Hi guys,
something weird I stumbled across when studying a small martingale. I used to think that if the house has an edge of p (say 1/100), than on average it will earn from a bet 1/100 of that bet.
However, lets see the following simple case, where the bankroll is 1 clam, the player is playing with 1 clam at a time, with 0.505 probability of losing his/her clam. The player plays two times in a row. Then, with probability 0.505*0.505 the bankroll will become 3 clams and with probability 0.505*0.495 it will stay to 1. All other times the bankroll is taken by the player. The "expected bankroll" is thus 1.01505, which is "much" less than the expected 1.02!!!
What it seems happening is that the fact that the bankroll cannot recover from zero, and this cuts out some profitable plays from the player. I reworked the same example, but with a bankroll of 3, and its expected value is the "correct" 3.02.
I am quite surprised by this and I am not even sure I did a sound reasoning, I would really like to hear your opinion. If this was already a well established fact please forgive me, references would be appreciated.
You're saying the house has 1 CLAM, and is willing to risk it all on a single bet I think. We wouldn't actually do that, but for the sake of argument I'll let you have that.
So there are 3 possibilities:
1) the player wins his first bet, the house is bust, probability = 0.495, profit = 1
2) the player loses his first bet and wins his second, probability = 0.505 * 0.495, profit = 0
3) the player loses both bets, probability = 0.505 * 0.505, profit = -2
So the player's expected profit is:
>>> 1 * 0.495 + 0 * 0.505*0.495 + (-2) * 0.505*0.505
-0.01505
as you said.
If the player was always making 2 bets then his expected profit will be -1% of 2, or -0.02. So why is his expected profit greater in this case?
Well, the reason is because sometimes he's only making one bet. His expected risk is smaller, and so his expected losses (being 1% of his expected risk) are proportionally smaller.
What's his expected risk? Well, with p=0.495 he bets 1 unit, and with p=0.505 he bets 2 units. So his expected total stake is:
>>> 0.495 * 1 + 0.505 * 2
1.505
This matches precisely -100 times his expected profit, as expected.
So, what you are saying is you can decrease the house edge (in you example from 1% to 0.7525%) in your favor,
Not really. You can't change the house edge. You always expect to lose 1% of the amount you risk.
What he's saying is that by sometimes stopping after risking only 1 unit, your expected total risk is less than if you always risk 2 units, and so your expected loss is less.
Less risk = less expected loss.
