# Why The Martingale Betting System Doesn't Work

Last video I talked about a strategy of betting called the martingale system.
It’s a strategy where you always chase the total amount you’ve lost so far by betting
that loss plus 2.
So that when you eventually win, you win 2 guaranteed.
A lot of you saw the flaw in this system was that the amount of money you’d need to sustain
it would be infinite.
The issue is, even if you have a really huge amount of money, as long as it’s finite,
the expectation value from playing this game is 0.
That’s because, with a very small chance, you’d run out of money to keep betting and
you’d lose a lot.
I didn’t believe this until I did the calculation, because it seemed like you were so unlikely
to even lose many times in a row and so you didn’t really need to worry about getting
to the stage where you’d need to bet big, so you didn’t have to worry about folding.
But let’s do the calculation and see what happens.
Everyone has some number of losses, that would mean they will have lost enough money that
they wouldn’t be able to pay the amount they need to bet for the next round, so they’d
have to give up.
Let’s call this number of losses for a particular person, n. n can be very high or very low,
it doesn’t matter.
The person only has a 1 in 2^n chance of losing the maximum n times in a row.
If they don’t lose then they win 2- that’s most of the time, with a the probability (1-1/2^n).
But in the unlikely event they do lose, their total loss is this sum of losses so far, which
we know is 2^(n+1)-2.
So we put all of this together in the expectation value formula: the probability of not folding
times the gain plus the probability of folding times the loss... and it magically works out
to zero.
The way to explain this is, when you play this strategy, with very good probability
you win make a modest gain, but with small probability you will make a disastrous fall.
It averages out to 0 in expectation.
The higher your value of n, the less likely you’ll lose to you when you play.
as n approaches infinity, the chance you can fail goes to 0.
In this version, where the person can always afford to gamble more, they always make 2.
As a lot of you pointed out, the casino also has a maximum bet for this very reason- so
everyone has to stop eventually.
So this strategy clearly doesn’t work- if you play it lots of times you’ll end up
with 0 on average.
But what if you only played a few times?
And what if you capped your maximum bet at a very reasonable price- so that you really
wouldn’t mind too much if you did lose, even though it does mean you’d lose more
often.
Let’s try an example.
Say you set your maximum times of losing at 4.
The probability of that is 1 in 16.
So you’d think that you’d be pretty safe if you only played a small number of rounds
compared to 16, for example, 4.
You’d lose money unless you won every round.
But the probability of that is only around 77%- no were near a guarentee, especially
considering how much you stand to lose.
Plus the amount you can win is just 8.
In my opinion, you’d be way better off getting a job.
That solves the paradox entirely for the real world as far as I’m concerned.
But there’s still a paradox, in my opinion, mathematically.
Let’s look at the game where the person paying never runs out of money.
You might say, well that would require them to have infinite money already.
Not really.
Each time they play, they only bet a finite amount, and they can always borrow money to
play because they’re guaranteed to make it back.
You might say, well, there isn’t even an infinite amount of money in the world they
could borrow.
Sure, but maths doesn’t care about what sort of world we’re in -we could have been
in a universe with different physics that allowed the universe to be infinite, and the
maths would be the same.
So this is the paradox, mathematically, for me.
This random red or black game should have expectation zero.
You just feel like that must be true.
There’s no way to on average make money from a game of complete chance.
But what’s the expectation of the martingale?
Well it’s 2.
You always win 2.
Or using the mathematical formula, we can see the same thing.
In fact, I can make this number anything I want- for example, say I bet 50 more than
my total losses each time- then the expectation is 50.
But I can make it worse again, by using this strategy, say you’ve lost a few times so
far and it’s the nth round- then bet all that you’ve lost plus 2^n.
Look at the expectation value.
It’s infinite.
So the real paradox for me is that you can some how take a game that you think should
have expectation 0, and then actually make that expectation whatever you want, including
infinity, yet when you restrict to only ‘realistic’ cases, you get 0.
huh?
This paradox showed me that my intuition about expectation values is wrong- there are situations