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?
Well, let's do the maths here.
The odds of losing on the next roll is still 8%, because each roll is independent.
But if you look at it before you do the rolls, then two L's next to each other has a probability of 0.64%. However, if you have already hit a L on 92%, you can't say that the next roll I do only has a 0.64% chance of losing. It still has a 8% chance of losing.
It is fairly common for 3 reds or even 4, 5 reds to occur on 92%.
In the end it doesn't matter because the house will always win. Some day your luck will run out, and you hit on the house edge.