Win, Lose or Draw: 7 Games of Chess

1.) Two chess players, X and Y, play a 7-game series of chess. For each game, there are three possible outcomes for a player, win, lose or draw. A win nets the victor one point and the loser zero points. In the case of a draw, each player is awarded a half-point. How many ways are there for one of the players, say, player X, to get an overall result of 3 wins, 2 draws and 2 losses?

Solution: A simple calculation here... We multiply the number of ways a player can win 3 out of the 7 games, the number of ways that player can get a draw or a loss in 4 of the seven games, and the number of ways that player can get a draw or a loss in 2 out of the 7 games. The factors are:
$\binom{7}{3}, \binom{4}{2}, \binom{2}{2}$

The answer is $\binom{7}{3}* \binom{4}{2}* \binom{2}{2}$ = 210

2. How many possible outcomes for the games are there with X ending up with 4 points and Y ending up with 3 points?

Solution: One way is to have X win 4 out of the 7 games and lose 3: $\binom{7}{4}*\binom{3}{3}$. Another way is to have X win 3, draw 2, and lose 2: $\binom{7}{3}* \binom{4}{2}* \binom{2}{2}$. Another way is to have X win 2, draw 4 and lose 1: $\binom{7}{2}* \binom{5}{4}* \binom{1}{1}$. And, finally, another way is to have X win 1 and draw 6: $\binom{7}{1}* \binom{6}{6}$

Answer = $ \binom{7}{4}*\binom{3}{3} + \binom{7}{3}* \binom{4}{2}* \binom{2}{2} + \binom{7}{2}* \binom{5}{4}* \binom{1}{1} + \binom{7}{1}* \binom{6}{6}$
= 35 + 210 + 105 + 7 = 357.

3. How many possible outcomes for the games are there where one player ends up with 4 points, the other with 3 points, and the full 7 games are played?

Solution: The 7th game must be played, which means that for one player, say, X to end up with 4 points at the end, he or she cannot lose that 7th game. Therefore, the only possibilities for X to end up with 4 points are either a full-point awarded in the seventh game, or only a half point. If the full point is awarded in the 7th game, that means X must get 3 points in the first 6 games. If only a half-point is awarded in the 7th game, then X must get 3.5 points in the first 6 games.

Let scenario 1 be the scenario where X gets a full point in the 7th game. This gives us for the first 6 games:

$\binom{6}{3}* \binom{3}{3}$ = 20 (3 wins and 3 losses)
$\binom{6}{2}* \binom{4}{2}* \binom{2}{2}$  = 90 (2 wins, 2 draws, and 2 losses)
$\binom{6}{1}* \binom{5}{4}* \binom{1}{1}$ = 30 (1 win, 4 draws, and 1 loss)
$\binom{6}{0}* \binom{6}{6}$ = 1 (0 wins and 6 draws)

Let scenario 2 be the one where X gets only a half point in the 7th game. This gives us for the first 6 games:

$\binom{6}{3}* \binom{3}{1}*\binom{2}{2}$ = 60 (3 wins, 1 draw, and 2 losses)
$\binom{6}{2}* \binom{4}{3}*\binom{1}{1}$ = 60 (2 wins, 3 draws, and 1 loss)
$\binom{6}{1}* \binom{5}{5}$ = 6 (1 win and 5 draws)

Answer = 20 + 90 + 30 + 1 + 60 + 60 + 6 = 267

Phone Numbers as Possible Combinations of 7 digits: Probability

1.) If the first digit isn't allowed to be a 0 or a 1, how many 7-digit phone numbers are possible?

Solution: For each digit in the phone number, there are 10 possibilities (0 - 9) if there are no restrictions. In this case the first number isn't allowed to be 0 or 1. For phone-number digits 2 - 7, there are $10^{6}$ possibilities: 10 * 10 * 10 * 10 * 10 * 10. Multiply this by the number of possibilities for the first digit (only 8 since two digits, 1 and 0, are excluded), and you get your answer:

No. of possibilities for the first digit * No. of possibilities for the remaining 6 digits

= $8*10^{6}$ = 8,000,000 possible phone numbers.

2.) If the above condition stays in place, along with the added condition that the first three digits of the phone number cannot be 911, how many possible phone numbers are there?

Solution: If the probability of an event A occurring is P(A), then the probability of event A not occurring is 1 - P(A). Consider the case where the event A is the choosing of the first three digits of the phone number to be 911:

The probability space of the first three digits is $8*10*10$ if we consider only the condition of part 1. The probability of the first three digits being 911 is $\frac{1}{8}*\frac{1}{10}*{\frac{1}{10}} = \frac{1}{800}$

So, the probability of the first three digits not being 911 is $ 1 -\frac{1}{800} = \frac{799}{800}$

Finally, the number of possible phone numbers with the first digit not being a 0 or a 1, and the first 3 digits not being 911, is
$799 * 10^{4} $ = 7,990,000.

(799 is the number of possibilities for the first 3 digits, and $10^{4}$ is the number of possibilities for the last 4 digits.)