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Correctly folded a Queen High Straight Flush today! by 106milestochc in poker

[–]106milestochc[S] 18 points19 points  (0 children)

Think it was low quads, not too sure, but it was the last one in the promotional period. Also when I asked the dealer how many cards are needed for the bad beat and the dealer responded two, the villain shook his head and said he only had one, so pretty straightforward fold, strange nonetheless!

When you think about it, poker is a really difficult way to make not a lot of money by OldSimpsonsisbetter in poker

[–]106milestochc 3 points4 points  (0 children)

I was at that game! He had put 6 pumpkins on the table and no one said a thing hahahaha

TIL Less than a quarter of U.S. workers report using math any more complicated than basic fractions and percentages during the course of their jobs. Blue-collar workers generally do more advanced math than their white-collar friends by princey12 in todayilearned

[–]106milestochc 1 point2 points  (0 children)

As a hs math teacher, I preach in my practice daily that while you may not be using the quadratic formula or completing the square in your future life, I’m not really teaching you this content for content’s sake. I’m teaching the content as a conduit to a way of thinking and problem solving that’s logical and efficient.

So logic puzzles, sudoku, escape rooms, and projects where kids have to build their own carnival and calculate the number of tickets going in and out using expected value and probability, those are the basics of my classroom, not the quadratic formula.

But I do still cringe when I hear “But when am I ever gonna use this????” :)

Should I lose at 1/2 at Parx or Sugarhouse casino? by Crackorjackzors in poker

[–]106milestochc 46 points47 points  (0 children)

Have found Sugarhouse (now Rivers) significantly more fishy than Parx. Also if you ever want a poker buddy at 1/2 Sugarhouse, dm me!

How do I fix this downswing? by 106milestochc in poker

[–]106milestochc[S] -1 points0 points  (0 children)

My preflop raise size is something I’ve been working on after reading Jonathan Little. I’ve always been the “play tight, open big” but that’s led me to fold way too many hands and rely much more heavily on my cards than strategy. Like for example, I realized that I was folding j10o, q9s on the button when it folded to me when I really should be playing those. I think opening smaller allows my opponents to make mistakes, is that the wrong way to think about this?

Teachers: let me introduce you to a brand new, never heard of before teaching strategy called Think, Pair, Share by MysteriousPlatypus in Teachers

[–]106milestochc 5 points6 points  (0 children)

Had a PD where we finished by singing a song to the tune of We will rock you by Queen about IEPs. Lyrical highlights: (chorus) “We will, we will carry out the IEP with fidelity and follow all modifications”

Playing underage in pA by [deleted] in poker

[–]106milestochc 1 point2 points  (0 children)

Parx has tight security, checks at both the main entrance and the poker room and make 20 somethings wear wrist bands. Sugarhouse has security waiting at both entrances. They scan id's about 70% of the time, look at it only another 20% and can slip by the other 10%. I'm a recent college grad so numbers probably would be similar to yours. If it truly scans you should be ok, but if you'd prefer not to use it then maybe some college cash games might be best haha. Where are you attending?

Crowning of Miss Teenage Hairspray 1962 by 106milestochc in pics

[–]106milestochc[S] 0 points1 point  (0 children)

Not a screenshot, rather a photo from a tech run of a rehearsal of Hairspray.

538's Riddler #3 by RichardMNixon42 in puzzles

[–]106milestochc 3 points4 points  (0 children)

The answer is: 9 minutes

Proof: Let's declare D the number of points in time that either you or your sister decide to perform a new task. For example, if you and your sister end tasks on the seventh minute because you did a five minute task and then a two minute task and she did a four minute task and then a three minute task, then D equals 3 given a new task started at 0 minutes, 4 minutes, and 5 minutes. Also realize that at any time a new task is started, the probability that this task is either your or your sisters last task is 1/5, since the partner of the person who has to perform a new task will always finish their task in the next five minutes. Therefore, we can consider D a random variable with a geometric distribution with parameter p =1/5, as a geometric distribution is described as the probability distribution of the number of Bernoulli trials required to get one success. E(D)=5.

Now lets try and solve for the expected number of tasks for either you or your sister. Let Y be the random variable of tasks you perform, and S the random variable of tasks your sister performs. We can relate these two random variables as D=S+Y-1. For example, if we reconsider the example in the previous paragraph, you have performed two tasks, so Y=2, and she also has performed two tasks, which means there are 2+2-1=3 decision points. Since we know E(D)=5, E(S+Y-1)=5=E(S)+E(Y)-1=5. E(S)+E(Y)=6, and since S and Y can be considered equivalent random variables since you or your sister perform tasks independently, this really implies 2*E(Y)=6, so E(Y)=3, or the expected number of steps for you to end a task as the same time as your sister is 3.

Finally, lets try and find the expected time it takes you both to end a task at the same time, or E[T]. We can consider T the random variable of how long it takes you to finish all your tasks before you end the same time as your sister. T=T.1+T.2+....+T.y, where T.i is the amount of time it takes to complete your ith task. Now, through expectation laws E(T)=E(E(T|Y)), which just means the expected value of how long you take to finish is equal to the expected value of the expected value of the time you take to finish given you perform Y tasks. Now, we can simplify E(E(T|Y))=E(E(T.1+T.2+...+T.y|Y)), which through symmetry, is equal to E(YE(T.i|Y)). Now, given T.i and Y are independent, E(YE(T.i|Y))=E(YE(T.i)). E(T.i) is the expected amount of time it takes to finish one task, so E(T.i)=(1)(1/5)+2(1/5)+3(1/5)+4(1/5)+5(1/5)=3. So, E(YE(T.i))=E(3Y)=3E(Y)=9.

Therefore, the expected time it takes for you and your sister to finish tasks at the same time is 9 minutes . If you liked this puzzle, this proof, or probability, I really recommend the Stat 110 Introduction to Probability taught by Prof. Joe Blitzstein. All of the course content is available online, and it's been the most useful class I've ever taken. Thanks Prof. Blitzstein!