I know I must be missing something very rudimentary but I can't find it.


On a roulette there are:

  • 18 black values
  • 18 red values
  • 2 values


Betting on either red or black yields a winning probability of 18/38 (obvious advantage to the house).

The odds on this kind of bet are 1:1 (You win a matching sum to what you bet, or you lose your bet).

Another type of bet, is betting on a specific number. This yields a winning probability of 1/38 (even more obvious advantage to the house). The odds on this kind of bet are 35:1 (if you win you get your dollar plus 35 if you lose, the dollar is gone.).

So far so good.

My question:

If I bet $1 on 9 black numbers (35:1 odds) my chance of wining is 9/38 (9 outcomes will be a win). I will At the worst case I will lose all my bets; best case, I will lose 8 dollars, and win 35 (if one of my numbers won). So I have a 9/38 chance of being profitable.

Now if in addition to betting on 9 black numbers, I also bet $10 on red to win, to me it seems like I would have a 18/38 chance of having my color win and an additional 9/38 chance of having my number win (of opposite color).

Therefore, I have a 27/38 (or just over 70%) chance to win (come out profitable) every turn?

That is,

  • I have a 30% chance of losing 19 dollars
  • 23% chance of having one of my numbers win: if I win a number I lose 18 dollars and get an additional 35 dollars
  • 47% chance of having my color win: If I win the color, I lose 9 dollars and win an additional 10 dollars

Could this be? Surely not! What am I missing?

  • 1
    $\begingroup$ Here's another way. Pick 35 numbers and bet $1$ on each number. If one of your numbers wins, you make a $1$ profit. So you have a 35/38 chance of making a profit! $\endgroup$ – bof Aug 4 '17 at 23:01
  • $\begingroup$ You have a 70% chance of winning somthing. And a 30% chance of losing something. The 30% of the something you lose is more than the 70% of the something you win so the house is happy because even the win more often than you lose, when you do lose you lose a lot more. $\endgroup$ – fleablood Aug 5 '17 at 2:05
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    $\begingroup$ Imagine a slot machine that if you put in a \$1 you have an 80% chance of winning back \$1.10. If 2,000 people play it in a day, on average 1,600 people will win 10 cents (and the house loses \$160) and 400 people lose a dollar (and the house gains \$400). So on the whole the house makes \$240$ despite the fact that 80% of the players are winners. $\endgroup$ – fleablood Aug 5 '17 at 2:09

Let us compute the expectation of you reward $R$:

$$\mathbb E(R)=\dfrac{11}{38}(-19)+\dfrac{9}{38}(35-18)+\dfrac{18}{38}(10-9)=\dfrac{-11 \cdot 19+9\cdot 17+18}{38}=-1$$

On average you will lose 1$ per game... Even if you have the sentiment to win in 70% of the cases. This is very pernicious! Actually when you lose, you lose quite a lot.

  • $\begingroup$ Could you please explain your calculation to me? The fractions I get, but the numbers in brackets confuse me. $\endgroup$ – Michael Seltenreich Aug 4 '17 at 23:03
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    $\begingroup$ @MichaelSeltenreich surely you have 70% of chance of winning. But in this 70% you count 48% of only winning 1\$ (not much). But you also have 30% of chance of losing 19\$ (this is a lot). Then on average you will lose 1\$. But if you play only once there is 70% of chance that you will go back with some money, it is true. The risk is that in 30% of the cases you will lose 19\$... $\endgroup$ – fonfonx Aug 4 '17 at 23:13
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    $\begingroup$ Or is it that I do have a 70% chance of winning, but I win very little in proportion to what I stand to lose that winning 3 games, and losing 1 still results in my losing more money than making. Is that it? $\endgroup$ – Michael Seltenreich Aug 4 '17 at 23:13
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    $\begingroup$ yes that is the idea! Look at this other example. You have a die. You roll it, if you get 1 you have to pay 10\$. Otherwise you win 1\$. You have 83% of chance to win 1\$, great! But you can still lose 10\$ And if you keep playing you will start losing money $\endgroup$ – fonfonx Aug 4 '17 at 23:16
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    $\begingroup$ And as far as the house cares They don't care if 83% of the customers win $\$1$ if 17% are losing $\$10$. So they don't care if you make one $$\19$ dolllar bet with a 70% chance of winning. So long as other people are making similar bets. Even if you play only once and assume you will win and do win, the house will play the expected averages. And they will win. $\endgroup$ – fleablood Aug 5 '17 at 1:49

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