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I would like to know the odds of winning at roulette when playing the dozens or columns. This means that out of a total of 36 possibilities, you place a bet on twelve of them. If any one of these twelve numbers hit, you win.Therefore if playing one time, the probability of winning is obviously 33% (not including zeros here which is the house commission) What is the probability of winning if I bet the dozen twice? Is it 66%? What is the probability of winning one time if I play the dozen three times? Certainly it can't be 100%. How about four times? Can anybody tell me how to calculate these probabilities? thanks

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  • $\begingroup$ If you bet the dozen twice, does that mean that you play the dozen on consecutive spins of the roulette wheel? $\endgroup$ – N. F. Taussig Aug 23 '17 at 17:05
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If you play the dozens, the probability of winning on any given spin is, as you state, 33%. Separate spins are independent; that is, the probability does not change based on previous results.

Now there are many ways to determine a probability of winning for multiple spins. But let's say you want to play the dozens until you win. So the probability of winning on the first spin is 33%. If you win, you quit and do something worthwhile with your money.

If you lose, you play again. The probability of winning on the second spin is the product of the probability of losing on the first spin (since if you win, you don't play the second spin) and 33%. The probability of losing on the first spin is roughly 67% ($1-33$%). The the probability of winning on the second spin is $\frac{2}{9}=22$%.

So the probability of winning on each subsequent spin decreases geometrically.

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  • $\begingroup$ I wouldn't agree with your last statement. What you showed is that under the premise that you play until you win a game, the probability that you dont win until the nth game is decreasing. This is different from the probability of winning on each subsequent spin which, since the events are independent, is always 33% $\endgroup$ – bthmas Aug 23 '17 at 20:08
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If I got your question correctly, you have 36 possibilities and your odds are $\frac{12}{36} = \frac{1}{3} = 33.33%$. You're asking what if you play twice (or more) what is the probability to win (at least once). So you can look at your question from a different angle: what is the probability of losing 2 times consecutively and the answer is: $(\frac{2}{3})^{2}$ so to answer your question, winning at least once is $1- (\frac{2}{3})^{2}=0.555$

for 4 times: $1-(\frac{2}{3})^{4}$ etc.

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    $\begingroup$ Since the probability of losing is $\frac{2}{3}$, the probability of losing twice is $(\frac{2}{3})^2=\frac{4}{9}$, so the probability of winning at least once in two spins should be $1-\frac{4}{9}=55\%$. $\endgroup$ – scott Aug 23 '17 at 18:22
  • $\begingroup$ @scott yep, correct $\endgroup$ – adhg Aug 23 '17 at 18:24

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