Descending Numerical Sequence???

[Sheridi]

Not sure if that is the correct forum or if this belongs in a math forum but here goes.

I am in a competitive club where payouts are made each week. When there are just 2 winners the payouts are 60% for first place and 40% for second place. If there are 5 places to be paid, 30%, 20%, 25%, 15% and 10%. I was handed down a formula sheet which goes out to 6 payout places.

My question is: Is there a way (or formula) to calculate payouts to x number of places where there is a gradual reduction from top to bottom, where first place obviously receives the lion's share and last place receives the least. Ideally I would like to input dollar amount and payees, and get a readout of how much each place should receive.

Thanks for any help or ideas.

 

[joeu2004]

When there are just 2 winners the payouts are 60% for first place and 40% for second place. If there are 5 places to be paid, 30%, 20%, 25%, 15% and 10%. [....] Is there a way (or formula) to calculate payouts to x number of places where there is a gradual reduction from top to bottom, where first place obviously receives the lion's share and last place receives the least.

No, not without something more specific than "lion's share" and "the least". For example, for 5 winners, why not 80%, 8%, 6%, 4%, 2%?

(BTW, I assume that you have a typo, and the current sequence for 5 winners is ...25%,20%..., not ...20%,25%.... Surely 2nd place should get no less, if not more, than 3rd place.)

The point is: for any "x" winners, there are many ways to distribute 100% so that each "place" gets no less, and perhaps more, than the next lower "place".

The number of "many" ways varies with the magnitude of "x". If there are 100 winners, there is only one way: each "place" receives 1%. If there are 99 winners, 1st place receives 2% and the remaining 98 "places" receive 1%. Etc, etc, etc. But for small "x", there can be very many ways.

 

[joeu2004]

No, not without something more specific than "lion's share" and "the least".

And "gradual".

If there are 100 winners, there is only one way: each "place" receives 1%.

And even then, I made the "brash" assumption that shares should be "integer" percentages (i.e. multiples of 1%).

If we allow "fractional" percentages (e.g. 1.5%), there are a nearly-infinite number of ways to distribute 100% for each "x" of winners.

 

[Sheridi]

Thank you for the reply, I thought that would be the answer but math is not my strong suit!