ok, so now say for the two scenario's there is a way to gain an advantage over your competitors. the question becomes, how is such a given advantage affected by the number of said riders (say the competitors all have equal skill relative to one another, except for our hero of course)?
such that:
A = hero's advantage over other riders
Y = value of the prize
X = cost of the ride
where Y>X
P = number of riders
fixed cost scenario
there is a merry go round with twelve horses and a brass ring that has value greater than the cost of the ride.
a lone rider will always experience plus ev, cause he is gonna get the brass ring.
but if the ride fills up with more riders and the value of the brass ring and the cost of the ride doesn't change, then there may reach a point where our hero will experience either zero ev or negative ev if the numeric value of the ratio of the value of the brass ring to the cost of the ride isn't high enough.
fill in the blank lol, ______________
here the ratio of the brass ring value to ride cost must have a greater numerical value than than actual number of riders inorder for play to be plus ev. ie. y/x > P
i think if that holds then the following general attributes about our hero's skill holds.
comparison of skill
say our hero is twice (A=2) as skillful as any other rider
so hero against 1 rider, hero win 2 times to their 1 so beat the field by 1 (note: A>z)
hero against 2 other riders, hero win 2 times to their 1 each so field wins 2 & tie (note: A=z)
hero against 3 other riders, hero win 2 times to their 1 each so field wins 3 field wins by 1 (note: A<z)
hero against 4 other riders, hero win 2 times to their 1 each so field wins 4 field wins by 2 (note: A<z)
hero against 5 other riders, hero win 2 times to their 1 each so field wins 5 field wins by 3 (note: A<z)
ect...
up to the highest number of riders, twelve
hero against 11 other riders, hero win 2 times to their 1 each so field wins 11 field wins by 10 (note: A<Z)
.....
say hero is three times (A=3) as skillful as any other rider
so hero against 1 rider, hero win 3 times to their 1 so beat the field by 2 (note: A>Z)
hero against 2 other riders, hero win 3 times to their 1 each so beat the field by 1 (note: A>Z)
hero against 3 other riders, hero win 3 times to their 1 each so field wins 3 & tie (note: A=Z)
hero against 4 other riders, hero win 3 times to their 1 each so field wins 4 field wins by 1 (note: A<Z)
hero against 5 other riders, hero win 3 times to their 1 each so field wins 5 field wins by 2 (note: A<Z)
ect...
up to the highest number of riders, twelve
hero against 11 other riders, hero win 3 times to their 1 each so field wins 11 field wins by 9 (note: A<Z)
ect. .........
time to interject a note here, as it is apparent the following relationships hold:
note: here z is the number of competitors
if A = z then hero ties the field
if A < z then the field beats our hero, (but this doesn't mean hero doesn't make value above his cost)
if A > z then our hero beats the field
so
say hero is 11 times (A=11) as skillful as any other rider
so hero against 1 rider, hero win 11 times to their 1 so beat the field by 10 (note: A>Z)
hero against 2 other riders, hero win 11 times to their 1 each so beat the field by 9 (note: A>Z)
hero against 3 other riders, hero win 11 times to their 1 each so beat the field by 8 (note: A>Z)
hero against 4 other riders, hero win 11 times to their 1 each so beat the field by 7 (note: A>Z)
hero against 5 other riders, hero win 11 times to their 1 each so beat the field by 6 (note: A>Z)
hero against 6 other riders, hero win 11 times to their 1 each so beat the field by 5 (note: A>Z)
ect....
up to the highest number of riders, twelve
hero against 11 other riders, hero win 11 times to their 1 each so field wins 11 & ties (note: A=Z)
so
say hero is 12 times (A=12) as skillful as any other rider
so hero against 1 rider, hero win 12 times to their 1 so beat the field by 11 (note: A>Z)
hero against 2 other riders, hero win 12 times to their 1 each so beat the field by 10 (note: A>Z)
hero against 3 other riders, hero win 12 times to their 1 each so beat the field by 9 (note: A>Z)
hero against 4 other riders, hero win 12 times to their 1 each so beat the field by 8 (note: A>Z)
hero against 5 other riders, hero win 12 times to their 1 each so beat the field by 7 (note: A>Z)
hero against 6 other riders, hero win 12 times to their 1 each so beat the field by 6 (note: A>Z)
ect....
up to the highest number of riders, twelve
hero against 11 other riders, hero win 12 times to their 1 each so beat the field by 1 (note: A>Z)
again:
note: here z is the number of competitors
if A = z then hero ties the field (still our hero realizes value above his cost)
if A < z then the field beats our hero, (still our hero realizes value above his cost)
if A > z then our hero beats the field
it's a good thing to be able to beat the field as often as possible.
diminishing cost scenario
but suppose that the carnival had a promotion where the cost X of the ride was reduced by dividing the original ride cost by the number of riders while keeping the value of the brass ring the same.
fill in the blank lol, ______________
same would be true here but it would be even more significant since the cost of the ride reduces while our hero gets more of the pie, lol.
maybe all that really happens for the most part for both of these scenarios is that our hero gets more plays faster, but definitely i think beating the field more often is a cost saver.
not even sure if this new stuff is correct, rushed to post, cause i'm off to do the real thing, lol.