Is there a mathematician in the house?
How do I analyse the odds of this strategy working?
He bets at 66%, which pays out 1.5x. He starts at 0.06 BTC (this first bet is not shown in the screenshot), doubles on loss, halves on win, never halves below 0.06.
So it's a random walk, which makes a net profit of 0.03 BTC each time it gets back to betting 0.06. Steps down are about twice as likely as steps up, so it seems unlikely to reach max bet and bust very quickly.
But the question is how do I calculate the probability of such a progression busting, given that he can afford to go N steps up the random walk?
Is it a Markov Chain thing? Or how do I analyse it?
How do I analyse the odds of this strategy working?
He bets at 66%, which pays out 1.5x. He starts at 0.06 BTC (this first bet is not shown in the screenshot), doubles on loss, halves on win, never halves below 0.06.
So it's a random walk, which makes a net profit of 0.03 BTC each time it gets back to betting 0.06. Steps down are about twice as likely as steps up, so it seems unlikely to reach max bet and bust very quickly.
But the question is how do I calculate the probability of such a progression busting, given that he can afford to go N steps up the random walk?
Is it a Markov Chain thing? Or how do I analyse it?
I think you can solve it like this. Consider the number of times the bet has been doubled as a simple random walk starting at zero. If we reach 1 we win and if we reach -N we have busted. Let P{k} be the conditional probability of reaching 1, given that we start at k. Then P{k}=pP{k+1}+(1-p)P{k-1} and we know that P{1}=1 and P{-N}=0. p is the chance of winning each bet (=0.66)
The characterisic equation x=px^2+(1-p) has solutions 1 and (1-p)/p (call it r), so then P{k}=c+dr^k=(r^(N+k)-1)/(r^(1+N)-1).
The answer is then given by P{0}=(r^N-1)/(r^(1+N)-1)