1

I have a simple game which works as follows:

There are 11 options where you can pick from, they all have a starting score of 100.

When a random option is chosen this score is lowered by X, the other options are raised by Y. This makes sure that popular options get a lower score and more unlikely options will receive a higher score.

But I am struggling a bit with the X and Y values.

Now I have for X: score * 0.99 and for Y: score * ((0.01 / 11-1) + 1) effectively making Y in this example 1.001.

When I run a simulation for 10000 iterations I get this result:

Multiplier: 1,001

0: 46,415482172560186982173898945 (974)
1: 66,838393287292917077894008614 (941)
2: 96,24742883459377976132240996 (908)
3: 121,38518290493490497771733027 (887)
4: 107,49245093808228436049973035 (898)
5: 161,78439604035762550303938848 (861)
6: 89,08343532143707485107456848 (915)
7: 81,54660657436701987059722203 (923)
8: 96,24742883459377976132240964 (908)
9: 103,98754295728204150478760963 (901)
10: 125,47647965274993316730236819 (884)

Where the number between brackets is the number of times that number is picked.

But when I raise the iteration count to 10000000 the numbers don't add up anymore:

Multiplier: 1,001

0: 0,0000000000000000000036583816 (909214)
1: 0,0000000000000000000000250626 (909665)
2: 0,0000000000000000000205077532 (909058)
3: 0,0000000000000000000000003767 (910424)
4: 0,0000000000000000000010153258 (909330)
5: 0,0000000038841252289686775249 (906708)
6: 0,0000000000000000410723804695 (908370)
7: 0,0000000000000000000121999845 (909105)
8: 0,0000000000000000000089536881 (909133)
9: 0,0000000000000000000000002027 (911122)
10: 0,0000000000000101904281714748 (907871)

Any suggestions how to tackle this problem?

Roger Far
  • 33

2 Answers2

3

The problem is that your adjustments decrease the total. Suppose there were 11 people, and each number got picked once. Then each number would be multiplied by $$1.001^{10}\cdot 0.99\approx 0.999944669$$

This is similar to the idea that raising a price by 10% and then lowering it by 10% will end up decreasing it by 1%.

If your goal is to have the numbers add up, I recommend lowering the number picked by $X$, and raising the others by $\frac{1}{10}X$, rather than any function of $Y$.

vadim123
  • 82,796
  • Yes ofcourse I forgot about this completely. Maybe I should add that number instead of multiplying them – Roger Far Jun 03 '13 at 19:35
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If you want to keep the total the same then the score of the other options must be increased by:

$$Y=\frac{\text{(Current Score of the chosen option)} \quad \text{Discount Factor}}{10}$$

where

$\text{Discount Factor}$ is how much you decide to discount the option picked.

response
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