this has been discussed before
here is the thread
https://bitcointalk.org/index.php?topic=939776.0
and this will be the best way to win 2 btc with 1 btc from the discussion of the thread
here is the thread
https://bitcointalk.org/index.php?topic=939776.0
and this will be the best way to win 2 btc with 1 btc from the discussion of the thread
Strategy 1:
You're always playing 9900x using a martingale sequence, i.e. your wager for each round is
Strategy 2:
You're constantly playing 20203 satoshi but increase the payout each round by exactly 1x starting at 4951x which will leave you with 1.00004850 BTC profit. You run out of funds after 4949 games and your last payout will be 9899x. You lose all of them with a probability of
I Hope the math is correct
You're always playing 9900x using a martingale sequence, i.e. your wager for each round is
Code:
ceiling((2 - PLAYER_BR)/9899,1e-8)
you run out of funds after 6861 bets. You lose all of them with a probability ofCode:
0.9999^6861 = 0.5035187605260536
or put differently you have a 49.64812% chance of winning.Strategy 2:
You're constantly playing 20203 satoshi but increase the payout each round by exactly 1x starting at 4951x which will leave you with 1.00004850 BTC profit. You run out of funds after 4949 games and your last payout will be 9899x. You lose all of them with a probability of
Code:
prod(1-0.99/i, i=4949..9899) = 0.5035283796177312
leading to a 49.64716% chance of winning.I Hope the math is correct
I think with any method the chances are always a little less than 50%.
So your math looks fine
