Quote from: arallmuus on March 11, 2015, 12:45:30 PM

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

Quote from: blockage on January 29, 2015, 04:24:18 PM

Strategy 1:
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 of

Code:

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 Tongue

Can you explain to me how it is possible to decrease the house edge overall by using such a strategy?