Again, probability does not mean that the outcome always is 50-50.
Here's some samples... http://www.mathgoodies.com/lessons/v...obability.html
Okay, since you offered this link, I am going to work on the assumption that you are actually interested in learning a bit about probability. If not, you may disregard the rest of this post.
In the examples through the link (or at least the first few, which are all I looked at), the various outcomes are all
equally likely. This is critical. For instance, with the four-colored spinner, if the four areas are each 1/4 of the disc and we make the usual assumptions about the game being fair, there is no reason for the spinner to "prefer" any one resting place over any other. So yes, the probability of any given color is 1/4.
But imagine that you changed that disc, so that the yellow area was the same size it presently is, but the other three areas -- red, green, and blue -- were all black instead. Now 1/4 of the disc would be yellow, and 3/4 would be black. The spinner would have a 1/4 probability of landing on yellow, the same as it did before. It would have a 3/4 probability of landing on black. Even though we are now down to two options, yellow and black, they aren't equally likely, so the probability isn't 1/2-1/2 (or, in jargon, 50-50, meaning percentages).
The six-sided die presents another case. All six numbers are equally likely, so the probability of rolling any particular one is 1/6. But let's say someone wanted to make a bet about rolling a multiple of 3 (3 or 6). There are two outcomes: You roll a multiple of 3, or you don't. But it's more likely that you won't. So that isn't a 50-50 proposition, even though there are only two outcomes. (The probability of rolling a multiple of 3 is 2/6, or 1/3.)
You are correct that, even when the probability
is 50-50, the outcome won't always
be 50-50. Here is an example. If you flip a fair coin twice, the most likely outcome is that you'll get one head and one tail. But you might get two heads, or two tails. Even though the probability of heads on any flip is 1/2, you might have 100% heads over two flips (or, for that matter, over three or more flips). If you flip the coin 100 times, the
most likely outcome is 50 heads and 50 tails. But in fact, you're quite
unlikely to get a perfect 50-50 split -- most likely it will be 53-47, or 48-52, or something like that. This is true even if the coin is perfectly fair.
Shooting percentages are like probabilities. There are two outcomes for a free-throw attempt: a make, or a miss. But the probability is not 50-50, because those outcomes are not equally likely. If Nash is a 90% free-throw shooter, then the probability of his making any one free throw is 90% -- he's much more likely to make it than miss it. (We could argue about how past performance is not necessarily a predictor of future results, but let's not.) That doesn't mean he
will make 9 out of his next 10, but it's the most likely outcome.
In a situation where there are many unknowns -- such as a shooter attempting a last-second three with a hand in his face -- there isn't really a way to assign probabilities to the outcomes. Hypothetically, you could set up the identical scenario a large number of times, see how many times the shot is made, and get a pretty good idea. Since that's impractical, all we can do is guess. But that's completely different from shrugging off the problem and calling it 50-50, because the outcomes aren't equally likely. My guess is that, under identical circumstances, Bell would make that shot much less than half the time.