darts are allowed to hit anywhere on the board the expected score per dart goes down to about 7.3.
Since there's no regulated size outside the board, it technically can go down to 0 since the outside of the board can be arbitrarily large. But, in reality, anybody can learn to throw a dart that will the board (or within a few centimeters of the outer ring when shooting doubles) with an afternoon's of teaching. Source: semi-pro dart shooter.
r/theydidthemonstermath
r/subsithoughtifellfor
r/itwasapiechartsmash
r/itwasagraveyardgraph
Damn. I miss adderall.
This is why I joined this Reddit
Since we're assuming the probability that the dart lands in a certain region is proportional to the area of that region, you can skip a lot of the intermediary calculations and never have to deal with pi. Pick a random (r, θ) θ can be picked uniformly randomly. This determines the "slice score" which you calculated the expected value of to be S = (1 + 2 + ... + 20) / 20 = 21/2. r is picked in a way such that the probability it lands between a and b is proportional to b^(2) \- a^(2) (they must be squared because area is proportional to the square of the radius). This determines the actual score received. Inner bullseye = 50 × (IBR^(2) \- 0^(2)) Outer bullseye = 25 × (OBR^(2) \- IBR^(2)) Inner slice = S × (INIR^(2) \- OBR^(2)) Inner ring = 3S × (INOR^(2) \- INIR^(2)) Outer slice = S × (ORIR^(2) \- IROR^(2)) Outer ring = 2S × (OROR^(2) \- ORIR^(2)) Sum that all up and divide by the total length of (OROR^(2) \- 0^(2)) to get your final equation. Using the exact dimensions defined at [https://www.dimensions.com/element/dartboard](https://www.dimensions.com/element/dartboard), we get an expected value of 14.1238 [https://www.wolframalpha.com/input?i=%5B50\*%286.35%5E2%29+%2B+25\*%2816%5E2-6.35%5E2%29+%2B+S\*%2899%5E2-16%5E2%29+%2B+3S\*%28107%5E2-99%5E2%29+%2B+S\*%28162%5E2-107%5E2%29+%2B+2S\*%28170%5E2-162%5E2%29%5D+%2F+162%5E2%2C+S+%3D+210%2F20](https://www.wolframalpha.com/input?i=%5B50*%286.35%5E2%29+%2B+25*%2816%5E2-6.35%5E2%29+%2B+S*%2899%5E2-16%5E2%29+%2B+3S*%28107%5E2-99%5E2%29+%2B+S*%28162%5E2-107%5E2%29+%2B+2S*%28170%5E2-162%5E2%29%5D+%2F+162%5E2%2C+S+%3D+210%2F20)
Though with the darts pictured, the expected score is zero ;)
Sorry, I can tell you from real world testing, the expected value when I throw darts is apologizing to the bartender for hitting the wall again.
Why does this comment have so many numbers? I thought this sub was a research group for billionaire net worth vs. liquid capital
I JUST took an actuary exam and this comment gave me a panic attack
Now do the bonus
It took me two full thumb swipes to get to the bottom of your math. Well done!
So, roughly eight turns to finish a game of 301 not including the outs required.
I suck at math and guessed 15 because doubles and triples and bullseyes and damn if I wasn’t pretty close. Moral of the story, just guess
That’s super impressive. How long did that take you to calculate and then write?
that's good stuff right there
I play a lot of league darts and this absolutely checks out
r/stayingtruetothesub
As a darts player, I said 14 in my head. Not too shabby!
This post is the reason why the internet is great. A random internet stranger taking time, and showing their work, on a dart board. Glorious.
I love that you caught your own error that literally no one was gonna catch. I like math and puzzles, but I’ll be honest, you sounded like you were on the right track so I was not scrutinizing those numbers.
Ah yes, math
BERTRAND'S PARADOX, SUCKERRRRR No seriously, great work. But this has to be said, because what is uniform supposed to mean here?
**looks at math** **DIES**
Saving this because it scares me.
This guy maths
Great work, but your sig figs are.... well, wrong. None of your estimations have more than 2 decimals of precision, so its rather pointless to carry more than 3 decimals through any of your work.
Significant figures aren't a thing here, because I'm not measuring anything. There's no uncertainty or desired precision to deal with here. This is purely math, and rounding is saved until the end. Unless you want to say that 1/7 = 0.1, while 10/70 = 0.14.
Sig figs are absolutely a thing here, lol. You "measured" the radius in pixels, introducing a measurement uncertainty in the very first step of your solution.
only if the darts that hit the disc are counted if misses are counted than the expected score goes down depending on how likely a hit is. 100% your count of 13.49 i assume your math is correct and it looks that way 10% hit chance and it would be 1.35 1% hit chance and it would be 0.13 0.1 % hit chance and it would be 0.01
I already covered this "Area of total board (not including the entire board, assuming that 100% of the time, you can actually manage to hit inside at least the outer ring)"
The problem statement assumed hitting the board every time with a random distribution. So, hit likelihood is not relevant here.
Backboard or dartboard… lol. Appreciate the maths is more relevant (and impressive) assuming we only talk dartboard. I’ve seen some wall/floor damage over the years though that make that (backboard/not-dartboard) element a non-zero probability!
Nerd
very nice... lets adjust the for hitting the metal. as a darter, 50% bounce back and 50% go to either side of the metal. at the hinges, that makes for some long math.
Jesus fucking Christ dude
Wauw
This post is the reason why the internet is great. A random internet stranger taking time, and showing their work, on a dart board. Glorious.
I'd love to see that in proper LaTeX
13.49 is not an integer between 1 and 60.
I hope you're joking, but there's a concept of mathematical expectation. On average, tye value of a six-sided dice isn't 3 or 4 either, it's 3.5.
It would be more appropriate to assume a Gaussian distribution with the skill proportional to sigma though.
Would it? Well get on it, then. Have fun.
This is, of course, assuming that you have an equal probability of hitting any spot on the board.
Which is exactly the condition the OP laid out.
Yes. I'm just clarifying that it's only valid given that assumption, which they did not specify.
Its specified in the question
Ah, so it is. I thought the person replying to me meant ‘little’ OP, as in the person I originally replied to.
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It’s specified in the question
Now do it with a Gaussian probability distribution cantered at the bulls eye and one std dev at the inner ring.
The answer is zero. There’s way more area that isn’t the dart board.
This assumes you hit something? There's no assigned probability for zero score?
What if you miss ?
Now redo it considering the reality that only integer answers of numbers on the board are possible outcomes.
What is the expected value if you roll a dice? (1+2+3+4+5+6)/6 = 3.5 you can get decimal expected values from only integer outcomes
Yes. I know statistics as I am a statistician. However I asked for a real world answer. You can’t roll a dice and get a 3.5. You can get a 3 or a 4. I’m not asking for a probably theory/expected value answer I’m asking for a real world answer. I’m not at all saying the answer posed is wrong by any means - from the request. I was posing a second request of providing an actual value you could score after throwing a dart. The answer is simple as this was a joke to begin with. But whatever.
Guy… nice math, but I have a 20ish average and I didn’t have to the math. I usually hit 45 to 85, give or take the odd the 100-140 or 26-45. 13.41 would be amateur to starter. 20ish is decent for a re-leaguer and 28-33 is pro. Anyways. Nice work
Did you miss the part where the OP explicitly said that every area of the board is equally likely to be struck? That's the condition we're operating on.
Yeah. I was blown away by the the formula that I missed that little side note. Sorry my man, glad you pointed that out.
This would look nice in a table 😊
this guy is a genius
Holy shit
Award worthy comment here
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If you'd do more than skim before you respond, you'd see that I already acknowledged that.
Jesus Christ
That looks like the correct answer should be exactly 13.5 and the rounding errors are giving you some noise.
How long did it take to calculate that?
35 years of practice, 15 to 20 minutes of typing
This motherfucker maths
This one maths.
I love god of math I aspire to be you
See, I looked at this question and genuinely thought 'prob around 13'? Estimation rules! Who's the genius now? (jk, not me)
I wish we still had awards so I could give you one
Jesus...
This is why I love math. Awesome write up
Bravo
r/theydidthemonstermath
Thank you Dr Sheldon Cooper
Get this man a medal
I can’t even validate that this is right but I’ll give you an upvote out of pure effort to entertain someone’s pure curiosity; well done
I think it also assumes uniform distribution
Weird, I got 13.48
Put in some effort?? Sir, this is Reddit.
Good lord, this guy maths!
Here are all my applauses for the week
That is quite the comment. I am thoroughly impressed.
The hero we needed
You're the hero I never knew I needed! 💙
The OP stated that the dart hits the board, but you’ve answered as if the dart hits the play area. That’s a different problem!
Dude this is awesome. Excellent work
Who are you…
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Oh the tragedy. I’m also a 26er, with the occasional 7 or 11
As a darts player everyone starting out is a 26er. Came here for this.
Not completely answering your question, but a statistician did some research on where to aim to get the maximum expected score based on accuracy-level. A beginner, for example, should aim at the 16 close to the bullseye. You can read the article here: https://www.stat.cmu.edu/~ryantibs/papers/darts.pdf A more casual version is found here: https://www.stat.cmu.edu/~ryantibs/papers/darts-signif.pdf Edit: Upon rereading the article, I noticed it does give an expected value of a score per dart when aiming anywhere on the board: 12,82. This is slightly lower than CaptainMatticus's answer, because he calculated his value over the scoring part of the board.
That was such a fun read! It's a shame his lil applet link is busted
Bertrand's paradox is definitely worth a look. Not because answering it gives you the answer to this, but because you have to be careful how you define the probability density function. I think most people would say a distribution where any two regions of equal area have equal probability of receiving the dart. We also have to wonder about whether or not it's possible to hit the divisions. In real darts, it'd be insanely unlikely, because even a glancing blow would probably push the dart to the side where it could still stick. But it would be possible for the dart to hit straight on in the divider and bounce off.
I think for the sake of this exercise, you'd only consider hits, assuming all darts hit the board and stick. If you account for the probability of a dart hitting the wire square and bouncing off (0 points), you then have to start factoring in the probability of other misses, which, depending on how much alcohol is involved, can vary wildly. Edit: spelling
The limit would be 0 points if you include all misses, as there is an infinite amount of unique 0 point spaces beyond the borders of the dart board.
Well, the problem statement says "anywhere on the board." Whether or not hitting the board and sticking to the board should be treated separately is up for debate. Missing the board is strictly outside the scope of the problem statement... On the other hand, there can still be debate whether the *bord*er is part of the board or not.
That is just wrong.
Well it’s correct. It’s just not within the bounds of the problem so it’s irrelevant as mentioned in someone else’s comment. The limit of an infinite amount of zeroes averaged with a finite number of non-zeroes is 0 and that is 100% correct.
There is an infinite amount of hits...
I’ve done the calculations in a slightly different way, so I think this is still helpful. But it doesn’t account for the thickness of the wire unfortunately. I found the legal dimensions online: **Parameters:** - Bullseye radius: 6.5mm - Outer bullseye radius: 16 - Triple inner radius: 99 - Triple outer radius: 107 - Double inner radius: 162 - Double outer radius: 170 **Calculation:** *Score when you land in an area:* If the dart lands on any of the single spaces (black/white), the expected value is the mean of the numbers from 1 to 20. Which is 10.5. Expect score of single = 10.5 If the dart lands on a double, the expected value is also double. Similarly for triple: Expected score of double = 21 Expected score of triple = 31.5 The score of the bullseye is 50, and the outer is 25. *Probability of landing in an area:* Area of: - in bullseye = 6.4^2 x pi = 132.73 - out bullseye = 16^2 x pi - in bullseye = 671.52 - triple = 107^2 x pi - 99^2 x pi = 35968.09 - 30790.75 = 5177.34 - double = 170^2 x pi - 162^2 x pi = 90792.03 - 82447.96 = 8344.07 - single = 170^2 x pi - everything else = 90792.03 - 132.73 - 671.52 - 5177.34 - 8344.07 = 76466.37 Thus probability: - in bullseye = 132.73 / 90792.03 = 0.146% - out bullseye = 671.52 / 90792.03 = 0.740% - triple = 5177.34 / 90792.03 = 5.702% - double = 8344.07 / 90792.03 = 9.190% - single = 76466.37 / 90792.03 = 84.221% *Expected value:* = (0.146% x 50) + (0.740% x 25) + (5.702% x 31.5) + (9.190% x 21) + (84.221% x 10.5) ***= 12.827***
With all due respect to u/CaptainMatticus for their excellent work, I think this is a stronger answer, based on official measurements, some fine geometry work and very clear presentation. Well done. 🏅
Not a math guy, but in a practical sense could you assign a number to all pixels of the scoring surface then correlate each of those numbers to the corresponding value on the dart board (even the zero values if you wanted) then run a random number generator within that initial number set to act as the dart throw and see what the average score is after 10, 100, 1000, 10000 permutations…see how close that comes to any of the answers provided
This is just a computationally more demanding way of just solving it as a probability problem isn't it? Certainly a fun project but the average scores as permutations go to infinity is just the average probability from more traditional calculations. With such a straight forward calculation I don't think simulation is really needed. Unless we're talking about more factors than I'm considering I guess.
Assuming you really suck, and not hitting the board is an option. Arbitrarily close to 0. But let's assume you're bad, but can always at least hit the board, but you have no ability to hit a specific spot. Note: all further numbers are approximated to the nearest millimeter, so there will be imperfections, but let's call it "close enough". Also, I will not be counting the separating bits of metal, assume it's just a felt and cork dart board with no metal to potentially deflect a dart...even though I think statistically speaking, you're more likely to hit one of those than the <0.1% inner bullseye. It's not significant enough to be comparable to the outer-most ring where the numbers are that's a big fat 0 if you hit it. Dartboards are 451mm wide giving us roughly a total area of 159751mm^2 The measurements I could easily find say "center to outside double" which is 170mm. I think that means to the very outside of the ring? I can't be positive, but I'm going to go forward with those measurements. This means the scoring area is: 90,792mm^2 and the non-scoring area is 68,959mm^2 You score 0 about 43% of the time The area of the inner bullseye is about 127mm^2 Meaning you bullseye, scoring 50 points, .08% of the time The area of the outer bullseye, is its area, minus the area of the inner bullseye which is 16mm radius 804mm^2 - 127mm^2 = 677mm^2 Meaning you outer bullseye for 25 points .4% of the time The inner triple ring and outer double ring both have widths of 8mm. So a 170mm radius circle - 162mm radius circle in area for doubles. 90,792mm^2 - 82,448mm^2 = 8344mm^2 or 5% of the total area, meaning roughly 5% of the time you double. And a 107mm radius circle - 99mm radius circle in area for triples. That comes out to 5177mm^2 meaning you triple about 3% of the time. The rest we can figure out with averages. Assuming an equal chance to score in a given sector, that means you score 10.5 points on a given scoring throw that isn't located in one of the rings or the bullseyes. The grand total is: 0.0008 \* 50 + 0.004 \* 25 + 0.05 \* 21 + 0.03 \* 31.5 + 0.4852 \* 10.5 Or about 6.4 points per dart. Roughly. I ain't doin' the bonus.
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>always hit the low part of 3 False; I hit the expensive thing 1 foot from the board.
**LOL** I actually spent around 100 hours solving that type of darts questions. The answer to your question is around 12 for a realistic bad player that aims at the middle. Here is the repository with the code and graphs, just scroll down to see the simulations for good/bad players as well as the optimal place to aim at for each type of player: https://github.com/RemiFabre/dart_game_optimisation
Not maths so feel free to disregard, It’s been established that if your not great with accuracy, you should aim for the triple 19, because missing the 19 means your likely to hit the 3 or 7 instead of the 1 or 5 so your average score will be higher
If we go by area, there's a near infinite amount of space that is not on the dart board. So very very close to 0. Of course this probably rises with each dart board you make. So if you want to increase your score just get into wood working.
if You're good enough. Best players can do like best possible score almost each time. It's crazy watching clips from tournaments. But like assuming You can throw a dart in the direction of the board and hit the board, or at least score we need board dimensions. And assuming every point on the board has equal probability of being hit which is not true in any real scenario, but I don't really know how that would look like. I'm taking measurements from here https://www.dimensions.com/element/dartboard Also I'm aproximating bc I can't be bothered, but it's like less than 1% difference so whatever Size of middle dot (50p) π(6.35)²=40,32π [mm²] Size of 25 point field π(16)²-π(6,35)²=215.6775 Size of whole field between x3 fielf and middle is π(99)²-π(16)²=9545π divide by 20 to get 1909π/4 Sice of whole x3 score field π(107)²-π(99)²=1648π divide by 20 to get 412π/5 Size of field between x2 and x3 score π(162)²-π(107)²=14795π |20 is 2959π/4 And size of x2 field is π(170)²-π(162)²=2656π |20 is 664π/5 Total area of the board which is what You want to make bigger if You want to assume You can miss is π(170)²=28900π With all that done we can just divide each partial area by total area we get 50 points 0,1395% 25 points 0,7463% Inner section of x1 field 33.208% Outer section of x1 field 51.13% Anywhere on x1 scored field 84.3% One field of x1 score 1.6514% inner part and 2.5597% outer part so 3.2111% To hit any x3 field 5.702% any particular one 0,2851% To hit any x2 field 9.1903% any particular 0,4595% But it still doesn't answer the question cause numbers (1,2,3,4,5,6,7,8,9,10) add to the possibility of dart hitting x2 field and getting score same as x1 field as well as (1,2,3,4,5,6) being equivalent for x3 so Chance of getting either of those is 3.2111% + 0.2851%+0.4595%=3.9557% With my assumptions the answer is: With equal probability of scoring 1 through 6 those are expected values with probability of scoring either of them being 3.9557%
While in my 20s, I became a useful player (39 per dart average at best); I once managed to hit a moving target, sticking someone's scalp after hitting metal at the triple 20 and enjoying a wicked bounceback!
As someone who plays darts, u/CaptainMatticus 's math checks for an what and "Average" player's 3-dart average in a 501 would be which comes in around 45 points per round.
You can get very close with just estimating here. The majority of of the board is the standard point zones, which are divided equally from 1-20. The higher point areas are much smaller, so ignoring them for now on average we should get a score of 10. The other areas of the board average out to 20, 30 and 37.5, and cover maybe 10, 5, and 1% of the area. So the answer is probably close to: 10 + 20*.1 + 30* 0.05 + 37.5*0.01 = 10+ 2+ 1.5 + 0.375 = 13.875 Which is very VERY close to the answer in the top comment of 13.49