Martingale: the maximum possible chain of continuous losses/profits - page 16

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Reshetov:
And it is clear to a drunken hedgehog that if we place N bets, the maximum possible series of losses in duration is N, because there is no way we can lose more than N times in a row. The probability of a maximal series of losses for N bets is (19/36)^N.
I'm certainly not a drunken hedgehog...but it's not clear how the total number of bets can be equal to the value of the maximum series. After all, the probability of the next outcome is 50%.(with a normal distribution of bets).....May I ask where this information comes from?
 
ivandurak:
You are confusing the probabilities of tails (if the coin is not bent it is 50%) and the probability of the next eagle being struck is in no way dependent on the previous series. With the probability of the series appearing. So the probability of a series of heads, heads, heads.... 20 heads, exactly the same as heads, tails, heads, tails,.... in that sense any series is unique. So it makes absolutely no difference whether you bet on heads or tails when using a martin.
It's not..the series has the same distribution as the data or do you mean to say that the seriesheads, heads, heads.... 20 heads falls as many times as the series heads, heads?
 
Hello, gentlemen martingale enthusiasts. I have some thoughts too, if you don't mind...
 
Andrey-F:
Hello, gentlemen martingale enthusiasts. I have some thoughts too, if you don't mind...
We look forward to hearing your thoughts.
 
Andrey-F:
Hello, gentlemen martingale enthusiasts. I have some thoughts too, if you don't mind...

"Sorry - my fault - of the villagers I am a savage" (c) :-), but I also don't mind...
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