In a box, there are 51 red balls & 49 black balls.
You pick randomly.
Pick the red ball, you win.
There are four payout options:
games.png
Intuitively, the 4 games are the same, but game 4 has the highest EV. Also, game 4 has the highest variance.
My second question is, how to calculate the standard deviation of game 1?
It should be close to 1, but what exactly is it? What's the formula for the calculation?
We usually express EV as a percentage of the initial bet and variance as the squared units of the initial bet. When expressed this way, the EV and variance of each game is the same.
[edit] The above assumes that you bet $1 in the first game and $2, $5 and $100 respectively in the other games. If you bet $1 in each case, then yes, you are correct that each subsequent game has higher EV and variance.
Last edited by Gronbog; 11-21-2018 at 07:58 AM.
Once you accept that the EV is 0.02 units or 2%, then the variance is:
(0.51 x (1 - 0.02)^2) + (0.49 x (-1 - 0.02)^2) = 0.9996
The standard deviation is sqrt(0.9996) = 0.99979998
[edit] this calculation is correct for game 1 assuming you bet $1 and is correct for the other games assuming you bet $2, $5 and $100 respectively.
Last edited by Gronbog; 11-21-2018 at 07:59 AM.
If these are expressed as a percentage of your initial bet then they are exactly the same, assuming you bet game1 $1, game2 $2, game3 $5, and game4 $100 for the payouts shown. Plus you should factor in BR.Originally Posted by San Jose Bella
The standard SD formula. First calculate the average. The take the square root of the sum of square of the differences from the average for each point and divide the square root by the number of data points. Hopefully memory didn't fail me. Feel free to search the internet to make sure I got it right. You probably should have done that to begin with.
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