Can anyone come up with the right formula?

[David Melis]

Condition. 11 players participate in online Formula 1 race. They each pay a $10 entry fee creating an overall cash pool of 110$. According to their result the money is distributed as follows:

Player 1: $20
Player 2: $18
Player 3: $16
Player 4: $14
Player 5: $12
Player 6: $10
Player 7: $8
Player 8: $6
Player 9: $4
Player 10: $2
Player 11: $0

As you can see the difference in the cash earnings between each player is the same = $2. The best performer gets the amount twice the entry fee & the worst performer the amount of $0

Question: What would be the formula that would keep the distribution in the same order (the difference in the earnings is the same) no matter how many players participate? The formula should allow to simply enter the different figure for the number of players participating each time.

 

[Deleted member 4993]

Condition. 11 players participate in online Formula 1 race. They each pay a $10 entry fee creating an overall cash pool of 110$. According to their result the money is distributed as follows:

Player 1: $20
Player 2: $18
Player 3: $16
Player 4: $14
Player 5: $12
Player 6: $10
Player 7: $8
Player 8: $6
Player 9: $4
Player 10: $2
Player 11: $0

As you can see the difference in the cash earnings between each player is the same = $2. The best performer gets the amount twice the entry fee & the worst performer the amount of $0

Question: What would be the formula that would keep the distribution in the same order (the difference in the earnings is the same) no matter how many players participate? The formula should allow to simply enter the different figure for the number of players participating each time.

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[Ishuda]

Does each player always pay $10?

To answer your question as I understand it (and as a hint to the OP), if you have n players then
Player 0 = worse player gets $0
Player 1 = 2^nd worse player gets $2
Player 2 = worse player gets $4
...
Player n-2 = 2nd best player gets $2(n-2)
Player n-1 = best player gets $2(n-1)

and the win pool is comprised of n fees of x a piece from each player.

 

[Ishuda]

Huh? What about 3 players:
winner: 20
2nd : 10
loser : 0

My understanding of th question "Question: What would be the formula that would keep the distribution in the same order (the difference in the earnings is the same) no matter how many players participate?" is
For n players what is the entry fee if the worse player gets $0 and, in order, each player gets two dollars more than the one he was just better than. So for three players the prize pool is
0 (for worse player) + 2 (for second best/second worse) + 4 (for best) = 6
Three players means a fee of $2=6/3

 

[David Melis]

I will try to be more specific.

Things that are always the same in every race:
1. Entry fee per player = $10
2. Prize for the best performer = $20 (x2 entry fee)
3. Prize for the worst performer = $0
4. Each player gets the same $ amount (let`s call it DIFFERENCE AMOUNT) more than the one he was just better than (except the worst performer who always gets $0)

Things that change with every new race:
1. The number of Players participating
2. The total cash pool

What has to be calculated by a FORMULA/FORMULAS in every new race
1. The DIFFERENCE AMOUNT in earnings between players. This amount is always different depending on the number of players.



NOTE: As I saw in a discussion, Denis totally nailed it for three players:
In this case the distribution is as follows:
winner: 20
2nd : 10
loser : 0

Here the overall cash pool is $30 & the DIFFERENCE AMOUNT is 10

We also established for 11 players:Player 1: $20
Player 2: $18
Player 3: $16
Player 4: $14
Player 5: $12
Player 6: $10
Player 7: $8
Player 8: $6
Player 9: $4
Player 10: $2
Player 11: $0

Total cash pool is $110 and the DIFFERENCE AMOUNT is 2.So how do we calculate the DIFFERENCE AMOUNT when, let`s say, 4 players participate or 12 players, or 55 or 273 or 79475........
Ideally the formula would allow to enter the Number of players participating in the race and would give the DIFFERENCE AMOUNT as a result

I hope that cleared things a bit. I would be very impressed if someone solves it.

Enjoy

David

 

[David Melis]

Hi David, Bravo, you made it look quite simple.

a little twist: What if evrything stays the same but the website keeps 10% of all the entry fees & therefore 10% of the cash pool.

how do we insert it to the formula you gave 'd = 20 / (n - 1)' or is there another approach?

David

 

[David Melis]

Hey Denis,

I am back as my question was, in fact, not that simple.
I need a different resolution.
You solved it correctly for the situation when the winner gets X2 times the fee amount.

But let's imagine that the winner gets X10 times the fee amount.

In our example let's sy 41 person participate in a game.
Each one puts in $10 entree fee. That creates a cash pool of $410.
Now if we apply your formula, then the distribution of winnings is as follows:

d= $100 (41-1) = $2.5 - the difference between consecutive terms
1. $100
2. $97.5
3. $95
4. $92.5
5. $90
6. $87.5
7. $85
8. $82.5
9. $80
10. $77.5
11. $75
12. $72.5
13. $70
14. $67.5
15. $65
16. $62.5
17. $60
18. $57.5
19. $55
20. $52.5
21. $50
22. $47.5
23. $45
24. $42.5
25. $40
26. $37.5
27. $35
28. $32.5
29. $30
30. $27.5
31. $25
32. $22.5
33. $20
34. $17.5
35. $15
36. $12.5
37. $10
38. $7.5
39. $5
40. $2.5
41. $0

BUT THERE IS ONE BIG PROBLEM!!! Our cash pool is only $410, so we cannot pay all the players the amounts from the above breakdown as we're limited to $410.

So, the question is: How do we manage to award the winer $100 (entry fee is $10) and have all the remaining players with consecutive results receive awards with the same difference between the terms.
Remember that we can only use $410 cash pool. That's where money have to come from.
Is that even possible?

 

[Ishuda]

Hey Denis,
...

In our example let's sy 41 person participate in a game.
Each one puts in $10 entree fee. That creates a cash pool of $410.
Now if we apply your formula, then the distribution of winnings is as follows:

d= $100 (41-1) = $2.5 - the difference between consecutive terms
...
BUT THERE IS ONE BIG PROBLEM!!! Our cash pool is only $410, so we cannot pay all the players the amounts from the above breakdown as we're limited to $410.
...

You have n players with fee f for a cash pool of n*f. The winner gets twice the entry fee 2*f which is the difference d times n-1. So
d = 2 f / (n-1)
place j gets a prize of (n-j)*d so pay out P is
P = d \(\displaystyle \Sigma_1^n\, (n-j)\, =\, d\, \Sigma_1^{n-1}\, j\, \)
= \(\displaystyle d\, \frac{n\, (n-1)}{2}\, \frac{2\, f}{n-1}\, =\, \frac{n\, (n-1)}{2}\, = n\, f\)

If there is a 'house cut', notice that d can also be written as
d = 2 P / [n (n-1)]
where P is the payout which would be the n*f minus the house cut.

Now to the equation by Denis. The difference between my equation and his is that he used an S and I used a P. You applied his incorrectly. Assuming no house cut,
d = 2 * 410 / (40 * 41) = 20 / 40 = 0.50

 

[David Melis]

Ishuda, Thank you for the breakdown, but you did not notice one thing.
This time I was asking how to solve it if the winner gets x10 TIMES the entry fee 10*f (NOT 2*f)

Then I believe it's impossible to apply the rule of arithmetic sequence & keep it within the $410 pot, isn't it?

 

[Ishuda]

Ishuda, Thank you for the breakdown, but you did not notice one thing.
This time I was asking how to solve it if the winner gets x10 TIMES the entry fee 10*f (NOT 2*f)

Then I believe it's impossible to apply the rule of arithmetic sequence & keep it within the $410 pot, isn't it?

Following the arguments and work it out. If you insist on an equal step prize amount, i.e. place j gets a prize of (n-j) d, you can not arbitrarily set the winners amount. The winners prize W is
W = (n-1) d = 2 f / (1+p)
where p is the fractional amount of the house cut H;
H = p P
and P is the total payout. This assumes the fee f is composed of the payout and House cut, i.e.
f = (P + H) / n