#*which is a misnomer because they DO talk to each other. a bit. mostly via an elaborate pen-pal system that took a while to get up and goin

[thydungeongal]

Alright, back on my numbers bullshit. This time I want to talk about dice pool systems, specifically of the World of Darkness and Shadowrun varieties. These two types of games are mostly interchangeable for the purpose of this discussion: the differences are mostly in die size and what counts as a success.

So you've got a system where you want players to grab a bunch of dice, commonly assembled by taking a number of dice equal to you Stat (or attribute or ability) and Skill (or talent or aptitude). Sometimes things like damage might be modeled via this same system, so you'll instead be taking a number of dice equal to the damage value of your attack plus any extras (a very common way of handling this is that any successes from the attack in excess of what was needed to succeed get added into the dice pool to represent particularly good hits).

After you've assembled your pool you roll them bones and look for certain numbers that count as successes. "Success" is a bit of a misnomer here, because usually just one success won't be enough for a given roll to succeed: instead you're looking for a number of successes equal to a difficulty threshold.

(Historical note: back in the nineties World of Darkness games actually modeled difficulty by changing the target number you needed to roll on each die. Multiple successes were generally only required for extended tests and opposed checks where the side who gained the most successes was the one who won the contest. Theoretically you could have a contest where one side was rolling 3 dice against a threshold of 6 and the other was rolling 5 dice against a threshold of 8. While not impossible to calculate especially with the tools I like to use, this adds an extra layer of math that is for the most part unnecessary, so most modern WoD games instead models difficulty through requiring a certain number of successes while the success threshold on an individual die stays the same.)

Individual games obviously vary when it comes to what types of dice are used (Shadowrun uses d6 as does Burning Wheel, World of Darkness and Chronicles of Darkness as well as the other Storytelling System games utilize d10s) and what counts as a success, but ultimately the math in a dice pool game is not that difficult provided you have the right tools. If you're rolling a pool of d6 and looking for a result of 4 or higher, then the chances of getting a success on a single die is 0.5. Incidentally, in this example the chances of not getting a success on a single die is also 0.5 so jot that down.

No matter how large your pool is it's always really easy to calculate how many successes you are likely to get on average: you simply take the number of dice and multiply by the probability of success on an individual die. Using Shadowrun as an example: assuming a character with a dice pool of 9 (not unheard of but not exactly common), counting results of 5 and 6 as successes, so a 1/3 chance of success per die, you will on average be looking at 3 successes.

And with smaller dice pools you can get the full picture very easily as well. Assuming the aforementioned game which uses d6 dice pools and results of 4, 5 and 6 count as successes. As I've said in my previous posts, as long as you keep yourself to just two dice, you can easily calculate all the various probabilities using our friend the table.

Probability of getting 0 successes is 9/36, or 1/4. Probability of getting 1 success is 18/36, so exactly 1/2. Probability of getting 2 successes is also 1/4.

Now if our target is at least 1 success we can simply add together the probabilities of getting 1 and 2 successes. The chance of getting at least 1 success is 3/4.

Things obviously get a lot more complicated once you add more dice, but there's still one easy trick you can do: you can always find out your chances of getting at least 1 success as long as you know the size of the dice pool and the target number via finding out the probability of getting NO successes.

Assuming a Chronicles of Darkness game where the threshold for a success is 8, 9 or 10. Each individual die has a 3/10 chance of yielding a success. Therefore, the chance of yielding no success is 7/10.

You can always find out the probability of getting no successes on a dice pool by taking the probability of failure on an individual die to the power of the size of the dice pool. This is actually evident from our table as well: the probability of a failure on an individual die was 0.5 and the probability of no successes on two dice is exactly 0.5*0.5=0.25. This applies to larger pools as well: returning to our CoD example where results of 8, 9 and 10 yield successes, our chance of getting no successes is 7/10³, or 343/1000, or 0.343. Expressed in percentages, the probability of getting no successes on 3 dice is 34.3%.

And as I said, we can use this to calculate the chance of getting at least 1 success: every other possible event that is not covered by that 34.3% will have at least 1 success. Therefore the probability of at least 1 success is 1-0.343=0.657 or 65.7%.

But what if we wanted to calculate the probability of getting exactly one success? That's not exactly hard either, but the fact that we have three individual dice somewhat complicates it, because we now have to consider all three dice individually.

For a pool to have exactly 1 success, one of the dice needs to have a result of 8, 9 or 10, and the two other dice need to have a result outside of that range. That's easy enough to calculate: 0.3*0.7*0.7=0.147.

BUT there are three different ways to achieve that result because there are three different dice. Where S means success and F means failure we have {S,F,F} {F,S,F} or {F,F,S} all as unique scenarios we need to consider. The probability of each of those events is 0.147 so how do we get the probability of any one of those events happening? Simple. We add the probabilities together (this is the same logic by which we could add together the probabilities of getting 1 and 2 successes earlier to get the probability of at least 1 success: the probability of getting 1 OR 2 success is the total of the probabilities of those events happening in isolation).

The probability of getting exactly 1 success on three dice is thus 0.147+0.147+0.147=0.441 or 44.1%. And yes, this can be scaled up to any number of successes and any number of dice, but at that point I do recommend another solution.

Okay so we now know how to calculate these probabilities so we can start to do analysis. But there is actually a better way. Next time I'll talk about my favorite online dice probability calculator and finally get down to business analyzing some actual examples and what they mean.

#long post #maths #diceposting

[fbitennis]

Command of Match (COM) and Lost Opportunity Score (LOS)

I’ve been watching the Davis Cup tie between Borna Coric and Frances Tiafoe.  

The first set, won by Tiafoe in a tiebreak, is largely marked by Borna Coric showing an incredible lack of touch around the net (he cannot volley at all*) and the near impossibility of keeping his forehand in the court.  

*Separate research project: Check The Match Charting Project to see if Coric is the worst volleyer in the Top 50.  Starting hypothesis is that he is.

Then when the second set kicks off, Frances Tiafoe can’t win a game, and barely any points.  His energy level plummets and his serve is about as bad as you will see from a male professional tennis player.  He loses the second set 1-6, and proceeds to carry that over into the third set.  In an 11 game stretch, he wins only 1 game, and only 14 of 53 points, resulting in a 0-4 deficit in the third set.  Although his aggressiveness waned, I think the most important factor is that Coric stopped making those horrendous unforced errors, at which point Tiafoe’s weaknesses were all brought to the fore (or is it “foe”?).

But at 4-0, Coric started making horrendous errors again.  I mean horrendous.  At one point, he makes 7 in a row, and voila, Tiafoe’s energy returns.  Tiafoe does not play great in the rest of the third set, but Coric is so bad that Tiafoe comes back and wins the second set tiebreak.  (In fairness, the tiebreak itself is fairly well-played by both players).

As I’m writing this, I have not started the fourth set.  I know Coric wins the next two sets because I know Croatia is in the finals, but I actually haven’t looked at the set scores for the final two sets.  Based on what I’ve seen, it seems almost certain that if Coric can keep his forehand in the court, he will win, and win easily.  For all the Twitter talk of how Tiafoe was a warrior in the match (apparently forgetting that 11 game stretch), he shows no sign he can control the points. His backhand is merely steady, and his wack-a-doo forehand stroke just rolls the ball around the court.  In other words, this match is not on his racquet.

I paused in my viewing of the match, partially because I needed a break and partially because I wondered if there’s anything in the statistics that would tell someone who didn’t watch the match that the match is entirely on Coric’s racquet.   And looked at another way, if Coric had lost, could you look at his stats and know just how bad the loss is, because the match truly was on his racquet, and only he could blow it?

There are probably several ways to do this, and what I’m presenting here is perhaps the most back-of-the-envelope way to do it, primarily because I’m starting it on a whim at 11 pm while trying to stay interested in this match.  So I think this is just a toy stat, although as I have posted before, I think toy stats have their own kind of value.

But “stat” is the wrong word for the two things I’m proposing here.  “Status” is probably a better word...toy status(?).  Both COM and LOS seek to identify particular matches, rather than producing a statistic for every match.  

I’m doing LOS first, because I was initially motivated by wondering how horrendous it would have been if Coric had lost this match when he was in total control of the match.  

Lost Opportunity Score (LOS) 

I’m using the acronym LOS for this concept, but it is a bit of a misnomer because it isn’t really a score. Nevertheless, the acronym is so apropos that I can’t drop it.  LOS should indicate when the match a player lost was almost entirely on his/her racquet and he/she blew it with too many errors.

Command of Match (COM) 

We already have Carl Bialik’s Dominance Ratio (available for every match on Tennis Abstract), which indicates how much a player dominated the match statistically, but we don’t know when that dominance is attributable to the winning player playing great, and when it is attributable to the losing player playing horribly.  

COM is trying to identify when the winning player was in control, even when the other player did not play poorly.  In other words, COM isn’t designed to measure how in command one player is (though I suppose you could use it for that), but rather, to identify those relatively rare matches where the match a player won was almost entirely on his/her racquet even thought his/her opponent may have played reasonably well.

Calculating LOS and COM

The fundamental basis for both LOS and COM is the same.  For each player, calculate this number:

(1-(OppUEs/Points Won)) - (UEs/Points Played)

The first part of the formula determines what percentage of points won by the player were not gifts from the opponent.  Some of those points may be unusual situations, but most of them will be winners or FEs caused by the player, and therefore within the player’s control.  The second part of the formula indicates what percentage of overall points were gifts given away by the subject player.

Conceptually, if your first number is high, you were controlling the match to a significant degree, but if your second number also is high, you gave away a lot of points in a match.

To calculate LOS and COM, you need just one more step.  

Lost Opportunity Score (LOS)

Divide the losing player’s number by the winning player’s number.  If the losing player’s quotient is greater than 1.10 (in other words, 10% higher), it’s a lost opportunity (LOS).  In other words, the losing player had the match on his/her racquet, but made so many unforced errors that he/she gave the match away.  The 10% buffer is to capture only the most egregious of these situations.  It is approximately 1 standard deviation away from the average loser quotient. 

Here’s an example from the first round US Open match between Sam Stosur and Caroline Wozniacki, won by Wozniacki.  From the match score (6-3 6-2) it appears to be an easy win, and Woz’s dominance ratio was 1.56. Stosur won 45 out of 110 points.  She made 34 UEs and Wozniacki made just 12.  

Stosur’s number via the formula above is (1-(12/45)) - (34/110) = .424

Woz’s number via the formula above is (1-(34/65) - (12/110) = .368

Then, .424/.368 = 1.15 (greater than 1.10), so Stosur gets a Lost Opportunity (LOS) “award.”

Looking at the first part of the formula, Stosur’s points won were largely because of good things she was doing (73.3%), and Woz’s points won were mostly about Stosur doing bad things (47.7%). The second part shows Stosur made unforced errors on nearly 31% of points played, and Woz, typically, only 10.9%.  That’s in keeping with what we know about their respective styles.

Bottom line:  Stosur controlled the action in the match, but due in large part to the high number of UEs, lost the match.  I suspect this is not uncommon for Wozniacki opponents.  (See Caveats at the end).

Command of Match (COM)

Subtract the losing player’s number from the winning player’s number.  If the winning player’s difference is greater than 0.13 for men, or 0.17 for women, the winning player had command of the match (COM).  In other words, the gap between how much control the winning player had, and how much control the losing player had, is so significant that we say the winning player was in command via his/her own efforts.  Significantly, you can get a COM even if your opponent played reasonably well.

You might wonder where the 0.13 and 0.17 come from.  Using US Open matches as the measuring stick, these numbers are 1.5 standard deviations from the mean differences between the players, so we are only capturing relatively rare matches with COM.  I tried it with 2 SDs, but the list was far too thin.

Here’s an example from the first round US Open match between Simona Halep and Kaia Kanepi, since most of us saw at least some part of that match and know there wasn’t much Halep could do in that match.  The score alone (6-2 6-4) gives us some indication of Kanepi’s level, and the dominance ratio was 1.36.  Halep won 47 out of 107 points, not that much different than in the Stosur example.  Unlike Stosur, she made only 9 UEs and Kanepi made 28.  

Halep’s number via the formula above is (1-(28/47)) - (9/107) = .320

Kanepi’s number via the formula above is (1-(9/60) - (28/107) = .588

Then, .588 - .320 = .268 (greater than .17), so Kanepi gets a Command of Match (COM) award.

Going back to our concept with the first part of the formula, Halep’s points won were largely because of bad things Kanepi was doing, with Halep controlling only 40% of those points.  She didn’t hurt herself with errors obviously.  And because of that, only 15% of Kanepi’s successful points were due to her opponent’s mistakes.  

Bottom line:  Kanepi controlled the action in the match, to such a degree that even her significant number of errors, and Halep’s lack of errors, could not stop her.  

Caveats

This is not scientific, so let’s get that out of the way.  I haven’t tested it on gobs and gobs of data.

Also, only 13 hours have passed since I first thought of the idea (and 7 of them were spent sleeping), so I reserve the right to make adjustments (or even scrap LOS and COM altogether).

I initially see three issues with LOS and COM:

1.  UEs are not official statistics of the ATP and the WTA.  They are typically recorded for the grand slams, although I noticed the IBM Slamtracker didn’t bother with many lower profile matches at the US Open.  Only 178 of the 254 US Open main draw matches had meaningful UE statistics.  In the other 76 matches, IBM Slamtracker reported UEs, but they are clearly understated by vast amounts, so I’m not sure why they even list them (or winners). For example, Andrey Rublev had only 5 winners and 13 unforced errors in a four set match, while his opponent Jeremy Chardy also had only 13 UEs?  High-risk player Nikoloz Basilashvili had only 7 UEs in a five set match against Aljaz Bedene?  I don’t think so.  

So, LOS and COM are good for only Grand Slams, matches that have been charted, or matches you are watching on TV that flash the summary numbers at the end of sets or matches.  I don’t feel too badly about this.

2.  UEs are extremely subjective.  Anyone who has charted a match and then seen the on-screen statistics from the TV broadcast knows the number of differences in judgment that can arise as to whether a player should have made the shot or not.  Hopefully some of that is taken care of by the 10% buffer in the LOS calculation and the 1.5 standard deviation buffer in the COM calculation.

3.  Aggressive players are far more likely to get a LOS or COM than steady players.  It’s not necessarily a bad thing in and of itself, so long as no one says “Wozniacki has 0 COMs in 2018″ (if in fact she does have zero) and uses that as a stick to bash her with. 

As a corollary, recognize that aggressiveness is just one way to measure who had control of the match.  Steady play with few errors is arguably just as valid a way to keep the match on your own racquet, though it is a lot more subtle.  Perhaps a player should get automatically get a COM if his or her opponent gets a LOS, but I’m not yet convinced that’s the right approach as it presumes the LOS players errors were mostly attributable to the steadiness of the opponent. 

Since this one is so long, I’ll do another post with the list of LOS and COM awards from this year’s US Open.

#LOS #COM #Lost Opportunity Score #Command of Match #US Open #Toy Stats