so today i was completely wiped out by the dice. i was playing the "1/8 bet" address:
lessthan 8000 1dice6YgEVBf88erBFra9BHf6ZMoyvG88 12.2070% 7.910x 3.000% 97.000% 0.0010 25.0000
after 75 consecutive losses i was wiped out, because i was placing relatively large bets. i have decided not to play this game any more.
i must be very unlucky or because if i'm not mistaken, the probability of getting 75 consecutive losses on a 12.20% dice are 1/(7/8)^75 . so basically this occurs once in every 22,356 turns, assuming a house edge of 0%. if the house has an edge of 3%, the occurrence is slightly more frequent 1/(0.87875^75) = once every 16,221 turns... still, i must be one unlucky bastard...
The house edge is determined by the payout when you win. The chance of winning is always the same - it picks a number between 0 and 65535 and you win if the number is less than 8000. Your calculation is off a little because 1 - 8000/65536 isn't 7/8, it's a little more than that.
I make it a one-in-17399 chance of losing all 75 bets, so still really unlucky. But then again 1-in-17k shots do happen. About once every 17k bets. And there have been over 600k bets.
>>> 1 ((1 - 8000/65536.0) ** 75)
17398.822919111808Here's the same for losing a 50/50 (less than 32768) bet 3 times in a row to sanity-check the calculation:
>>> 1 ((1 - 32768/65536.0) ** 3)
8.0and as you would expect, that's a 1-in-8 chance.
What's puzzling me is the payout for the SatoshiDice bets. A 0% house edge would pay out 65536/8000 = 8.192x. And so a 3% house edge should pay out 3% less than that, or 7.94624x. So why does the SD website say it pays out 7.910x? 7.910 * 8000/65536 is 0.965576171875, ie. a house edge of 3.44%.
Not that it makes any difference to you, since you only get the payout if you win a bet...