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Big Brother and online Hunger games.

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Aug 13, 2014 by Ilikebugs
Generally, you can take a look at the number of players, the numbers of rounds, and the cut-off and guesstimate the number of losses you can absorb before landing out of contention. Experience helps in this regard. Or you can use this plug-and-play formula to really nail down the numbers, which simplifies the model by assuming no draws, and then you can figure in draws just by simple reasoning.

Where n is the number of players in the tournament, r is the number of rounds and P subscript l is the number of players with l losses. The function symbol is the product symbol, so for 64 players in a 4-round tournament, the number of players with 0, 1, 2, 3 and four losses would be:

P0 = (64/2^4) = 4
P1 = (64/2^4)(4) = 16
P2 = (64/2^4)((4)*(3/2)) = 24
P3 = (64/2^4)((4)*(3/2)*(2/3)) = 16
P4 = ((64/2^4)((4)*(3/2)*(2/3)*(1/4)) = 4

This means that in this particular tournament, to guarantee yourself top 8, you'll have to either have no losses or be in the top quarter of the players with one loss on breakers. Players attempting to draw into top 8 will make it harder for players with one loss to get in based on breakers, but there's no easy formula to figure out how hard.

More sensibly, a 64-man Swiss tournament would usually have six rounds. So plugging that into the formula:

P0 = (64/2^6) = 1
P1 = (64/2^6)((6)*(5/2)) = 15

At this point, we can see that we have found our top 8 players - the one undefeated player, and seven of the 15 players with one loss. Again, players attempting to draw in to the top 8 makes it more difficult for players with one loss to get into the top 8.

Last example. Let's say it's a 1,436 man tournament with 10 Swiss rounds, cut to the top 128.

P0 = (1436/2^10) = 1.4 (This partial figure means that random pairings can result in a number of undefeated players between 1 and 2)
P1 = (1436/2^10)((10)*(9/2)) = 63.1 (Somewhere between 63 and 64 players)
P2 = (1436/2^10)((10)*(9/2)*(8/3)) = 168.3

At this point, we've found that 0 or 1 loss in this tournament guarantees you top 128. Somewhere between 62 and 64 of the 168 to 169 two-loss players will get into the top 128 cutoff, and whenever a player attempts to go 8-1-1 to get into the top 128, that makes it more difficult for the 8-0-2 players to get in. Incidentally, we can also see that more than one loss puts you out of contention for the top 64, and more than zero losses mean you'd have to fight your way into the top 32.

So all you have to do is plug in the appropriate numbers and do a bit of quick math on a calculator until you've found enough players with each bracket of losses to fill your cutoff and you can make your decisions from there.

Hope this helps!

Comments

P l = ( n/2^r)= l
Sent by Ilikebugs,Aug 13, 2014

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