Keep all-in a game with 0% EV, money kept going down? why???
Game rule:
50% chance win 5%, 50% change lose 5%.
So EV=50%*5%+50%*(-5%)=0
If Starting bankroll=$100, no matter win or loss, I keep all-in for 200 times.
In average, I would win 100 times, lose 100 times.
$100*(1.05^100)*(0.95^100)=$77.855704
I assumed I would end up with $100 because this game has 0% EV, but the more times I all-in, the more I lose. Why is that?
Quote: ReyGarciaI'm really confused, someone please enlighten me.
Game rule:
50% chance win 5%, 50% change lose 5%.
So EV=50%*5%+50%*(-5%)=0
If Starting bankroll=$100, no matter win or loss, I keep all-in for 200 times.
In average, I would win 100 times, lose 100 times.
$100*(1.05^100)*(0.95^100)=$77.855704
I assumed I would end up with $100 because this game has 0% EV, but the more times I all-in, the more I lose. Why is that?
Let's look at two outcomes only for illustration:
Event 1 Win/Lose
You bet $100 and win, your total is now $105.
You bet $105 and lose 5%, now your total is $99.75.
Event 2 Lose/Win
You bet $100 and lose, your total is now $95.
You bet $95 and win, your total is now $99.75.
The inevitable thing is that, as your bankroll goes up, the 5% that you risk losing becomes a larger and larger amount relative to the amount that you are betting. As you lose, the 5% that you stand to gain on the next attempt is smaller than it was before you incurred the most recent loss.
If you could simply replenish your total bet to $100 every time you lose and reduce your total to $100 every time you win (pull winnings out) then your result after 100 wins and 100 losses would be that you would have $100. You'd always win or lose $5.
No matter how I varying my bet, the total amount of bet result/number of bets=$100, but if I win half of the time, lose half of the time, my ending result would certainly be less than $100.
Quote: ReyGarciaNo matter how I varying my bet, the total amount of bet result/number of bets=$100, but if I win half of the time, lose half of the time, my ending result would certainly be less than $100.
How do you get "the total amount of bet result/number of bets = $100"? If your first bet wins, your next bet is $105, so the total amount of bet result/number of bets after two bets is $102.5, isn't it? Similarly, if your first bet loses, your second bet is $95, so the average is $97.5.
So in the first round you land up on $32 or $96; then continue to game for a total of four rounds.
.. .. .. .. 64
.. .. .. 32 .. 96
.. .. 16 .. 48 .. 144
.. 8 .. 24 .. 72 .. 216
4 .. 12 .. 36 .. 108 .. 324
1 4 6 4 1
You can see that the most likely event is to get to $36 but if you were lucky enough to have a winning streak you could win a fair bit. This offsets the lower values of having an average or losing streak - if you add up 1x$4 4x$12 ... 1x$324 you get an average of $64!
Quote: ReyGarciaYou're right, but it seems to imply "betting systems work", that I can steer a 0% EV game to become a negative EV game by varying the bet amount, which I don't understand.
No matter how I varying my bet, the total amount of bet result/number of bets=$100, but if I win half of the time, lose half of the time, my ending result would certainly be less than $100.
You can do the reverse also. Just bet less when you win and more when you lose. And end up a winner if you hit 50/50.
Quote: unJonYou can do the reverse also. Just bet less when you win and more when you lose. And end up a winner if you hit 50/50.
In fact, isn't that one of the "selling points" of the D'Alembert system?
Quote: ThatDonGuyIn fact, isn't that one of the "selling points" of the D'Alembert system?
Yup.
The OP system is sort of a watered-down reverse martingale where you bet more when you win and less when you lose. It won’t change the EV, but it will change the distribution of outcomes around the EV, with more frequent losses that are small and less frequent wins that are large.
It’s the same concept as the martingale system leading to frequent small wins and infrequent large losses.
Quote: charliepatrickThis offsets the lower values of having an average or losing streak
Thanks, that's a very clear explanation.
A 50/50 chance game doesn't mean guaranteeing win 50 times and lost 50 times.
(Winning streak excess)=(losing streak deficit)+(same number of win/lose deficit)
Quote: ThatDonGuyHow do you get "the total amount of bet result/number of bets = $100"? If your first bet wins, your next bet is $105, so the total amount of bet result/number of bets after two bets is $102.5, isn't it? Similarly, if your first bet loses, your second bet is $95, so the average is $97.5.
Charliepatrick explained it very well.
The EV is still zero. What you've stumbled on is the difference between the arithmetic and geometric mean. The linear growth is zero while the exponential growth rate is negative, and the latter is more relevant here because you're wagering percentages rather than fixed dollar amounts. This is where the Kelly Criterion comes in handy. It maximizes the exponential growth, in turn maximizing the median outcome. In this case, it would tell you not to play at all.Quote: ReyGarciait seems to imply "betting systems work", that I can steer a 0% EV game to become a negative EV game by varying the bet amount, which I don't understand.
If you're not familiar with the Kelly Criterion, here's a great paper on it - http://www.eecs.harvard.edu/cs286r/courses/fall12/papers/Thorpe_KellyCriterion2007.pdf
