Expected value for a roulette wheel problem?

Question

A roulette wheel is divided into $38$ pockets. $18$ are numbered red pockets, $18$ are numbered black pockets and $2$ are green pockets.

You can place bets on specific red or black pockets with odds of $35:1$. That is, if you are correct you win $35$ dollars (plus the dollar you bet), and if you are incorrect, you lose the dollar you bet. A ball is spun into the wheel, and we will assume it lands in all pockets with equal probability. Your friend has developed what he describes as a “foolproof system” for winning. It involves him betting on ${$1}$ on $7$ red again and again. You try to discourage him, but he is insistent.

What are his expected winnings after $36$ spins of the wheel?

Any help would be appreciated.

I know it's the $$\text{(probability of red)} \times \text{(value of red)} + \text{(probability of not red)} \times \text{(value of not red)}$$ but don't know how to extend that onto $36$ spins.

Answer 1

The probability of getting a $7$ red is $\frac{1}{38}$. Thus, the expectation of one game is:

$$\mathbb{E} = 35\cdot\frac{1}{38} -1\cdot\frac{37}{38} = -\frac{2}{38}$$

Now, realize that what happens in one game does not influence any subsequent games.

Thus, after $36$ spins, your friend will lose \$ $\dfrac{72}{38}$.

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Your friend seems to be suffering from Gambler's Fallacy.