If 60% chance of picking winner of a game, and 32 games played, what are odds of picking 12 of 16 correctly?

gistfile1.txt

Update: All of this turns out to be nothing more than a classic
Binomial Distribution: http://en.wikipedia.org/wiki/Binomial_distribution

Jim asked: "If 60% chance of picking winner of a game, and 32 games
played, what are odds of picking 12 of 16 correctly?"  (I'm going to
ignore the "32 games played" and assume you meant 16 played, since I
don't understand the 12 of 16 of 32 statement as it stands.  I.e., how
do you pick the 16 out of the 32?)

Assumption: every game is independent of the other and the only
possible outcomes are losses and wins (no ties!).  I.e., the results
of any game cannot influence the results of any of the other games.

I'm pretty much going to use the binomial theorem here, so quick
refresher at http://en.wikipedia.org/wiki/Binomial_theorem

Let's simplify to 3 independent games, where you have a fixed
probability 'x' of winning each one.  For the sake of convenience, let
'y' denote the probability of losing a given game.  So, y = 1 - x,
since there are only two possible outcomes here (no ties!) and the
probabilities must add up to 1 for the two outcomes.

Let's also use the notation P(W, L, W) to indicate the probability of
winning the 1st, losing the 2nd, and winning the 3rd.

Q: So, what's P(W, L, W), assuming x = 0.6?
A: Given that these are independent games, the probability is 
x * y * x = x^2 * y, where y = 1-x = 0.4.

Q: How many ways can we win 2 and only 2 games? 
A: Counting them off there's 3 of them: (W, W, L), (W, L, W), (L, W, W).

Q: What's the probability that we win exactly two games out of three?
A: That's the sum of the probabilities of each of the independent
outcomes above:
  P(W, W, L) + P(W, L, W) + P(L, W, W) 
= x * x * y + x * y * x + y * x * x
= 3 * x^2 * y

Let's denote the above quantity, the probability that you win exactly
'k' of 'n' games, by Q(n, k).

So, we just computed Q(3, 2) = 3 * x^2 * y.

Q: What's the probability that we win *at least* 2 games of 3?
A: That's the probability that we win exactly 2 games + the
probability that we win all 3 games (x * x * x).  Using our Q
notation, this is Q(3, 2) + Q(3, 3).

I.e., x^3 + 3 * x^2 * y.

If you look at the Binomial Theorem page linked above, you'll notice
this is the sum of the first two terms in the expansion for (x + y)^3.

This isn't surprising, since another way of looking at the coefficient
of x^k * y^(n-k) in the expansion of (x + y)^n is that it is the
number of ways of picking 'k' x'es and (n-k) y's when you multiply it
all out.

This binomial coefficient for x^k * y^(n-k) is 
C(n,k) = n!/(k! * (n-k)!).

Going back to the original problem and plugging things in, the number
of ways of winning *exactly* 12 games of 16 is
  C(16, 12)
= 16*15*14*13/1*2*3*4
= 1820.

Our x'es here are the probability that the game was won (0.6) and the
y's are the probability it was lost (1 - 0.6 = 0.4).  So the
probability of the outcome being any one such sequence of12 wins and
4 losses is:
  0.6^12 * 0.4^(16-12)
= .0000557256278016

Pretty low! But that's for a given sequence of 12 wins and 4 losses.
There are C(16, 12) such sequences, so the overall probability that we
win exactly 12 of 16 is
  C(16, 12) * x^12 * y^(16 - 12)
= 1820 * .0000557256278016
= .1014206425989120

I.e., you will win *exactly* 12 games a little over 10% of the time.

This is Q(16, 12) by our earlier notation.

Q: What's the probability that you'll win *at least* 12 of 16?
A: This is the proability of winning exactly 12 + probability of
winning exactly 13 + ... + probability of winning exactly 16.
= Q(16, 12) + Q(16, 13) + Q(16, 14) + Q(16, 15) + Q(16, 16)

Using good old Unix bc with some helper functions (code below), this
comes out to:
  q(16,12,0.6)+q(16,13,0.6)+q(16,14,0.6)+q(16,15,0.6)+q(16,16,0.6)
= .16656738435072000000

or just over 16.6%

Like I said, probability is not something I'm good at, so I tested the
above with something I felt I had an intuitive answer for: the
probability of winning at least half of 17 games if the probability of
winning each is 50%.

I used 17 because there's an even number of outcomes, 18, including the
"0 games won" outcome and so we can split the outcomes neatly.

By the formula above, this is:
  q(17,9,0.5) + q(17,10,0.5) + q(17,11,0.5) + q(17,12,0.5) + q(17,13,0.5)
+ q(17,14,0.5) + q(17,15,0.5) + q(17,16,0.5) + q(17,17,0.5)
= .50000000000000000000

I.e., 50% of the time we win at least half of 17 games, which is nice
and intuitive and makes me feel better about this exercise.

Sorry about the length of this!  While the final destination may be of dubious
certainty, I had fun getting there and figured I'd map out the journey.

-Sudish

bc functions used (the rest of this can be cut'n'pasted
as is into the Unix bc calculator):

define f (x) {
  if (x <= 1) return (1);
  return (f(x-1) * x);
}

define c (n, k) {
  return (f(n) / (f(k) * f(n-k)));
}

define q (n, k, p) {
  return (c(n,k) * (p^k) * ((1-p)^(n-k)));
}

c (16, 12)

scale = 10

q (16, 12, 0.6)

q(16,12,0.6)+q(16,13,0.6)+q(16,14,0.6)+q(16,15,0.6)+q(16,16,0.6)

q(17,9,0.5) + q(17,10,0.5) + q(17,11,0.5) + q(17,12,0.5) + q(17,13,0.5) + q(17,14,0.5) + q(17,15,0.5) + q(17,16,0.5) + q(17,17,0.5)