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Thursday, 7 August 2014

chapter 2 - Example of Rolling a Dice

Scenario-1:
You are playing a game with your friend, which involves rolling a single die once. The rules of the game are that you’ll win if an even number shows up while your friend will win if an odd number shows up.
It should be immediately apparent to you that both you and your friend are equally likely to win this game. Why? Because you win in three possible cases, i.e, if one of the numbers $2$$4$ or $6$ shows up, while your friend wins if one of $1$$3$or $5$ shows up. Since for a (fair) die, each of the six faces is equally likely to show up, we can safely state that since both of you have $3$ favorable cases to your credit respectively, both of you are equally likely to be the winner.
Can we somehow quantify this discussion? That is, can we somehow assign numerical values to the various chances involved? It turns out that we can, and in a way that is very intuitively appealing, as follows:
Technically, we term any incident, an event (we will define events more precisely later; for the time being, just think of an event as an incident). Now, if an event $E$ is sure to occur, we say that the probability of the occurrence of $E$is $1$, and we write this as
 $P\left( E \right) = 1$
On the other hand, if an event $F$ is sure not to occur, we say that the probability of the occurrence of $F$ is $0$, and we write this as
 $P\left( F \right) = 0$
For any event $G$ that is likely to happen, we should then have
 $0 \le P\left( G \right) \le 1$
which means that the probability of $G$ occurring must lie between $0$ and $1$; it can be $1$ at the most which implies that $G$ is sure to occur; it can be $0$ at the least which implies that $G$ is sure not to occur.
Here are some simple examples:
 Event $E$ : The sun will rise in the east tomorrow Event $F$ : The sun will rise in the west tomorrow Event $G$ : You’ll obtain a “Heads” upon tossing a fair coin
Thus, we must have,
 $P(E) = 1$ ; $E$ is sure to occur $P(F) = 0$ ; $F$ cannot occur under any circumstances $0 < P(G) < 1$ ; You can’t say for sure that you will definitely obtain a “Heads”. Thus, $P(G)$ cannot be $1$. Similarly, $P (G)$ cannot be $0$ either since you can’t say for sure that you will definitely not obtain a “Heads”. $P (G)$ therefore must lie somewhere between $0$and $1$; where exactly we’ll soon understand!
Coming back to the die-game you and your friend were playing, let us assign a numerical value to the chance of any of the six faces showing up.
This should be easy! There are six faces, and it is easy to observe that each of these six faces is equally likely to show up, which means that there is no reason why we should believe one face to be more likely to turn up than any other.
Thus, if we let $P(i)$ denote the probability of the $ith$ face turning up, we must have,
 $P\left( 1 \right) = P\left( 2 \right) = P\left( 3 \right) = P\left( 4 \right) = P\left( 5 \right) = P\left( 6 \right)$ $\ldots(1)$
What we’ve stated till now is just that each of the six faces is equi-probable; we still haven’t obtained the numerical values to these probabilities.
To do so, observe that one of the six faces must show up:
Thus, we can intuitively assert that the probabilities of the six faces showing up must sum to $1$, because one face must show up. This means that
 $P\left( 1 \right) + P\left( 2 \right) + P\left( 3 \right) + P\left( 4 \right) + P\left( 5 \right) + P\left( 6 \right)$ $\ldots(2)$
From $(1)$ and $(2)$, we have
 $P\left( 1 \right) = P\left( 2 \right) = P\left( 3 \right) = P\left( 4 \right) = P\left( 5 \right) = P\left( 6 \right) = \dfrac{1}{6}$
So far, so good! However, we’ve still not given any rigorous justification for $(2)$. For now, you’ll have to accept the validity of $(2)$ on faith.
Now, the game’s rules were that you’ll win if an even number turns up. Thus, if we let $E$ denote the event of you being the winner, we can safely state (according to what we’have been doing till now) that,
 $P\left( E \right) = \,\,\,P\left( 2 \right) + P\left( 4 \right) + P\left( 6 \right)$ $= \,\,\,\dfrac{1}{6}\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\dfrac{1}{6}\,\,\,\,\, + \,\,\,\dfrac{1}{6}$ $= \,\,\,\dfrac{1}{2}$
Similarly, if we let $F$ denote the event that your friend wins, we should have
 $P\left( F \right) = \,\,\,P\left( 1 \right) + P\left( 3 \right) + P\left( 5 \right)$ $= \,\,\,\dfrac{1}{6}\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\dfrac{1}{6}\,\,\,\,\, + \,\,\,\dfrac{1}{6}$ $= \,\,\,\dfrac{1}{2}$
Thus, both you and your friend have a probability $\dfrac{1}{2}$ of winning the game. This should have been otherwise obvious also! Since $E$ and $F$ must be equi-probable, and we also have
 $P\left( E \right) + P\left( F \right) = 1$ we must have $P\left( E \right) = P\left( F \right) = \dfrac{1}{2}$
Well, the preceeding discussion was too long for too simple an example. But it was necessary to get you tuned to the flavor of this subject.