### Learning Goals

You should know the definition of probability and compute probabilities for sample spaces with equally likely outcomes.

### Concept Check

Q: Are the lecture slides posted on the course website?

A1:  Working on it right now. :)

Q: is this similar to hamming distance?

A1:  absolutley is.. very good

Q: (n-2)!

A1:  yes!

Q: does R mean all real numbers?

A1:  Yep, R is the set of all real numbers

Q: Lecture slides still aren’t up right? Just want to make sure I’m not looking in the wrong place :)

A1:  Not yet. Chris has a build system that I’m, um, learning right now. It’s not just editing HTML, unfortunately.

Q: I see the IntroProb tab on the website now! But it’s taking me to Object Not found. Is this happening to others as well?

A1:  http://web.stanford.edu/class/cs109/lectures/3-IntroProbability/3-IntroProbability.pdf

Q: why do we say "sample spaces have equally likely outcomes", why isn't it event spaces?

A1:  In thiis case, we’re talking about sample spaces where all events are equally likely. In those cases, probabilities can just be computed from counting and event space sizes.

Q: so would HHHHHHHHH be equally likely as THTHTHTTTHHT because they are all independent?

A1:  absolutely (assuming the coin is fair)

Q: nvm i get it now

A1:  yay

Q: why if thinking sample space as sum is not equally likely?

A1:  I’m not sure I fully understand the question. We’re currently constraining all outcomes to be equally likely so we can count them and use those counts to construct probabilities.

A2:  We can do that when the outcomes aren’t equally likely, but we need additional information about relative likehihoods.

Q: How is unordered indistinct different from ordered indistinct? Since they are already indistinct, wouldn’t the order not matter anymore in both cases since they’re just the same thing?

A1:  Some of the objects are different from one another. In an unordered world you and {banana, mandarin}, which is the same as {mandarin, banana}. In an ordered world, those two subsets are different.

Q: What’s the connection between the unorderd and ordered strategy that makes the math end up the same?

Q: will both the unordered and ordered probabilities usually/always be equal?

A1:  generally yes, as per Chris’s answer in lecture. :)

Q: feed two birds with one scone 😂 love it

A1:  :)

Q: are we counting, e.g., King, Ace, 2, 3, 4, as a straight?

A1:  good question! not here… there’s no wraparound in standard poker.

Q: Sorry, do straights all need to have the same suit?

A1:  nope… suits don’t matter with poker straights

Q: Is Ace low, high, or both?

A1:  And I was wrong. :) Chris just clarified that he intended Aces to be either high or low. Sorry about that.

A2:  assume it’s only high. Otherwise the problem is a little more complicated.

Q: how did we get 10?

Q: why is 10 the lowest number?

Q: how did we get the 5 if the first card is already chosen?

A1:  Basically, there are 40 different cards than can be the lowest of a straight, but Chris decomposed it to be 10 (the number of ranks that can be the lowest) times choose(4, 1)

A2:  You have to choose the suit of the first card as well.

Q: why did we use (4 choose 1) ^ 5 for the straight probability vs. just 4 choose 1 for the straight flush probability?

A1:  (4 choose 1) ^ 5, because even the lowest rank card can be any one of four ranks, so it’s really 40 * (4 choose 1) ^ 4, which is the same as 10 * (4 choose 1) ^ 5.

A2:  It’s 4 choose 1 for the straight flush for the same reason… there are fundamentailly 40 different possible straight flushes: 4 for each of the 10 possible ranks that can be the lowest rank of the straight.

Q: Why is it (1_C_4)^5? Why isn’t it to the power of 4?

A1:  Because even the lowest card of the straight can be one of four different ranks. Another way of thinking about it is that the lowest card of a straight can be any one of 40 different cards.

Q: In P(E^c), what does the c mean?

A1:  it means not!

Q: Isn’t this or technically an xor? Should it be denoted with a delta rather than a cup?

A1:  xor and or are equivalent if you’re told the two event spaces don’t overlap.

Q: OR is the union right? Not the intersection?

A1:  that’s correct

Q: Why do we take it to the power of n? At the end of the n shuffles there are still only 52! distinct possible orderings, right?

A1:  thats for all orderings of all time (52!) is only one ordering

Q: how did we get (52!-1)^n again?

A1:  52! -1 are the ways that one shuffling is not yours. Probability that n are not yours is (52! -1)^n

Q: what happens as n goes to infinity?

A1:  probability will eventually become 1… but its got to get much larger before it gets close :)