My math brain is broken

[rabbidrabbit]

OK,

We are having an event where first prize is 1500, second is 750, third is 500.  Now, we are offering a side bet where the pot will be split by the same ratio after we take a 5% cut.  The question is, how do I maintain this ratio at any given amount in the pot.  My brain is taking a vacation tonight.

[DMBEAR]

Assuming your pot was paid out in full to 3 prizes, then your example would show the following:

1st prize is 54.55% of the pot = 1500

2nd is 27.27% of the pot = 750

3rd is 18.18% of the pot = 500

If you took out a cut of 5% before splitting the pot...

1st would be 1425.12

2nd is 712.43

3rd is 474.15

You maintain this ratio at any given amount in the pot by coming back from vacation and giving me 1/2 of the cut which would be a mere $68.75, and I'll do it 4 ya. 

[eagl]

It's easier done by ratio.

It's a 6-3-2 ratio...  Divide the pot by 11, apportion by the ratio, ensuring that 2nd place always gets half of first place, third place always gets 2/3 of second place and 1/3 of first place.

Example - If the pot was 1100, the winnings would be 600 300 200.  The ratio works.

[Nilsen]

Jetboi is onto somthing here wabbit. Its the easyest way to calculate it imo too.

[rabbidrabbit]

Ya, that's where I ended up as well.  Basically, the total of the 3 pots was 2750 and all pots were divisible by 250 so I calculated there were 11 parts total.  I kept thinking there was a solve for X in this formula somewhere but I was too tired to figure it out.

It's a charitable Tuna fishing tournament I'm putting on.  The pots aren't huge but the point is helping folks out, not getting rich.
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