conditions where there is no max? What are the odds that you would get bankrupt if playing a Martingale system where the Min bet is $1 and the Max bet is. $10000. How often will you lose 20+ hands in a row playing perfect BS?
conditions where there is no max? What are the odds that you would get bankrupt if playing a Martingale system where the Min bet is $1 and the Max bet is. $10000. How often will you lose 20+ hands in a row playing perfect BS?
Norm, I posted it because someone claimed that using that system and losing it all is unlikely at a min. $1. Max $10,000 game as one might lose 50 or 75 or whatever hands in a row it takes to reach that $10k Max las a less than .01% probability likely. Thus, the player at such a game is risking little.
Your friend is an idiot. Work it out for yourself. You queried 20 straight losses at $1 min start bet. You will lose approx $. 2,093,024. You'll go over max on, I think, the 14th loss. And if you happen to win the 13th - congratulations - you're up $1. Rough calc as I did it in my head vs a spreadsheet.
I did put it on a spreadsheet. At the 13th loss, and on his 14th try, his bet will be 8192, with a cumulative investment at that point totalling 16383. If he finally wins, the princely sum of $1 may be added to his net worth. After the 20th consecutive loss, assuming no table max, his wager will be $524288, with a total investment of $1048575. If that wager should be won, he may then add the princely sum of $1 to his net worth.
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