Rethink2m6

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.

2M6. Imagine that black ink is heavy, and so cards with black sides are heavier than cards with white sides. As a result, it’s less likely that a card with black sides is pulled from the bag. So again assume there are three cards: B/B, B/W, and W/W. After experimenting a number of times, you conclude that for every way to pull the B/B card from the bag, there are 2 ways to pull the B/W card and 3 ways to pull the W/W card. Again suppose that a card is pulled and a black side appears face up. Show that the probability the other side is black is now 0.5. Use the counting method, as before.

Lösung

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): \(Pr(2b|1b)\).

Bayes-Box to the rescue:

Whenever the probability of all paths (in a tree diagram) is the same, we do not need to write down the probability of the path for the likelihood. It is enough to write the number of paths.

Hyp	Prior	Likelihood	unstand_post	std_post
bb	1	2	2	0.5
bw	2	1	2	0.5
ww	3	0	0	0.0

Categories:

probability
bayes
bayesbox
rethink-chap2
string

---
extype: string
exsolution: NA
exname: rethink2m6
expoints: 1
categories:
- probability
- bayes
- bayesbox
- rethink-chap2
- string
date: '2023-11-08'
slug: Rethink2m6
title: Rethink2m6

---



```{r libs, include = FALSE}
library(tidyverse)
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.


2M6. Imagine that black ink is heavy, and so cards with black sides are heavier than cards with white sides. As a result, it’s less likely that a card with black sides is pulled from the bag. So again assume there are three cards: B/B, B/W, and W/W. After experimenting a number of times, you conclude that for every way to pull the B/B card from the bag, there are 2 ways to pull the B/W card and 3 ways to pull the W/W card. Again suppose that a card is pulled and a black side appears face up. Show that the probability the other side is black is now 0.5. Use the counting method, as before.


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</br>
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# Lösung


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*): $Pr(2b|1b)$.


Bayes-Box to the rescue:

```{r echo = FALSE}
d <-
  tibble::tribble(
  ~Hyp, ~Prior,
  "bb",     1L, 
  "bw",     2L,   
  "ww",     3L, 
  ) %>% 
  mutate(Likelihood = c(2,1,0),
         unstand_post = Prior*Likelihood,
         std_post = unstand_post / sum(unstand_post))
```


Whenever the probability of all paths (in a tree diagram) is the same, we do not need to write down the probability of the path for the likelihood. It is enough to write the number of paths.

```{r echo = FALSE}
d %>% 
  gt::gt()
```





---

Categories: 

- probability
- bayes
- bayesbox
- rethink-chap2
- string