For a zero edge game...
Probability of Success = (Starting Bankroll)/(Starting Bankroll + Target Profit)
For a low house edge game, it gives a decent probability estimation if you wager a modest proportion of bankroll on each wager, especially so with a progressive such as Marty or d'Alembert
I acknowledge that it's far more complex, and beyond me to derive an equation where there is a house edge.
Actually, for flat betting, there is a formula for the probability of reaching the target before losing the entire initial bankroll:
(1 - ((1-p)/p)^B) / (1 - ((1-p)/p)^(B+T))
where p is the probability of winning the initial bet (assuming p < 1/2), B is the initial bankroll (in terms of the number of bets), and T the target profit (again, in terms of the number of bets).
For p = 1/2, the probability = B / (B + T), as you state in your blog post.
Can someone help me? I just want to post my own form questions but I don't know how.
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I'm not saying it's a long run winner
I mean for playing for fun
as far as I know it doesn't have a name
I think it's somewhat similar to what John Patrick (no I don't recommend him) calls 𝙪𝙥 𝙖𝙣𝙙 𝙥𝙪𝙡𝙡 but it is not the same
the progression is 4 bets and if you lose any one or win all you return to the beginning
assuming you win your first 3 bets the progression is:
𝐛𝐞𝐭 𝟏............𝐛𝐞𝐭 𝟑............. 𝐛𝐞𝐭 𝟐...................𝐛𝐞𝐭 𝟒
the worse thing (other than long losing runs) is if you lose your 2nd bet - 3 - which means in the first 2 bets you will lose 2 instead of breaking even if you flat bet
if you win the 2nd bet and lose the 3rd you will win 2 - whereas if you flat bet you would win only 1
if you win the 3rd bet but lose the 4th you will win 2 - which is the same as if you flat bet
if you win all 4 bets you will win 𝟏𝟎 𝐮𝐧𝐢𝐭𝐬...........................𝘵𝘩𝘢𝘯𝘬 𝘺𝘰𝘶 𝘑𝘦𝘴𝘶𝘴...................which means you've won 6 more than if you flat bet