Quote:
Originally Posted by Yoshi63
The variance on the number of wins will be highest on a 50% bet. However game 2 will experience higher $-variance because the payout for each win is much larger.
This. Variance is the sum of squared distances from the mean divided by the number of events. The (dollar) mean in both cases is 0. In the coin flip, the distance is always the same (1) and 1 squared is also 1. It should be easy to see that the dollar variance per flip is 1.
In the dice version, it's 1 in 9/10 cases, and 9 otherwise. So for every 9 instances of -1 (which contributes 1^2 to variance) you get 1 instance of +9 (which contributes 9^2 = 81). So think of like a canonical streak of 10 events, with 9 losses and 1 win. Your variance would be
(1+1+1+1+1+1+1+1+1+81)/10 = 90/10 = 9
