statistics/10_days/11_day3drawingmarbles.ipynb

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   "source": [
    "Solution to: [Day 3: Drawing Marbles](https://www.hackerrank.com/challenges/s10-mcq-6/problem)"
   ]
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   "source": [
    "<h1 id=\"tocheading\">Table of Contents</h1>\n",
    "<div id=\"toc\"></div>\n",
    "\n",
    "\n",
    "- Table of Contents\n",
    "- Math Solution\n",
    "    - Facts\n",
    "- Monte Carlo Solution\n",
    "    - Imports\n",
    "    - Constants\n",
    "    - Auxiliary functions\n",
    "    - Main"
   ]
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    "%%javascript\n",
    "$.getScript('https://kmahelona.github.io/ipython_notebook_goodies/ipython_notebook_toc.js')"
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   "cell_type": "markdown",
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   "source": [
    "This script contains 2 sections:\n",
    "1. Math solution to the problem\n",
    "2. Monte Carlo simulation of the problem"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Math Solution\n",
    "A bag contains 3 red marbles and 4 blue marbles. Then, 2 marbles are drawn from the bag, at random, without replacement. \n",
    "If the first marble drawn is red, what is the probability that the second marble is blue?\n",
    "\n",
    "\n",
    "## Facts\n",
    "- 7 marbles in the bag\n",
    "- 1st one is always red\n",
    "- P(B)?\n",
    "\n",
    "If the first one is always red, we don't calculate the probability of the 1st draw.\n",
    "Thus, we use the 6 remaining marbles.\n",
    "\\begin{equation}\n",
    "\\large\n",
    "P(B) = \\frac{4}{6} = \\frac{2}{3}\n",
    "\\end{equation}"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Monte Carlo Solution"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Imports"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "lines_to_next_cell": 2
   },
   "outputs": [],
   "source": [
    "from typing import List\n",
    "import random"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Constants"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {},
   "outputs": [],
   "source": [
    "MARBLE_DICT = {\n",
    "\t'r'\t:\t3,\n",
    "\t'b'\t:\t4\n",
    "}\n",
    "FIRST_MARBLE = 'r'\n",
    "SECOND_MARBLE = 'b'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Auxiliary functions"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {},
   "outputs": [],
   "source": [
    "def create_marble_bag(marbles: dict) -> List[str]:\n",
    "\t\"\"\"Returns list of marbles to draw from.\"\"\"\n",
    "\tbag = []\n",
    "\tfor k, v in marbles.items():\n",
    "\t\tm = [k for _ in range(v)]\n",
    "\t\tbag += m\n",
    "\treturn bag"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {},
   "outputs": [],
   "source": [
    "def remove_first_marble(bag: List[str], marble: str) -> List[str]:\n",
    "\t\"\"\"Returns bag after removing marble.\"\"\"\n",
    "\tbag.remove(marble)\n",
    "\treturn bag"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {},
   "outputs": [],
   "source": [
    "def check_second_marble(bag: List[str], marble: str) -> bool:\n",
    "\t\"\"\"Returns boolean if sample from bag is the marble.\"\"\"\n",
    "\treturn random.choice(bag) == marble"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 7,
   "metadata": {},
   "outputs": [],
   "source": [
    "def get_ratio(bag: List[str], marble: str, iterations: int) -> float:\n",
    "\t\"\"\"Returns ratio of times sample from bag is marble.\"\"\"\n",
    "\twas_marble = 0\n",
    "\tfor _ in range(iterations):\n",
    "\t\tif check_second_marble(bag, marble):\n",
    "\t\t\twas_marble += 1\n",
    "\treturn was_marble / iterations"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Main"
   ]
  },
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   "cell_type": "code",
   "execution_count": 8,
   "metadata": {
    "lines_to_next_cell": 2
   },
   "outputs": [],
   "source": [
    "def main():\n",
    "\tbag = create_marble_bag(MARBLE_DICT)\n",
    "\tbag = remove_first_marble(bag, FIRST_MARBLE)\n",
    "\n",
    "\titerations =  1000000\n",
    "\tratio = get_ratio(bag, SECOND_MARBLE, iterations)\n",
    "\tprint(ratio)"
   ]
  },
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   "execution_count": 9,
   "metadata": {},
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    {
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     "text": [
      "0.665724\n"
     ]
    }
   ],
   "source": [
    "if __name__ == \"__main__\":\n",
    "\tmain()"
   ]
  }
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