Today the Indiana Pacers won the first game in the 4 out of 7 series with the Miami Heat. Assuming that either team has a 50% chance of winning each of the remaining games, what is the probability of the Heat winning 4 of the remaining games?
There are $6$ remaining games. The desired criteria is that Heat wins at least $4$, when given that Heat lost the first 1. This is a binomial distribution; so named because of the use of the binomial coefficient to count number of permutations of outcomes that match the desired criteria.
The probability of exactly $k$ successes in $n$ trials with probability $p$ of success in any trial, is: $${n\choose k}p^k(1-p)^{n-k} \;=\; \,^n\mathrm{\large C}_k\;p^k(1-p)^{n-k} \;=\; \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$
So: $\mathbb{\large P}(\text{win at least }4\text{ more of }6) = {6\choose 4}\left(\frac 12\right)^4\left(\frac 12\right)^2+{6\choose 5}\left(\frac 12\right)^5\left(\frac 12\right)^1+{6\choose 6}\left(\frac 12\right)^6\left(\frac 12\right)^0$. $$\therefore \mathbb{\large P}(\text{win at least }5\text{ more of }6) = \frac 1{2^6}\left(\frac{6!}{4!2!}+\frac{6!}{5!1!}+\frac{6!}{6!0!}\right)$$
The probability of some event happening is a mathematical (numerical) representation of how likely it is to happen, where a probability of 1 means that an event will always happen, while a probability of 0 means that it will never happen. Classical probability problems often need to you find how often one outcome occurs versus another, and how one event happening affects the probability of future events happening. When you look at all the things that may occur, the formula (just as our coin flip probability formula) states that
probability = (no. of successful results) / (no. of all possible results).
Take a die roll as an example. If you have a standard, 6-face die, then there are six possible outcomes, namely the numbers from 1 to 6. If it is a fair die, then the likelihood of each of these results is the same, i.e., 1 in 6 or 1 / 6. Therefore, the probability of obtaining 6 when you roll the die is 1 / 6. The probability is the same for 3. Or 2. You get the drill. If you don't believe me, take a dice and roll it a few times and note the results. Remember that the more times you repeat an experiment, the more trustworthy the results. So go on, roll it, say, a thousand times. We'll be waiting here until you get back to tell us we've been right all along. Go to the dice probability calculator if you want a shortcut.
But what if you repeat an experiment a hundred times and want to find the odds that you'll obtain a fixed result at least 20 times?
Let's look at another example. Say that you're a teenager straight out of middle school and decide that you want to meet the love of your life this year. More specifically, you want to ask ten girls out and go on a date with only four of them. One of those has got to be the one, right? The first thing you have to do in this situation is look in the mirror and rate how likely a girl is to agree to go out with you when you start talking to her. If you have problems with assessing your looks fairly, go downstairs and let your grandma tell you what a handsome, young gentleman you are. So a solid 9 / 10 then.
As you only want to go on four dates, that means you only want four of your romance attempts to succeed. This has an outcome of 9 / 10. This means that you want the other six girls to reject you, which, based on your good looks, has only a 1 / 10 change of happening (The sum of all events happening is always equal to 1, so we get this number by subtracting 9 / 10 from 1). If you multiply the probability of each event by itself the number of times you want it to occur, you get the chance that your scenario will come true. In this case, your odds are 210 * (9 / 10)4 * (1 / 10)6 = 0.000137781, where the 210 comes from the number of possible fours of girls among the ten that would agree. Not very likely to happen, is it? Maybe you should try being less beautiful!
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