Binomial Distribution.ipynb

{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### Binomial is a specific type of a discrete probability distribution."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let's see an example question first, and then learn about the binomial distribution."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Example 1: Two players are playing basketball, player A and player B. Player A takes an average of 11 shots per game, and \n",
    "has an average success rate of 72%. Player B takes an average of 15 shots per game, but has an average success rate of 48%.\n",
    "\n",
    "Question 1: What's the probability that Player A makes 6 shots in an average game?\n",
    "\n",
    "Question 2: What's the probability that Player B makes 6 shots in an average game?"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "We can classify this as a binomial experiment if the following conditions are met:\n",
    "    \n",
    "    1.) The process consists of a sequence of n trials.\n",
    "    2.) Only two exclusive outcomes are possible for each trial (A success and a failure)\n",
    "    3.) If the probability of a success is 'p' then the probability of failure is q=1-p\n",
    "    4.) The trials are independent.\n",
    "    \n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The formula for a Binomial Distribution Probability Mass Function turns out to be:    "
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "$$Pr(X=k)=C(n,k)p^k (1-p)^{n-k}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Where n= number of trials,k=number of successes,p=probability of success,1-p=probability of failure (often written as q=1-p)."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This means that to get exactly 'k' successes in 'n' trials, we want exactly 'k' successes: $$p^k$$ \n",
    "and we want 'n-k' failures:$$(1-p)^{n-k}$$\n",
    "Then finally, there are $$C(n,k)$$ ways of putting 'k' successes in 'n' trials.\n",
    "So we multiply all these together to get the probability of exactly that many success and failures in those n trials!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "--------------------------------------------------------------------------------------------------------------------------------"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Quick note, C(n,k) refers to the number of possible combinations of N things taken k at a time.\n",
    "\n",
    "This is also equal to: $$C(n,k) =  \\frac{n!}{k!(n-k)!}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "### So let's try out the example problem!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 67,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " The probability of player A making 6 shots in an average game is 11.1% \n",
      " \n",
      "\n",
      " The probability of player B making 6 shots in an average game is 17.0% \n"
     ]
    }
   ],
   "source": [
    "# Set up player A\n",
    "\n",
    "# Probability of success for A\n",
    "p_A = .72\n",
    "# Number of shots for A\n",
    "n_A = 11\n",
    "\n",
    "# Make 6 shots\n",
    "k = 6\n",
    "\n",
    "# Now import scipy for combination\n",
    "import scipy.misc as sc\n",
    "\n",
    "# Set up C(n,k)\n",
    "comb_A = sc.comb(n_A,k)\n",
    "\n",
    "# Now put it together to get the probability!\n",
    "answer_A = comb_A * (p_A**k) * ((1-p_A)**(n_A-k))\n",
    "\n",
    "# Put the answer in percentage form!\n",
    "answer_A = 100*answer_A\n",
    "\n",
    "\n",
    "# Quickly repeat all steps for Player B\n",
    "p_B = .48\n",
    "n_B = 15\n",
    "comb_B = sc.comb(n_B,k)\n",
    "answer_B = 100 * comb_B * (p_B**k) * ((1-p_B)**(n_B-k))\n",
    "\n",
    "\n",
    "#Print Answers\n",
    "print ' The probability of player A making 6 shots in an average game is %1.1f%% ' %answer_A\n",
    "print ' \\n'\n",
    "print ' The probability of player B making 6 shots in an average game is %1.1f%% ' %answer_B"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "So now we know that even though player B is technically a worse shooter, because she takes more shots she will have a higher chance of making 6 shots in an average game!"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "But wait a minute... what about a higher amount of shots, will player's A higher probability take a stronger effect then?\n",
    "What's the probability of making 9 shots a game for either player?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 31,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      " \n",
      "\n",
      " The probability of player A making 9 shots in an average game is 22.4% \n",
      "\n",
      "\n",
      " The probability of player B making 9 shots in an average game is 13.4% \n",
      "\n",
      "\n"
     ]
    }
   ],
   "source": [
    "#Let's find out!\n",
    "\n",
    "#Set number of shots\n",
    "k = 9\n",
    "\n",
    "#Set new combinations\n",
    "comb_A = sc.comb(n_A,k)\n",
    "comb_B = sc.comb(n_B,k)\n",
    "\n",
    "# Everything else remains the same\n",
    "answer_A = 100 * comb_A * (p_A**k) * ((1-p_A)**(n_A-k))\n",
    "answer_B = 100 * comb_B * (p_B**k) * ((1-p_B)**(n_B-k))\n",
    "\n",
    "#Print Answers\n",
    "print ' \\n'\n",
    "print ' The probability of player A making 9 shots in an average game is %1.1f%% ' %answer_A\n",
    "print '\\n'\n",
    "print ' The probability of player B making 9 shots in an average game is %1.1f%% ' %answer_B\n",
    "print '\\n'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Now we see that player's A ability level gives better odds of making exactly 9 shots. We need to keep in mind that we are asking\n",
    "about the probability of making *exactly* those amount of shots. This is a different question than \" What's the probability that player A makes *at least* 9 shots?\"."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "#### Now let's investigate the mean and standard deviation for the binomial distribution"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "The mean of a binomial distribution is simply: $$\\mu=n*p$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "This intuitively makes sense, the average number of successes should be the total trials multiplied by your average success rate."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Similarly we can see that the standard deviation of a binomial is: $$\\sigma=\\sqrt{n*q*p}$$"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "So now we can ask, whats the average number of shots each player will make in a game +/- a standard distribution?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 43,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "\n",
      "\n",
      "Player A will make an average of 8 +/- 1 shots per game\n",
      "\n",
      "\n",
      "Player B will make an average of 7 +/- 2 shots per game\n",
      "\n",
      "\n",
      "NOTE: It's impossible to make a decimal of a shot so '%1.0f' was used to replace the float!\n"
     ]
    }
   ],
   "source": [
    "# Let's go ahead and plug in to the formulas.\n",
    "\n",
    "# Get the mean\n",
    "mu_A = n_A *p_A\n",
    "mu_B = n_B *p_B\n",
    "\n",
    "#Get the standard deviation\n",
    "sigma_A = ( n_A *p_A*(1-p_A) )**0.5\n",
    "sigma_B = ( n_B *p_B*(1-p_B) )**0.5\n",
    "\n",
    "# Now print results\n",
    "print '\\n'\n",
    "print 'Player A will make an average of %1.0f +/- %1.0f shots per game' %(mu_A,sigma_A)\n",
    "print '\\n'\n",
    "print 'Player B will make an average of %1.0f +/- %1.0f shots per game' %(mu_B,sigma_B)\n",
    "print '\\n'\n",
    "print \"NOTE: It's impossible to make a decimal of a shot so '%1.0f' was used to replace the float!\""
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "####Let's see how to automatically make a binomial distribution."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 45,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "7.92\n",
      "1.48916083752\n"
     ]
    }
   ],
   "source": [
    "from scipy.stats import binom\n",
    "\n",
    "# We can get stats: Mean('m'), variance('v'), skew('s'), and/or kurtosis('k')\n",
    "mean,var= binom.stats(n_A,p_A)\n",
    "\n",
    "print mean\n",
    "print var**0.5"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Looks like it matches up with our manual methods. Note: we did not round in this case."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "####We can also get the probability mass function:"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Let's try another example to see the full PMF (Probability Mass Function) and plotting it.\n",
    "\n",
    "Imagine you flip a fair coin. Your probability of getting a heads is p=0.5 (success in this example).\n",
    "\n",
    "So what does your probability mass function look like for 10 coin flips?"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 71,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "data": {
      "text/plain": [
       "array([ 0.00097656,  0.00976563,  0.04394531,  0.1171875 ,  0.20507813,\n",
       "        0.24609375,  0.20507813,  0.1171875 ,  0.04394531,  0.00976563,\n",
       "        0.00097656])"
      ]
     },
     "execution_count": 71,
     "metadata": {},
     "output_type": "execute_result"
    }
   ],
   "source": [
    "import numpy as np\n",
    "\n",
    "# Set up a new example, let's say n= 10 coin flips and p=0.5 for a fair coin.\n",
    "n=10\n",
    "p=0.5\n",
    "\n",
    "# Set up n success, remember indexing starts at 0, so use n+1\n",
    "x = range(n+1)\n",
    "\n",
    "# Now create the probability mass function\n",
    "Y = binom.pmf(x,n,p)\n",
    "\n",
    "#Show\n",
    "Y\n",
    "\n",
    "# Next we'll visualize the pmf by plotting it."
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "####Finally, let's plot the binomial distribution to get the full picture."
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 77,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
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       "fWxhvruAL+TpPwlslF9vCSyOiKuAH9Oavx9rVjcnWLOkWPM7F9ik8P4SUlJ7GBgKLOxgvurlReH1\n",
       "DGAiMJ50v/Mt4FLSb48+mL9G8zNSrbc4b/uFRjwPfBP4M/AwMCki/tDN7asedhRwfN6+yaSLCIDR\n",
       "pKbjScC8wvTfBT4uaTKpqbjSlL07cH++1/tt4Ow64jJrWf65OjMzsxK4BmtmZlYCJ1gzM7MSOMGa\n",
       "mZmVwAnWzMysBE6wZmZmJXCCNTMzK4ETrJmZWQmcYM3MzErwf/5jwxXBXuwnAAAAAElFTkSuQmCC\n"
      ],
      "text/plain": [
       "<matplotlib.figure.Figure at 0x15941a90>"
      ]
     },
     "metadata": {},
     "output_type": "display_data"
    }
   ],
   "source": [
    "import matplotlib.pyplot as plt\n",
    "\n",
    "# For simple plots, matplotlib is fine, seaborn is unnecessary.\n",
    "\n",
    "# Now simply use plot\n",
    "plt.plot(x,Y,'o')\n",
    "\n",
    "#Title (use y=1.08 to raise the long title a little more above the plot)\n",
    "plt.title('Binomial Distribution PMF: 10 coin Flips, Odds of Success for Heads is p=0.5',y=1.08)\n",
    "\n",
    "#Axis Titles\n",
    "plt.xlabel('Number of Heads')\n",
    "plt.ylabel('Probability')\n",
    "\n"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "That's it for the review on Binomial Distributions. More info can be found at the following sources:\n",
    "\n",
    "1.) http:https://en.wikipedia.org/wiki/Binomial_distribution\n",
    "\n",
    "2.) http:https://stattrek.com/probability-distributions/binomial.aspx\n",
    "\n",
    "3.) http:https://mathworld.wolfram.com/BinomialDistribution.html"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "Thanks!"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": null,
   "metadata": {
    "collapsed": true
   },
   "outputs": [],
   "source": []
  }
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