So, this is it. The first week of the study group. For this week I would like you to read Chapter 1 of the book. It is really a simple introduction to probabilities and in games of chance in general.
I have a couple of discussion points, please feel free to add to the discussion or keep the discussion going too, by replying to this post. Also, any questions you might have about Chapter 1 should be attached under "comments" in this post.
- The authors say that right now the decision with highest expected value will be the best one. As in many things in life, making the most money now, is not necessarily the best for making the most money in the future. For example a company might increase their profits in the short term by dumping poison into some lake, but the negative PR that comes with that would greatly offset their short term gains. Is it not the same with poker? Don't you think just focusing on one hand is narrow-minded and will not yield you the highest expectation in the long run?
- Do you think any game of chance with complete known information between two players is solvable?
Anyway, I warned you that this is not an easy book and there will be homework, so the first assignment is here. It is a PDF file, 20.5KB. Please send the completed homework to my e-mail address bellatrix78 at yahoo.com (replace the "at" with @). I don't care in which format you send it (Word, LaTeX or handwritten scanned), but please try to write out all formulas correctly with an editor (it gets really tedious looking at text formulae, e.g 10^5/2*3). I will go over any homework at any time, but if you want to keep up with the course, send it to me by Monday July 21st, which is when the next Chapter will be discussed and the next homework will be posted.
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19 comments:
Nice idea. Assignment link is broken though.
fixed link
For 2a, I'm guessing we're supposed to assume we're the only ones playing that week. Your EV goes down as more people play because you run into a higher chance of chopping.
Yes, for simplicity, assume that you are the only one playing.
Also for 2a, the pari-mutuel winnings are not stated and vary with the size of the pool. Unless there's more information that I'm not finding, then this doesn't seem to be calculable.
http://www.calottery.com/Games/MegaMillions/HowToPlay/
Hi Aaron!
That sucks, I didn't check that the parimutuels weren't showing. When we did the exercise in Germany they showed them on their website. Gotta be more careful, sorry.
Anyways, I updated exercise 2a) to some more realistic numbers. Hope this time it works out.
Thanks Bella for hosting this. I'm really looking forward to it.
I thought chapter 1 was a good refresher on probabilities and a good indicator that this is going to be a tough read (I had to spend some time on the last few paragraphs to really make sure I nailed it).
On discussion topic 1) I think the authors would agree that having the highest long-run expectation is best, so that giving up some positive expectation now that would pay off in future expectation would be a good decision. Reconciling this with the mathematics of poker, your actions now may effect the distributions of events later (by impacting what range of cards your opponent plays) or the value of those distributions (by impacting how much your opponent is willing to gamble).
On 2) I would say yes, any game with complete information is solvable, given sufficient computing power. It's just that the computing power required is monumental for complex games - and certainly beyond the reach of the human brain's capabilities.
BTW -- any reason for doing this on blogger rather than on the forums?
Can someone give a primer on sports betting (esp. moneyline) and how we might figure out the odds based on that? Also, I know the house takes a vig on sportsbets, but I don't know how much - if we're meant to include, can someone explain this as well?
Based on where we are in the text, I'm wondering if we're equipped to answer question 2a? From chapter 1, I would calculate the prob of getting 3 numbers using something like (6/49)*(5/48)*(4/47) = 0.0011 or 1 in 921. However this is way off from the published odds of 1 in 57. A quick look at wizard of odds tells me that the right answer has something to do with combinations (i.e. 6/49 chance of getting the first number is wrong because there are multiple opportunities to hit the first number) - however I don't know how to do the combination calculations.
So... should I go read up on combinations, or use the published odds chart, or am I missing something?
Counting combinations is something that you can do for small cases such as these by just making a quick list.
How many ways can you get 5 right out of 5? Just one because you can't miss anything.
What about 4/5? Use an X to represent a hit and an O to represent a miss:
XXXXO
XXXOX
XXOXX
XOXXX
OXXXX
That's 5 ways of hitting 4/5. You can do it again for 3/5, 2/5, and 1/5 and be done with that part in about 2 minutes.
There are more mathematical ways of doing it by using factorials and all that, but I don't think those will be necessary in the long run.
OK thanks Aaron, so to take it to the next step, if your 5-number lotto has 10 numbers in it, then there are 10x9x8x7x6 possibiilities = 30,240 - so you would have a 1 in 30,240 chance in getting all 5 numbers correct, a 5 in 30,240 chance of getting 4 correct and so on. Is that correct?
Nah crap that gives me about 10 billion to 1 to hit all 6 numbers in the 49 number game (ie. 49*48*47*46*45*44 = 10 billion). Help!
Hi CrMenace!
Regarding Sports Betting. The advantage for the sportsbooks comes from having a spread on two sides. The most obvious spread is the 20 cents they have on the -110/-110 bet, in reality it should be +100/+100 to make it equal.
In moneyline games if one team is favored as -180, if the sportsbook makes no money the other side will be +180. So normally they adjust it and make the line -200/+160. Anyways, this is not important for Exercise 3, just wanted to introduce the advantage that sportsbooks have.
The -180, means that if you pay the sportsbook 180$, if your team wins (moneyline just means that the team wins straight up, no point spread or something like that) then you get paid 100$. That means that teams with a "-" in front of them are favored. If the team has a "+" in front of them, then the team is an underdog. Say you put 100$ on a +150 team, then you get 150$ if they win. The most even money would be the +100. If your team wins, for every 100$ you bet you get 100$, so the odds on that bet are 1:1 and the probability of your team winning is 50%.
I hope that made some sense.
Crmenace, before you even try 2a) maybe you could mail me 1c), that helps get in the mindset of picking "something" out of a pool of something. :)
When you get AKs, it doesn't matter whether you get the ace first, then the king, or the king first, then the ace. You're looking at the final state of cards you see in front of you. You should make sure you got that one right, too.
The same thing happens with lotto numbers. You don't care what order things are picked in, so you need to make sure you adjust your numbers for that.
Bella, I assume for 2b) you're refering to a double-zero roulette, as that is what is played in Vegas. In Europe and Asia, some of the wheels have only 1 Zero. Of course, the double-zero would increase the house edge by a bit.
mav, doesn't matter which one you use. I'll know it from the result or because you write it out for me. There are therefore two possibilities.
Aaron,
Maybe using factorials is most appropriate for our assignment (lottery question). It can become tedious when solving a long equation with such small fractions. eg:
(6/49)(5/48)(4/47)...
This is just to determine the probability of certain combinations. Once all probabilities are determined, the expected value formula becomes even more tiresome.
I struggled with the bonus 3 card poker question. I believe I found the correct number of combinations for straight flushes, (non-flush) straights, (non-straight) flushes, and sets.
When I calculated the probability of pairs, the result appeared far too small, though I can't immediately find the flaw.
There are 13 ranks of cards, so:
13(4/52)(3/51)(48/50)
The first two terms equal 1, which makes sense, because once you see a first card, you then must match it exactly one time in your next two cards. Although, this may not quantify the times when the second two cards are a pair. Though this is a combination, not permutation. Prease halp!
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