Jan 15th, 2021

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

Combinatorics, Equal Likelihood

https://www.gradescope.com/courses/226051/assignments/934322

**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?

**A1: ** live answered

**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?

**A1: ** live answered

**Q:** why is 10 the lowest number?

**A1: ** live answered

**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 :)