TWRoO said:
nordlead said:
It was easier to find this than I thought 
Tbone said:
Nord.
If someone bets on a game VG$1k for a game selling 50k and his prediction was 35k does he earn more money than someone who predicted 45k and bet VG$500. |
It depends on what the average bet is. The equation goes something like this
(1-(diff/actual))/(average accuracy)*bet
so using your example we get the following scenarios
| average Accuracy |
1k bet |
|
500 bet |
| 100 |
-300 |
|
-50 |
| 90 |
-222.222 |
|
0 |
| 80 |
-125 |
|
62.5 |
| 70 |
0 |
|
142.8571 |
| 60 |
166.6667 |
|
250 |
| 50 |
400 |
|
400 |
| 40 |
750 |
|
625 |
| 30 |
1333.333 |
|
1000 |
| 20 |
2500 |
|
1750 |
| 10 |
6000 |
|
4000 |
| 1 |
69000 |
|
44500 |
However, I am yet to see someone make more than 50% off of a bet, so I have a feeling there is a cap, but I haven't seen the equation yet. Also, you would have to get very lucky for everyone to bet so far off, and you be so close.
As you can see, the better you do, and the worse the average does, the more likely you are to win VG$.
|
Hang on... what's going on there.
If I do the equation you said on the 50% accuracy row, I get 1,400 for the 1k bet, which is $400 profit..... but for the 500 bet it comes to 700, or a $200 profit.
In fact your whole $500 bet column doesn't add up.... as long as everything else is the same, if you bet double you should get double the profit/loss..... so both "0"s should be on the same row.
|
you have to read the example up above. the person who bet 1k only had a 70% accuracy, while the person who bet 500 had 90% accuracy. You can easily see this by where they earn VG$0.
So to break even or make a profit the person who bet VG$1k needs the average accuracy to at or below 70%, while the person who bet VG$500 needs the average accuracy to be at or below 90%.
I could create a graph where all 3 variables change, but that would just get really confusing
