So I tried this strategy at a dice site and I'm thinking of the odds of winning
So I set the chance of winning to 92% and start with minimum bet of 1000 Satoshis with profit of 76 Satoshis
Now, I could win like 9-11 times in a row, sometimes higher and lose
After losing, what would be the chance that I could lose again twice in a row?
I'm thinking of if I lose with that settings, I could go all-in with higher bet since my chance winning is higher.
What do you guys think?
I know what you are talking about. This is a known strategy which
must work as many gamblers think and I myself also was part of them. What are the odds of losing two times in a row playing with 92% winning chance? Yes, we all know that the outcome of the previvous bet has no impact on the outcome of the consecutive one, but still it seems the chances of two reds in a row are extremely small. So after losing we increase our base bet 10 times and, in most cases, we recover the previvous loss. But the thing is that it
is possible to lose two, three or even four times in a row playing with 92% winning chance. And when this happens we lose it all.