How many field combinations played?

[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