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So I know the house always win and someone has already figured this out. But I can't see the catch for my self

I was watching this video Best System for Roulette

Basically progressive betting on a 2-1 bet of the groups of 12 numbers

SpinbetwinProfit
1100200200
2100200100
3150300100
420040050
530060050
645090050
7675135050
81025205075
91525305050
102300460075
113450690075
125138102761
15413

So you are risking 15,413 to win about $73 on average. That would mean you have to win 212 times for each loss. I calculate the odds of a group of 12 not being hit 12 spins in a row as ~1 in 95 chance (26/38)^12 = 0.0105. So clearly it is quite possible you go bust.

Now here is the catch to the video. The user says wait until there are 5 non wins thus now the wheel would have to have 17 spins in a row with out a win. So I calculate those odds at ~1 in 633 chance (26/38)^17 = 0.0015.

Now it seems reasonable, but what am I missing I know each spin is independent of the history of spins so why does it look like the chances of not busting have gone up?

I can reason that Person A is playing 12 spins and 12 spins only regardless of if Person A knows the history of the spins or not.

Where did I make a math mistake?

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It does not matter how many spins have happened in the past, as you say each time is different.

So wait until you have 5 non-win spins. What is the probability of other $12$ non-wins? It's still $\left(\frac {26}{38}\right)^{12}$

Before you started spinning, having $17$ non-win in a row is unlikely (well, more unlikely than $12$). After you already have 5 non-wins, you have additional information: you know that the first 5 already are non wins, so your probability changes.

To make it a little bit more formal there is difference between

$$P(\text{ $17$ non wins}) = \left(\frac {26}{38}\right)^{17}$$ and

$$P(\text{$17$ non-wins}\mid\text{ $5$ non wins already})$$ ($P(A \mid B)$ means probability of $A$ given $B$, given the additional information that $B$ has happened)

and in fact $$P(\text{$17$ non-wins}\mid\text{ $5$ non wins already}) = P(\text{$12$ non wins})$$

As a last example consider $P(\text{tomorrow rains})$ and $P(\text{tomorrow rains} \mid \text{tomorrow I can see the sun})$

Clearly the latter is equal to $0$ since if I can see the sun, then it's not raining. But in general the probability that tomorrow rains (the former one) is not $0$. With additional information probabilities change

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