Cricket Probability question My understanding of probability is little weak, and it was since high school I am struggling with it.This question struck me yesterday while watching IPL.
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In a normal game of cricket, there are 10 fielders, 2 batsman, 1 bowler and 2 empires, (total 15) actively present on the field which can be oval or circular spanning across $90$ mts in radius.
Now, often it happens that when a fielder catches the ball, he throws it up in the air showing enthusiasm or being too happy. 
Here is where it struck me, that the ball "can" hit any of the person on the ground while ball descends. 
So, if I were to calculate the probability that it will definitely hit someone, will it be $\frac{1}{15}$? 
If I am correct, then a fielder standing at the slip (position near batsman) and at long on (position near boundary) both will fall under same probability?
I imagined that there are only 2 fielders present on the ground and one of them throws the ball up in the air.The probability that the ball will land up on others head will be (going by the same logic) 1/2 irrespective of the distance. That means 50 times the fielder will get hit out of 100 throws. Seems weird, because I am not considering the "external factors" like the distance between the fielders, wind speed direction etc.
My question is simple : 
1) Does the probability remains the same ($\large\frac{1}{15}$). If not, then what will be the probability that ball will hit some one while it descends!
2) Do we need to include external factors that can (actually) influence the outcome to calculate probability of occurrence of any event. 
I need people to enlighten me! :)
Thanks
 A: You're right to think that these predictions are strange! The error you seem to be making is to assume that there is only one probability distribution (the uniform one) for a particular set of possible outcomes. If you want to describe some experiment probabilistically, you need two things. First, you need some set of possible outcomes, say $\{x_i\}$. Then you need a probability distribution $p$, which is a function that assigns to each outcome $x_i$ a non-negative real number $p_i$ such that $\sum_ip_i=1$. These are the probabilities - you say that $p_i$ is the probability of outcome $x_i$ occurring.
In your example, the set of outcomes might be written $\{n,x_1,\ldots,x_{15}\}$, where $n$ represents nobody getting hit and $x_i$ represents player $i$ getting hit. In order to come up with a probability distribution that accurately describes the situation, you need to create some kind of mathematical model. A (very rough) first approximation might be to divide the field into squares with the same area as a player's head, and assume that the ball is equally likely to fall in any of these squares, and that if a player is standing in the square the ball hits, he or she will be struck. If there are $N$ such squares, then the probability of a particular player (say player $i$) being hit will be $p_i=1/N$.
This model is of course far too simple to be very accurate. You could next start to think about the distribution of angles and speeds of tosses in order to figure out the probabilities of the ball going various distances in different directions. Wind speeds could deflect the ball (and might depend on height off the ground). Players might also react to seeing the ball in the air, running towards or away from it. There's not really a limit to the number of factors you could include, and as you include more and more, your model will increase in complexity!
A: You're considering the classical theory of probability where every event is considered equally likely, try reading axiomatic approach to probability. If you provide more data such as ball speed air speed and direction along with other useful things (like if the players are considered point sized or occupy some area.), it is easy to calculate using the probability as a fraction of favorable outcomes upon total outcomes. Try reading Mathematical Modelling too. 
A: There really is a lot of factors that affect this. In general, if a ball is thrown into the air "at random", then the probability that it will hit a person on the field is much less than 1/15.
What you're thinking is that if the ball DOES hit a random person, there is a 1/15 chance that it would be a particular person.
So let's say that a ball is thrown vertically up into the air. The position of the ball's origin will determine how dense that part of the pitch is. For example, if the ball was thrown from the middle of the pitch, the immediate area is densely populated by the umpire, the wicket keeper, the batsmen, etc. There is more chance that someone will be hit than if the ball was caught on the boundary.
For the purpose of a random throw though, we can also assume that the ball can be thrown from anywhere in the pitch, with equal probability of landing anywhere else in the pitch (discounting all the throws that will land outside the pitch) - for this, we map out the area of the pitch that is covered by players. 
If we give a generous estimate, assuming most players are flailing about, celebrating their catch, each player can take up to 1 square meter of space, out of the 17000 or so square meters of a cricket ground with a 75 meter radius. This means that there's some 15/17000 chance that the ball will land on someone's area!
However, if the ball was travelling at a flat enough angle, it could cross a larger portion of the pitch at player level, increasing the chance of hitting someone in the face or body.
So there, I hope that clears it up. In short, the origin of the throw, the angle of the throw and the randomness of the throw will affect this probability, but however you calculate it, the probability of a random person getting hit would be much lower than 1/15.
