I got the latest numbers from dooglus post and tried the theoretical formula for consecutive losses in a row and came up with this.
bets p-win q-lose loss-row
604294 0.5 0.5 18.20489587
1118230 0.48828 0.51172 19.71766168
244037 0.36621 0.63379 24.99911792
243386 0.24414 0.75586 39.27274907
94075 0.18311 0.81689 48.22841691
182495 0.12207 0.87793 76.89902593
So, those playing the 50% game can expect to lose 18 or 19 times in a row if they played that game since the beginning. Those playing the other games can expect to lose much more.
2. p is probability of winning the game you're playing, q is the probability of losing the same game, q = 1 - p
3. Then expected losses in a row = -log(bets*p + 1)/log(q)
Did I do this correctly?
It looks fine, as long as you have the win/lose probabilities right.
You can also calculate the expected number of rolls before n losses in a row:
n = -log(bets*p + 1)/log(q)
n = expected losses in a row
bets = number of bets
p = probability of winning the game
q = 1 - p
So
-n*log(q) = log(bets*p + 1)
(1/q)^n = bets*p + 1
Expected number of bets before n losses in a row = (q^(-n) - 1) / p
I've seen a few methods of proving this. The one I have to hand uses Markov chains and is a bit dense, and I can't find the other which is more inuitive to follow. If I find it I'll post it.