175,833,708 had positive profit
179,380,878 had negative profit
2,722,802 had zero profit (and so I can't immediately tell if they won or lost, being 0 BTC bets)
>>> 175833708 * 100.00 / (175833708 + 179380878) = 49.5007
49.5007% of the non-zero bets at 49.5% won!
Wow , that seems really accurate to the odds.
Mathematically , what would the average error % of the odds be ? That seems too close to the given odds to be true.
The
law of large numbers says that the more bets we have, the closer the actual percentage win rate should get to the expected percentage win rate. We've had over 300 million bets at 49.5% so it's not surprising that it's close.
Of the 355214586 non-zero bets at 49.5%, we would expect to see 175831220.07 wins and 179383365.93 losses, but we saw 175833708 wins, which is 2487.93 too many. So in a sense we're not close to expectation - we saw over 2 thousand too many wins. But in another sense we're very close - because that's only 0.0007% too many.
As time goes on, we don't expect that 2487 number to get any smaller. We just expect the total number of bets to get bigger, which causes the 2487 number to become a smaller percentage of the total.
I know that doesn't answer your question, but that's because I don't know the answer.
A cheap answer would be "the average would be 0% error, because all the positive errors cancel out all the negative errors". But you're probably asking for the average of the absolute value of the error, which I don't know how to calculate.