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Essentially I am confused why I am getting a positive answer for the martingale working.
Chance of losing = 1/256 aka 2^8
Chance of winning = 255/256
((255/256) x 5) - ((1/256) x 640) = 2.48
(chance of winning five dollars) - (chance of losing 640dollars (which is how much youd be risking if you used the martingale)) = 2.48 expected win?!!
I KNOW the martingale doesn't work. What am I doing wrong? Is my math just terrible?!
Follow up question: if there was a $1 minimum table with no maximum bet, aka you can bet infinite, would the martingale work?
Than you so much
Quote: CrapLostInCRAPsFollow up question: if there was a $1 minimum table with no maximum bet, aka you can bet infinite, would the martingale work?
Than you so much
Yea, if you're the Gazillionaire.
By the way, check out his act under Caesars tent, he's a riot. Best show on the strip.
Quote: terapinedYea, if you're the Gazillionaire.
By the way, check out his act under Caesars tent, he's a riot. Best show on the strip.
Loved Absinthe - great show! Not a bad seat in the tent!
Quote: CrapLostInCRAPs
Follow up question: if there was a $1 minimum table with no maximum bet, aka you can bet infinite, would the martingale work?
If you can bet infinite, that means you have an infinite bankroll.
So, if you have an infinite bankroll, why gamble at all? You cannot increase infinite.
Quote: CrapLostInCRAPsSay we are playing roulette, 1 to 1 payout. $5-1000 bets.
Essentially I am confused why I am getting a positive answer for the martingale working.
Chance of losing = 1/256 aka 2^8
Chance of winning = 255/256
You are not always losing or winning exactly 1 unit for any given sequence of 8 events. You have to cycle over all sequences of 8 events XXXXXXXX where X is either a win or loss, each of which has a different win/loss amount. The probability and total payout of each sequence will vary according to the specific game you are considering. If you do that with the usual spread sheet combinatorial analysis, you will get the right answer, no matter the game.
Quote: CrapLostInCRAPs
Chance of losing = 1/256 aka 2^8
Chance of winning = 255/256
((255/256) x 5) - ((1/256) x 640) = 2.48
First of all, chances of losing is not 1/2^8 (that would be on a fair game), it is (19/37)^8 or (20/38)^8 on a single or double zero wheel.
Second, after losing the 8th try, you lost 1275 (your 640 is the 8th bet, but you lost 1st to 7th bet already).
The correct formula should be
(1 - (19/37)^8) * 5 - (19/37)^8 * 1275 = -1.19 (single zero)
or
(1 - (20/38)^8) * 5 - (20/38)^8 * 1275 = -2.54 (double zero)
Wonderful and simple answer! I'm ashamed of never having thought of it that way! Kudos, Keyser!Quote: KeyserSoze
So, if you have an infinite bankroll, why gamble at all? You cannot increase infinite.
Inf + 1 = Inf, thus you CANNOT win!!!
Quote: KeyserSozeIf you can bet infinite, that means you have an infinite bankroll.
So, if you have an infinite bankroll, why gamble at all? You cannot increase infinite.
I beg to differ:
https://www.youtube.com/watch?v=2VSYmGSJtCA
Quote: KeyserSozeSo, if you have an infinite bankroll, why gamble at all? You cannot increase infinite.
Actually, infinite comes in different sizes.
Hence "You cannot increase infinite" is the wrong explanation why a player should never play a martingale even with infinite bankroll. Because if this explanation where true, the casino should never offer such a game (which is clearly wrong).
But inf + Inf is still Inf, so no avail. You would need to have an infinity of players. And yet this still would not make your countable infinite a continuous one, so, no again.
Anyway, we are talking "player with an infinite bankroll", and he is playing against one casino. We didn't say he would be playing an infinity of rolls or in an infinity of casinos.
Sorry, I can't see what you deem "clearly wrong"?