Rethink2m5

probability

bayes

bayesbox

rethink-chap2

string

Published

November 8, 2023

Aufgabe

This question is taken from McElreath, R. (2020). Statistical rethinking: A Bayesian course with examples in R and Stan (2. Ed.). Taylor and Francis, CRC Press.

2M5. Now suppose there are four cards: B/B, B/W, W/W, and another B/B. Again suppose a card is drawn from the bag and a black side appears face up. Again calculate the probability that the other side is black.

Lösung

The only difference to the question 2M4 is that we now have two bb cards, rendering the prior plausibility for balck twice as high. The rest is the same as in 2M4.

Let’s label the cards bb (black on both sides), bw (black on one, white on the other), and ww (both sides are white), respectively.

Wanted is the probability that the second side of the card is black (2b), given one side is already identified as black (1b): $Pr(2b|1b)$.

library(tidyverse)

d <-
  tibble::tribble(
  ~Hyp, ~Prior,
  "bb",     2, 
  "bw",     1,   
  "ww",     1, 
  ) %>% 
  mutate(Likelihood = c(2,1,0),
         unstand_post = Prior*Likelihood,
         std_post = unstand_post / sum(unstand_post))

Hyp	Prior	Likelihood	unstand_post	std_post
bb	2	2	4	0.80
bw	1	1	1	0.20
ww	1	0	0	0.00

Tables like this are sometimes called “Bayes Box” or “Bayes Grid”.

Alternatively, we can write out the two black cards not in one, but in two rows, so that each black card gets it own row.

Hyp	Prior	Likelihood	unstand_post	std_post
bb	1	2	2	0.40
bb	1	2	2	0.40
bw	1	1	1	0.20
ww	1	0	0	0.00

The second Bayes Box yields the same probability as the first, which is reassuring. However, it feels a bit akward (at least to me) to write down the same hypothesis twice.

Categories:

probability
bayes
bayesbox
rethink-chap2
string

--- extype: string exsolution: NA exname: rethink2m5 expoints: 1 categories: - probability - bayes - bayesbox - rethink-chap2 - string date: '2023-11-08' slug: Rethink2m5 title: Rethink2m5 --- ```{r libs, include = FALSE} library(gt) ``` ```{r global-knitr-options, include=FALSE} knitr::opts_chunk$set(fig.pos = 'H', fig.asp = 0.618, fig.width = 4, fig.cap = "", fig.path = "", message = FALSE, warning = FALSE, out.width = "100%", cache = TRUE) ``` # Aufgabe This question is taken from McElreath, R. (2020). *Statistical rethinking: A Bayesian course with examples in R and Stan* (2. Ed.). Taylor and Francis, CRC Press. 2M5. Now suppose there are four cards: B/B, B/W, W/W, and another B/B. Again suppose a card is drawn from the bag and a black side appears face up. Again calculate the probability that the other side is black. # Lösung The only difference to the question 2M4 is that we now have two *bb* cards, rendering the *prior* plausibility for balck twice as high. The rest is the same as in 2M4. Let's label the cards *bb* (black on both sides), *bw* (black on one, white on the other), and *ww* (both sides are white), respectively. Wanted is the probability that the *second* side of the card is black (*2b*), *given* one side is already identified as black (*1b*): $Pr(2b|1b)$. ```{r} library(tidyverse) ``` ```{r} d <- tibble::tribble( ~Hyp, ~Prior, "bb", 2, "bw", 1, "ww", 1, ) %>% mutate(Likelihood = c(2,1,0), unstand_post = Prior*Likelihood, std_post = unstand_post / sum(unstand_post)) ``` ```{r echo = FALSE} d %>% gt() %>% fmt_number(columns = 5) ``` Tables like this are sometimes called "Bayes Box" or "Bayes Grid". Alternatively, we can write out the two black cards not in one, but in two rows, so that each black card gets it own row. ```{r echo = FALSE} d2 <- tibble::tribble( ~Hyp, ~Prior, "bb", 1, "bb", 1, "bw", 1, "ww", 1, ) %>% mutate(Likelihood = c(2, 2, 1, 0), unstand_post = Prior*Likelihood, std_post = unstand_post / sum(unstand_post)) ``` ```{r echo = FALSE} d2 %>% gt() %>% fmt_number(columns = 5) ``` The second Bayes Box yields the same probability as the first, which is reassuring. However, it feels a bit akward (at least to me) to write down the same hypothesis twice. --- Categories: - probability - bayes - bayesbox - rethink-chap2 - string