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build_quarto_site(unfreeze = FALSE) for release 1.2.0
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| "markdown": "---\ntitle: \"Conditional Probability\"\nauthor: \"John Benninghoff\"\ndate: '2024-03-26'\ndate-modified: '2024-03-26'\ncategories: notes\norder: 105\noutput:\n html_notebook:\n theme:\n version: 5\n preset: bootstrap\n css: assets/extra.css\n pandoc_args: --shift-heading-level-by=1\n toc: yes\n toc_float:\n collapsed: no\n smooth_scroll: no\n---\n\n\nAn exploration of conditional probabilities in R, inspired by a 2015 blog [post](https://statmodeling.stat.columbia.edu/2015/07/09/hey-guess-what-there-really-is-a-hot-hand/) on the hot hand.\n\n\n::: {.cell}\n\n```{.r .cell-code}\n# no libraries\n```\n:::\n\n\n# Background\n\nI recently stumbled across a blog post from 2015,\n\"[Hey - guess what? There really is a hot hand!](https://statmodeling.stat.columbia.edu/2015/07/09/hey-guess-what-there-really-is-a-hot-hand/)\"\nThe article had some R code in it that was intriguing, exploring the following proposition the post\nquoted from a [paper](https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2627354) it cited:\n\n> Jack takes a coin from his pocket and decides that he will flip it 4 times in a row, writing down\n> the outcome of each flip on a scrap of paper. After he is done flipping, he will look at the flips\n> that immediately followed an outcome of heads, and compute the relative frequency of heads on\n> those flips. Because the coin is fair, Jack of course expects this conditional relative frequency\n> to be equal to the probability of flipping a heads: 0.5. Shockingly, Jack is wrong. If he were to\n> sample 1 million fair coins and flip each coin 4 times, observing the conditional relative\n> frequency for each coin, on average the relative frequency would be approximately 0.4.\n\nWhat? OK, so let's follow along with the R code. The first block runs the simulation:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nrep <- 1e6\nn <- 4\ndata <- array(sample(c(0, 1), rep * n, replace = TRUE), c(rep, n))\nprob <- rep(NA, rep)\nfor (i in 1:rep) {\n heads1 <- data[i, 1:(n - 1)] == 1\n heads2 <- data[i, 2:n] == 1\n prob[i] <- sum(heads1 & heads2) / sum(heads1)\n}\n```\n:::\n\n\nThe second block naively calculates the average:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nprint(mean(prob))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] NaN\n```\n\n\n:::\n:::\n\n\nThis doesn't work since, as the post points out, \"sometimes the first three flips are tails, so the\nprobability is 0/0.\" Discarding these gets us the correct average, which is approximately 0.4 and\nnot 0.5 as the quote predicts:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nprint(mean(prob, na.rm = TRUE))\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.4050585\n```\n\n\n:::\n:::\n\n\nReading this and trying the code myself led me to ask, What the heck is going on here?\n\n# Huh?\n\nLet's follow along with the R code and try to work out why the conditional probability is 0.4.\n\n## Simulation\n\nLooking at the first part of the code:\n\n```r\nrep <- 1e6\nn <- 4\ndata <- array(sample(c(0, 1), rep * n, replace = TRUE), c(rep, n))\n```\n\nThis code simulates flipping the coin 4 times in a row 1 million times and stores the results in a\nmatrix:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nhead(data)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n [,1] [,2] [,3] [,4]\n[1,] 1 1 0 0\n[2,] 1 1 0 1\n[3,] 1 1 1 0\n[4,] 0 0 1 1\n[5,] 0 0 1 0\n[6,] 0 1 0 0\n```\n\n\n:::\n:::\n\n\nBy convention, 1 is heads and 0 is tails. If the coin is fair, we should expect the proportion of\nheads to be about 0.5.\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmean(data)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.5000165\n```\n\n\n:::\n\n```{.r .cell-code}\nround(mean(data), 2)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.5\n```\n\n\n:::\n:::\n\n\nWhile there is some expected variance, the proportion is approximately 0.5.\n\n## Calculation\n\nLooking at the second part of the code:\n\n```r\nprob <- rep(NA, rep)\nfor (i in 1:rep) {\n heads1 <- data[i, 1:(n - 1)] == 1\n heads2 <- data[i, 2:n] == 1\n prob[i] <- sum(heads1 & heads2) / sum(heads1)\n}\n```\n\nThis counts the relative frequency of heads immediately after heads, by finding heads in positions\n1-3 (`heads1`), comparing to heads in positions 2-4 (`heads2`), and calculating the proportion of\nheads after heads (`prob[i]`). To see how this works in practice, we can test all possible\ncombinations of heads and tails:\n\n\n::: {.cell}\n\n```{.r .cell-code}\ncalc_prob <- function(flips) {\n heads1 <- flips[1:(n - 1)] == 1\n heads2 <- flips[2:n] == 1\n sum(heads1 & heads2) / sum(heads1)\n}\n\ntest_data <- expand.grid(0:1, 0:1, 0:1, 0:1)\ntest_prob <- rep(NA, nrow(test_data))\n\nfor (i in seq_len(nrow(test_data))) {\n f <- test_data[i, ]\n input <- paste0(\"c(\", toString(f), \")\")\n test_prob[i] <- calc_prob(f)\n print(paste0(i, \": \", input, \" = \", test_prob[i]))\n}\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] \"1: c(0, 0, 0, 0) = NaN\"\n[1] \"2: c(1, 0, 0, 0) = 0\"\n[1] \"3: c(0, 1, 0, 0) = 0\"\n[1] \"4: c(1, 1, 0, 0) = 0.5\"\n[1] \"5: c(0, 0, 1, 0) = 0\"\n[1] \"6: c(1, 0, 1, 0) = 0\"\n[1] \"7: c(0, 1, 1, 0) = 0.5\"\n[1] \"8: c(1, 1, 1, 0) = 0.666666666666667\"\n[1] \"9: c(0, 0, 0, 1) = NaN\"\n[1] \"10: c(1, 0, 0, 1) = 0\"\n[1] \"11: c(0, 1, 0, 1) = 0\"\n[1] \"12: c(1, 1, 0, 1) = 0.5\"\n[1] \"13: c(0, 0, 1, 1) = 1\"\n[1] \"14: c(1, 0, 1, 1) = 0.5\"\n[1] \"15: c(0, 1, 1, 1) = 1\"\n[1] \"16: c(1, 1, 1, 1) = 1\"\n```\n\n\n:::\n:::\n\n\nThere are 16 possible combinations of 4 coin flips, with 5 possible outcomes: 0, 1/2, 2/3, 1, and\n0/0 (`NaN`). \n\n## How it works\n\nLooking at the permutations, the conditional probability starts to make sense. Calculating the\nconditional relative frequency for all permutations gives us:\n\n\n::: {.cell}\n\n```{.r .cell-code}\nmean(test_prob, na.rm = TRUE)\n```\n\n::: {.cell-output .cell-output-stdout}\n\n```\n[1] 0.4047619\n```\n\n\n:::\n:::\n\n\nWhich is approximately 0.4. As we repeat coin flips, the frequency approaches this value.\n\nRecall that the [conditional probability](https://en.wikipedia.org/wiki/Conditional_probability)\nis: $\\large{P(A \\mid B) = \\frac{P(A \\cap B)}{P(B)}}$ (I had to look it up).\n\nIn this case, we are trying to calculate the probability of heads ($A$) given heads occurring at\nleast once ($B$), which is equivalent to `mean(test_prob, na.rm = TRUE)`.\n", | ||
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