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IGoUnseen t1_is14fre wrote

Reply to comment by [deleted] in Roulette simulation: 10'000 players each betting 5'000 times (74% chance of having a negative return ) [OC] by nautilus_red

That's only because in your example he's betting twice as much. A two unit bet on black is the same house edge as two one unit bets on different dozen bets.

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GoodellsMandMs t1_is14v1r wrote

youre right, its not a strategy to lose more money, its just a strategy to lose money faster lol

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IGoUnseen t1_is17qg5 wrote

Not really, only if you are betting twice as much. If you're average bet would be normally 10 dollars, and you instead bet 5 dollars on each of those, it's equivalent to one 10 dollar bet.

This applies more generally to all bets in roulette. Every single bet in roulette has the exact same house edge and can often be replicated by multiple bets of a different one. I.E. one double number bet has the same odds as two single number bets of the same amount. Making two dozen number bets is equivalent to a hypothetical 24 number bet that just isn't normally offered at the table.

Basically, how much you lose at roulette is 100% based on the volume of your overall bet. Which numbers/bets you select does not matter. The only thing it matters towards is variance. Some people want to have rare large wins and frequent small losses, some people want frequent small wins and rare large losses. The selection of numbers you pick will affect that, but again not the expected amount lost.

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