Expected gain from a game of roulette

Question

I am trying to understand the concept of expected value and got confused by example 2 given in Wikipedia:

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I do not understand. Why is the expected gain negative?

Answer 1

Expected value means the "average outcome" (note, not the most likely outcome). More concretely, it's the sum of outcomes weighed by their probability.

In the said Roulette bet, one's bet has a $1/38$ chance of winning $35\$$. However, otherwise the bet is lost and the outcome is $-1\$$ (a loss of a dollar). The expected value is then calculated as: $(37/38) \times (-1\$) + (1/38) \times 35\$ \approx -0.05\$$.

Intuitively, this means that for very long series of bets, you're most likely to lose approximately five cents per bet made. Note that losing five cents on a single bet is not a possible outcome when betting a dollar - it's an average per bet that is likely to come true in the long run.

Answer 2

The expected gain is negative because there is more chance for the player to lose money than to win some. In English words, you can picture expected value for discrete probabilities as "the average amount of money (or something else) I win at each turn"

Answer 3

Expected value is just like a weighted average. You just multiply the outcomes of the random variable by the respective probabilites of the outcomes occuring. Here the random variable is 'gain of your bet'. If you place a bet and lose, you lose your dollar. This corresponds to an output of -1 on your random variable and occurs with probability 37/38, as you bet on 1 of the 38 numbers. Similarly, if you win, you receive $35 profit. But this only happens with probability 1/38. So even though the payout is relatively large, it happens with such a small probability that the expected profit remains negative.