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We have a strategy that has won us a fair bit of money on the roulette table and I want to try and understand the odds a bit better. The strategy is as follows:

  • Select either red or black and always bet the same color
  • Bet the table minimum, let's say \$10, on the first bet
  • If we win the bet we double up for a total profit of \$10
  • If we lose, then we triple the next bet to \$30
  • Every time we win we collect our winnings and start again at the minimum bet (\$10)
  • Every time we lose we triple the bet until we win again (eg. bet 1: \$10, bet 2: \$30, bet 3: \$90, bet 4: \$270 etc)

The table we're playing is an American Roulette table which has a 47.4% probability of winning a red/black bet. There is no table limit on the table and so the only way I see this being a losing strategy is if you're unable to cover your triple up (eg. you don't have sufficient bank roll) and you lose X number of times in a row.

The reason we triple up when we lose is to ensure we cover our lost bet and always make a profit from a winning bet.

The only time we would not triple up is when we get back to our starting stack (let's say we start with \$150 and have won \$300 for a total of \$450 then if our triple up was going to take us below our starting point then we would just start again at the minimum bet of \$10 - this further reduces the risk of the strategy).

To help with this I'd be keen to know what the odd's are of losing 2, 3, 4, 5, and 6 times in a row if we select the same color. I'm also keen to know if there is anything that we haven't thought of that could trip us up here.

Thanks folks.

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    $\begingroup$ Let's be absolutely clear: There is no long-term winning strategy at Roulette. It's simply not possible when all bets are losing propositions to find a combination of them that wins. I realize that is not what you are asking, just noting that. $\endgroup$ – Thomas Andrews Oct 11 '15 at 23:21
  • $\begingroup$ You haven't actually expressed the question here. " want to try and understand the odds a bit better." $\endgroup$ – Thomas Andrews Oct 11 '15 at 23:36
  • $\begingroup$ Also, this strategy, as stated, means if you lose the first bet, you still bet the minimum? The first round would have to be all "random" until you got more than your starting point. $\endgroup$ – Thomas Andrews Oct 11 '15 at 23:39
  • $\begingroup$ Hey @ThomasAndrews - Would it be true to say that if we started with \$150 and tripled up on these losses we would therefore need to lose the first 3 in a row on a regular basis for this not to work a reasonably high percentage of the time? And if that is true then what would be the odds of losing 3 times in a row? $\endgroup$ – bwash70 Oct 11 '15 at 23:46
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The strategy you're discussing is well-known and ever-popular Martingale, except the best-known version doubles the stakes after a loss instead of tripling them. No, it's not a particularly good strategy: your outcome is either a small win with high probability or a huge loss at a small probability, with the expected value being negative. Casinoes earn their living from gamblers who think they've got an unbeatable strategy but underestimate the probability of the worst outcome happening.

Basically, since your probability of winning is $18/38$ when playing black or red, but the payoff is only $1:1$, you won't be making certain money with the roulette wheel. The order of your bets placed doesn't matter - you're going to lose a few percent of them on average.

Since you ask for probabilities of consecutive losses: it's $(10/19)^n$ for $n$ consecutive losses, with the total cost being $10 \times 3^n$ dollars. The probabilities are quite high: $(10/19)^6 \approx 2\%$, so for every fifty people trying your strategy, about one will start the game with a fine 7290 dollar losing streak. Whether you play the same color, different color, odd or even every time doesn't matter.

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  • $\begingroup$ Hi @kviiri - Thanks for the answer, that's what I was looking for. I can now happily call this a bit of fun as opposed to an ultimate winning strategy :) $\endgroup$ – bwash70 Oct 12 '15 at 21:39

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