The idea of probability is that you don't just consider the cases you want to study. You look at all possible outcomes and determine how often the outcomes that interest you will come up.

First, let's discuss the law of total probability. For any outcome you want to study, you want to make sure that the odds of some outcome occurring are 100%. For example, if there is a 57% chance to win and a 17% chance to lose, that adds up to only 74%. What happens the other 26% of the time? Is it a draw? Is the game canceled? Since the team wins 57% of the total 100%, you need to know what the 100% means.

Without additional information, we assume the team wins 57% of the time and loses 100-57= 43% (for a total of 57+43=100%).

The probability that the team will lose the first game is 43%. The probability the team will lose the second game is 43%. What about the total number of outcomes for two games?

You can have win-win, win-lose, lose-win, and lose-lose. How do you figure out the breakdown of probabilities for each outcome?

Because winning or losing a game has no direct bearing on winning or losing the next game, these events are called independent. Additionally, you cannot win the first game and lose the second but at the same time lose both games (if you win the first game, you did not lose both games) so these outcomes are mutually exclusive or discrete. The sum of their probabilities should be 100%. Additionally because the individual games are not affected by previous games, we can apply something called the product principle. It means for independent events, the probability that both happen is the product of their probabilities.

So win-win happens $57\% \times 57\%=32.49\%$ of the time.

Win-lose and lose-win each happen $57\% \times 43\% = 24.51\%$

And lose-lose happens $43\% \times 43\%=18.49\%$. If you add up these 4 probabilities, you get 100%: $32.49+24.51+24.51+18.49 = 100$

This is easily extended to more than two games through multiplication (or exponents). For example, three consecutive losses would have a probability of $(43\%)^3=7.9507\%$ and six losses would have a probability of $(43\%)^6\approx 0.63\%$