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# Supercross Can someone break the points down for the last four?

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I know anything can happen but if everyone stays healthy, could someone break down what it would take for each of the top five in the points to land in first or second. For example, what would it take for Dungey to win? Canard? does Stewart still have a shot, etc....."**_____ would have to finish in at least______the last four races." Haven't really studied how the points work for each race. Not sure if this is even possible to estimate it close with still four more this tight.

Appreciate the ideas though. The suspense is intense!

Edited by coastalrider

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Still too early to call, IMHO. However, Villopoto is clearly the best positioned to win this thing.

And yes, Stewart still has a shot. But not if he keeps riding the way he has the last several rounds, and not unless he has some help from the competition.

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100 pts for 4 wins.

25-1st

22-2nd

20-3rd

18-4th

16-5th

Its too early to call.

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Thanks. Going to be a great finish no matter who wins. I think most times it's already decided before the last race, but this may make it to the end finale.

Edited by coastalrider

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That is an astronomical number of data points. I started trying to plot it out in excel and then realized just how much of an idiot I was being for even trying...after spontaneously remembering high school math.

Each rider has 21 possible point variations per race over 4 races with 5 riders in the points hunt.

That's 194,481 possible outcomes per rider (21^4). When you combine the possible outcomes between the 5 riders that's 278,218,429,446,951,548,637,196,401 (194481^5). Combinations and permutations are a bitch!

Even if we assume the top 5 will only ever place in the top 5 and not under that's still 95,367,431,640,625 possibilities.

Take the number of possible outcomes per rider per race and put it in the "n" field. For SX this is 21 (1st-20th plus DNS). Then in the "r" field you place the number of races or 4. Then you copy the calculated number and place it in the "n" field. This is the total possible outcomes per rider. Then put the number of riders in the "r" field or 5. Calculate.

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".......so you're saying there's a chance!"...Dumb and Dumber.

LOL

I think I'll just watch and see the results like normal. Thanks

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That's 194,481 possible outcomes per rider (21^4). When you combine the possible outcomes between the 5 riders that's 278,218,429,446,951,548,637,196,401 (194481^5). Combinations and permutations are a bitch!

So, what's the problem? With only 278,218,429,446,951,548,637,196,401 combinations, given that it would take you one second for each possible outcome, that would only take you 8,822,248,523,812,517,397 years.

Get started!