Streaks are actually quite expected, but it does depend on how many times you rolled in that hour too. For example, the expected number of times to get a streak of size S in N rolls is roughly (1/2)^S * N. For 7-streaks, you'd "expect" one to show up about every 128 rolls.
(Ugh I hope I applied linearity of expectation correctly here. I'm getting rusty)
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If i remember correctly this all has to do with the "independent event" problem.. Marky does seem to be "unlucky" but with each roll being an independent event this streak could go on for quite a while.
I mean even if you tried to play a 100 games at 1% chance of winning, you still only have around a 60% chance of winning at least once in a 100 games....
I dont like those odds... Ill get some tokens from delegating but thats as far as im willing to go..
The games i did play, i won all of them so better to quit while im ahead. :D
It's true. However we can evaluate the probability of it happening in a given set of rolls. Definitely some funniness with the statistics of the rng. But it appears to be a known behavior. The real requirement is that the values cannot be predicted. We'll see if there's any flaw in the current setup. Possibly the formula shows that every 10 blocks the rolls within may be correlated in some way. Especially since the only different numbers are block num + client hash (controlled by user). Not for me to break though, I'm terrible at this kind of thing.
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Not as often as I see them. Within 5 minutes, I had 4 7+ losing streaks.
How many rolls in those 5 minutes though? But yeah, depends.
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It happens constantly, back to back.
Oh.... That does seem weird then...
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And I just realized now on a different interface I can see the string of losses together. Wow. And you are playing near-even... Can you tell me of the games you played with this configuration, how many of them were wins out of how many total? If you were playing these odds, this is definitely not what I'd expect. 11 wins out of 83 where the probability of winning is 44/100.
Let's see... Central limit theorem will tell us the distribution of number of wins (under certain conditions which is satisfied here) will be very close to normal distribution with mean np and standard deviation sqrt(np(1-p))...
Mean 83*.44=36.52 Stdev Sqrt(83*.44*.56) = 4.52You are about 5.62 standard deviations from the mean in this batch, and that's already very rare....
I know that random number generator isn't truly random, but hum.....