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Pascal’s Triangle and Independent Events

                                                                                                        
Simple probability problems are split into two categories, dependent and independent events.  Drawing cards from a deck without replacement, which is the standard way to play card games, is a situation where events are dependent; the probability changes as cards leave the deck; to get an ace on your first card, you have 4 chances of success out of 52; on your second card, the chances are either 4 chances out of 51 if you didn’t get an ace the first time, or 3 chances of 51 if you did.  Flipping coins or rolling dice are situations where the events are independent; if one coin turns up heads, it doesn’t effect the odds on the next coin turning up heads.

We saw in the section entitled “How many different poker hands are there?” that the binomial coefficients answer a lot of questions about the odds in games with dependent events; in games with independent events, the binomial coefficients still show up.  Let’s give an example.  We are going to play a game; I’ll throw four fair coins in the air, each one with a 50-50 chance of landing heads; you win the game if at least three of the coins end up heads, and I win otherwise.  What are the odds you will win?

Because every event is equally likely to happen, we can just list all possible outcomes and mark which ones win for you, then divide the second number by the first:

4 heads, 0 tails	HHHH	1 way
3 heads, 1 tail	HHHT, HHTH, HTHH, THHH	4 ways
2 heads, 2 tails	HHTT, HTHT, HTTH, THHT, THTH, TTHH	6 ways
1 head, 3 tails	HTTT, THTT, TTHT, TTTH	4 ways
0 heads, 4 tails	TTTT	1 way
		16 ways total

        Given the rules of this game, you have ⁵/₁₆ of a chance to win, or 31.25%.

The numbers 1,4,6,4,1 should look familiar to anyone who has wandered around this website for any length of time, because that is the 4th row of Pascal’s Triangle.  Just as in games with dependent events, the binomial coefficients turn up in calculating odds in games with independent events.

Consider a basketball player who makes, on average, 70% of his free throws.  If he takes 10 free throws, how many will he make?  The first answer that comes to mind is 7, since 70%×10 = 7.  But that is what we call the expected value; given any group of ten shots he might make more or he might make less than seven, or he might make exactly seven.  Since his chance of success on every shot is 70%, his chance of failure is 30%.  The formula for making k out of n shots, where  is  , where p is the chance of success and 1-p is the chance of failure.  For our 70% free throw shooter to make exactly 7 of 10 shots, the probability is ; while this may seem low, note that his chance to make 8 of 10 is about 23.34% and his chance to make 6 of 10 is about 20.12%, and 7 of 10 is the most likely event of all.  To explore this graphically, try our Java applet called BarGraph, which lets you change the percentage of possibility of success on two different events from 0% to 100% and the number of times the events will happen, from 1 to 25.