How many field combinations played?

[Hitchhiker 42]

Based on rough calculations, there have been 121 events as of the end of week 6. Each event features about 100 matches (80-ish quals, the rest in elims). Total, that is 12100 matches. According to above, about 90% (here: FRC Blog) different field combinations have been played at each event (out of number of matches). That'd make up 10890 combinations (probably some repetition, but this is a rough estimate). Out of 18432 possible combinations. Total, that makes up 59% played combinations (in reality, less due to many combinations being played over and over and different events). So 59% is a high estimate.

A low estimate would assume the 90% different fields at each event were the same combination. So that'd be about 90 different combinations or 0.49%.

Is there a way to get a better estimate than 0.49%-59%?

[kiasam111]

This is probably the most exciting math problem of ALL TIME

How did you get 18,432 possible combinations? As you'll read, I got a very different number, but I could be very wrong.

I'm trying to wrap my head around this, and this is how I've started approaching what I hope is the answer...

On one alliance side, there are 5 positions for defenses. The low bar is always there, so there are always 4 different positions for defenses. For my math, we'll disregard the audience selection and assume that defense no.3 is also selected by the alliance. Yes, this will make the number inaccurate but I don't know how to factor them in on such a large scale... help?

Based on this, for one alliance in one game, the possible combinations for selecting their defenses would be calculated as 1*8*6*4*2, as there is 1 possible defense for position 1, then there are 8 possible defenses to select for position 2. The defenses are paired in categories A, B, C and D. Let's assume a defense from category A is chosen for position 2, both of those defenses now CANNOT be selected for position 3. This is why there are 6 possible defenses to select for position 3, similarly 4 and 2 for positions 4 and 5 respectively.

Now for the math - 1*8*6*4*2 = 384

We'll square the number to find out the total possible defensive combinations, because there are two alliances.

384^2 = 147,456

I'm not entirely sure how to interpret the FRC Blog's statistic of 90%. I'm interpreting it (for now, I am probably wrong) that 90% of matches at an individual event will have a unique defense combination that isn't repeated. That means approximately 10% of defense combinations are repeated.

Using these figures, we can calculate the number of matches across the 121 events (out of a total of 12,100) that have had unique defensive combinations and repeated defensive combinations.

12,100*0.9=10,890
12,100*0.1=1,210

Based on these numbers, that means that 10,890 combinations have been played (as the remaining 1,210 are repeats of the 90%) out of the 147,456 possible combinations.

10,980/147456 = 7.39% (2dp)

Therefore, 7.39% of the possible combinations have been played (I think..)

What are your thoughts? I hope this working out makes sense, math on CD is difficult, and it doesn't help that this math is tricky. I could be completely wrong, I'm just keen to know what the final number is!!!

P.S. A bit of cross-cultural trivia for you, in Australia, we call it maths rather than math! I translated for you through this post

[tig567899]

If we take audience selection into account, then the defense at position 3 is limited to 2 options.

Taking this into consideration, we have 1*6*2*4*2.

For example, if category A was selected as the 3rd defense, then it would be unavailable for any of the other positions.

This gives us 96.

Squaring 96 gives us 9216... which is precisely half of the number presented. I don't know where that extra 2 came from. If anyone could care to enlighten me, that'd be welcome. From what I see if 10980 combinations were played we've played more combinations than are available. Just my take on the issue.

[kiasam111]

That's such a clean and easy way to include the audience selection - I love it It also means that your number is more accurate than mine #goteam

I've calculated my numbers again using 9,216 combinations (thanks!) and an Arithmetic Series, but this is based on a shaky interpretation of the 90% found in the FRC Blog.

An average event has 100 matches, and 90% of those have unique field combinations (90 matches).

Therefore Event 1 - 90/9216 = 0.98%

At Event 2, 90% of the field combinations have been used before (at Event 1) and 10% haven't. (I think..)

Therefore Event 2 - 0.98% + (0.98%*0.1) = 1.078%

Similarly, at Event 3, 90% of the field combinations have been used before (at Event 1 and 2) and 10% haven't.

Therefore Event 3 - 0.98% + (0.98%*0.1) + (0.98%*0.1) = 1.176%

From this, we can see a pattern is emerging. For each event, we add (0.98%*0.1), therefore we can create a generic formula to represent the projected percentage of defensive combinations used at any given event.

Therefore Defensive Combinations = 0.98% + (n-1)(0.98%*0.1)

This fits the generic form of an Arithmetic Progression, where Term n = a + (n-1)d, where a is our first term (0.98%), n is the event number, and d is the common difference (0.98%*0.1).

Using this, the 121st event would be term 121 (n=121).

Therefore, Defensive Combinations used by the 121st event would equal
0.98% + (121-1)(0.98%*0.1) = 0.98 + 120(0.098)
= 0.98 + 11.76
= 12.74%

So this means that after the 121st event, roughly 12.74% of the defensive combinations have been played. Does that make sense??? Happy to explain further. This is so much fun!!!

[Hitchhiker 42]

Originally Posted by kiasam111

This is probably the most exciting math problem of ALL TIME

How did you get 18,432 possible combinations? As you'll read, I got a very different number, but I could be very wrong.

Here's my math:

(low bar locked)*(8 possible audience selections***)* [6*4*2]^(2 alliances) = 18432

*** Explanation of 8 possible audience selections: In any round of 6-12 matches, depending on the event, there are only 2 choices. However, overall, there are 8 different possible defenses that can go into that 3rd slot, so I use 8 there. The rest of the numbers follow easily enough: 2 alliances have 6 defenses for slot 2, 4 (b/c of the groups) for slot 4, and 2 options for slot 5.

[Hitchhiker 42]

Originally Posted by kiasam111

An average event has 100 matches, and 90% of those have unique field combinations (90 matches).
.
.
.
At Event 2, 90% of the field combinations have been used before

The way I interpreted their numbers was that each event had 90% of the matches use different fields than any other in that event. So I'm not sure how to check how many are popping up across events, but if it's totally random which defenses get selected (which it's not), I'll assume each field combo has an equal shot at selection (100 matches per event, 121 events assumption), so, each event has a (however uniques matches played)/(total field combos) chance to repeat a match. I'll attach an excel file in a moment to explain.

[kiasam111]

I'm still not 100% sure on how you're getting that (I'm so sorry). When I factor out 8 for the audience selection, 6, 4 and 2 for the remaining defensive positions, I still have 48 as a factor. Why is this the case?

I'm also not sure on how you've factored in that there are 2 alliances. I understand that BOTH alliances have 6 defenses to choose from in position 2, but wouldn't you still need to square the number to take both alliances into account?

[Hitchhiker 42]

The excel file attached should show my calculations as well as my explanation for how I worked it out... I narrowed it down to (of course with all the simplifying assumptions) 43% matches possible played.

[kiasam111]

Originally Posted by Hitchhiker 42

The way I interpreted their numbers was that each event had 90% of the matches use different fields than any other in that event. So I'm not sure how to check how many are popping up across events, but if it's totally random which defenses get selected (which it's not), I'll assume each field combo has an equal shot at selection (100 matches per event, 121 events assumption), so, each event has a (however uniques matches played)/(total field combos) chance to repeat a match. I'll attach an excel file in a moment to explain.

This makes more sense! Not sure what I was doing before

The spreadsheet also means we can extrapolate!! Hooray!

Not sure how helpful this will be, but TBA has the total number of matches played as 11162 right now (found here)

[kiasam111]

The spreadsheet also means we can extrapolate!! Hooray!

Speaking of extrapolation, it would take 3918 events to play 90% of all combinations. As far as I can tell, at this point it becomes a limiting sum.

[Hitchhiker 42]

Originally Posted by kiasam111

This makes more sense! Not sure what I was doing before

The spreadsheet also means we can extrapolate!! Hooray!

Not sure how helpful this will be, but TBA has the total number of matches played as 11162 right now (found here)

Thanks. I can adjust the average matches/event to 92.25... That shifts it down to 40.9% field combos played.


Also, I edited my previous post explaining how many possible field combos there are (thanks for the catch, it should've been (6*4*2)^2).

[Van.Augur]

Time to play devil's advocate:

If you want to be exact, you might want to take into account that the rough terrain, according the rules, is a random layout of the steel cubes. There are an extremely large number of rough terrain defenses that could theoretically exist (keep in mind they aren't symmetrical when you rotate them 180 degrees), making the real number of possible unique defense layouts impractically large...
but still calculable

[EricH]

Originally Posted by Van.Augur

Time to play devil's advocate:

If you want to be exact, you might want to take into account that the rough terrain, according the rules, is a random layout of the steel cubes. There are an extremely large number of rough terrain defenses that could theoretically exist (keep in mind they aren't symmetrical when you rotate them 180 degrees), making the real number of possible unique defense layouts impractically large...
but still calculable

Nice try, go read the field drawings. All the rough terrain units were built the same. "Random" was in fact in quotes in the Manual. You only get one rough terrain defense, not one per combination...

[Hitchhiker 42]

Originally Posted by Van.Augur

Time to play devil's advocate:

If you want to be exact, you might want to take into account that the rough terrain, according the rules, is a random layout of the steel cubes. There are an extremely large number of rough terrain defenses that could theoretically exist (keep in mind they aren't symmetrical when you rotate them 180 degrees), making the real number of possible unique defense layouts impractically large...
but still calculable

Assuming, this, which it isn't (as per above post), you'd be completely limited by the number of built rough terrains.

[logank013]

Just to clarify the number of combinations possible, it was discussed in an earlier thread here

They seemed to agree on 18,432 possible combinations as well.