Quote from: nicolaennio on December 07, 2013, 10:39:47 PM

I reply since I was originally doubting it, maybe I can tell you how I see this point now.

The expected loss is less than 1.27BTC because on average you bet much less than 127BTC (in the above case it is roughly 7BTC on average). If you compute the average bet size, which is much smaller than the worst case one, you can see that the expected return is -1% and this is the only possible expected return. However, if you specify that your strategy can lose at maximum 127BTC and wins at maximum 1BTC, you can find different ways of betting the money with different average bet sizes and thus different expected losses. Betting all the 127BTC is probably the worst way :-)

That is not correct I think. It is true that you can never turn a negative EV into a positive, but you can make it less bad. The average bet size is irrelevant. The martingale strategy dooglus used in his example has only two outcomes: either you win 1 or you lose 127. The probability of losing is q_m=(1-p)⁷ and therefore the expected return is EV=1*(1-q_m)-127*q_m.

Just as dooglus pointed out earlier, the surprising thing is that q_m is much smaller than the probability of losing a single bet with the same return, i.e. the following two strategies have exactly the same payoffs, but the probabilities are not the same:

Single bet with bet size 127:
p=0,982265625
EV=p*1 - (1-p)*127 = 0,982265625 - 0.017734375 * 127 = -1.27

Martingale starting at 1
p = 1-0.505⁷ = 0,99162394
EV = p*1-(1-p)*127 = 0,99162394 - 0,0083760574 * 127 = -0,0721353 !!!!

The martingale is clearly much better. The expected value is much closer to zero than the single bet.

Quote from: SebastianJu on December 07, 2013, 11:01:43 PM

@nicolaennio... no martingale doesnt work. You cant beat luck this way. I hope you wont try it to make a fortune since you only will lose at the end. If you dont believe me ask google for explainations.

Prove me wrong.