# Segment 3 Sanmit Narvekar

## Segment 3

#### To Calculate

1. The slides used a symmetry argument ("relabeling") to simplify the calculation. Redo the calculation without any such relabeling. Assume that the doors have big numbers "1", "2", and "3" nailed onto them, and consider all possibilities. Do you still have to make an assumption about Monty's preferences (where the slide assumed 1/2)?

Nothing changes in this scenario. The assumption about Monty's preferences being 1/2 on picking door 3 when the contestant picked door 2 (which is also the prize door), is just dependent on how many total doors there are, and assumes Monty will randomly open one of the remaining doors (which both have goats).

#### To Think About

1. Lawyers are supposed to be able to argue either side of a case. What is the best argument that you can make that switching doors can't possibly make any difference? In other words, how cleverly can you hide some wrong assumption?

2. We stated the problem as requiring the host to offer the contestant a chance to switch. But what if the host can offer that chance, or not, as he sees fit? Then, when offered the chance, should you still switch?

The only thing that matters is whether Monty opens one of the remaining doors (revealing a goat) after you make your initial pick. If he does not show you anything, then each door still has a 1/3 probability of containing a car, and it doesn't make a difference whether you switch or not.

Now, assuming that he does open another door (and that door contains a goat), then you should always switch if you can. This reduces to the same scenario in the slides -- his choice of whether to offer you to switch or not doesn't change the probabilities, as it affects neither the prior **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(H_i)}**
nor the evidence **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(O_j|H_i)}**
, where **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H_i}**
is the event that the car is behind door **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i}**
and **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle O_j}**
is the event where Monty opens door **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j}**
to reveal a goat.

3. Mr. and Mrs. Smith tell you that they have two children, one of whom is a girl.

(a) What is the probability that the other child is a girl?

Assuming there is an equal probability of having a boy or a girl (this is the background information) and that the gender of each child born is independent of the previous ones, the probability that the second child is a girl **given that** the first is a girl is simply 1/2. Mathematically:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P(\text{2 Girls} | \text{1 Girl})= \frac{P(\text{Girl} ) P(\text{Girl} )}{P(\text{Girl} )} = P(\text{Girl} ) = 1/2 }**