This comment was posted to reddit on Jun 18, 2019 at 8:26 am and was deleted within 16 hour(s) and .

Lets have a look at a similar game with coins: If you throw n heads in a row you lose all your money. If you throw a tail before that happens, you win one unit.

This is essentially the same game as playing fair blackjack with his strategy. If he wins, he wins one unit. If he loses n times (n is unknown at the moment) in a row, he has no more money to bet.

So, what is n in his strategy? Lets assume that he has x units at the beginning and/or the betting limit is x units.

Because of his strategy he needs to double his bet every time he loses. Therefore he needs to bet 2^{n} units after n lost games in a row. That means with a budget or betting limit of x units he has lost the whole game after only floor(log2(x+1)) =: n lost games in a row.

Lets assume the coin is fair, then the probability of throwing n heads in a row is 1/2^{n} =: p. The probability of winning (or not throwing n heads in a row) is 1-p.

You can see that the probability for winning one such game is quite high for a decent value of x. But the problem is that you are only winning one unit, but lose x units when you korse just one time.

We win 1 unit with probability 1-p. We lose x units with probability p.

We can now calculate the expected amount of money that you can win in one such game:

1*(1-p) + (-x)*p = 1 - (x+1)*p

For easier calculations assume that x=2^{m-1} for some m>=0. Then n=floor(log2(2^{m-1+1))=m.} And the expected amount of money won after one game is:

1 - (2^{m-1+1)} * (1/2^{m)} = 0.

If we now assume the game is good for the bank, then our losing probability p is even greater and the expected amount of money won in one game becomes negative. Therefore we lose money on the long run when we use his strategy.