And more specifically, if we have 20 losing bets in a row, then the probability of the next 20 losing bets will be the same, right? But in this case these probabilities are different from the probability of 40 losses in a row, which seems paradoxical. Thus, in a sense, previous outcomes affect the probability of future rolls...
I don't see a paradox here:
If A and B are independent events then P(A then B) = P(A) times P(B).
A is "lose first 20 bets"
B is "lose next 20 bets"
P(A) is equal to P(B)
P(losing 20 bets) * P(losing 20 more bets) = P(losing 40 bets)
I see your point but it is not strictly the same. We know the probability of each loss in a row of 40 successive losses (so we know the area of a square beforehand), but with each roll made the probabilities for rolls yet to be made change, right? Therefore, it turns out that the outcome of each previous roll changes the probability of the next roll...
I don't find a fallacy in this logic, do you?
Before you start, it's 2^40 against that you will lose 40 times in a row.
If you lose your first bet, you only have 39 more to lose to reach your 'goal', so it's 2^39 against that you will lose 40 times in a row.
You've already 'made progress' and so have doubled your chances of 'success'.
Maybe reading about Bayes' theorem will help:
A = "we lose all 40 bets"
B = "we already lost the first 20"
The probability that (we lose all 40 bets), given that (we already lost the first 20) is:
P(A|B) = P(B|A).P(A)/P(B)
= (probability that we lose the first 20 given that we lost all 40) * (probability that we lose all 40) / (probability that we lose the first 20)
= 1 * 0.5^40 / 0.5^20
= 0.5^20
We had a bad case of this "I just lost 10 in a row so I must be due to win soon" syndrome in Just-Dice last night:
Jesus he went crazy, he could have started betting 1 btc not that much
He was betting CLAMs, not BTC, so it's about 200 times less bad than it looks.