Thursday, August 30, 2007

More on Bayes Rule, and Friday Musings

On Thursday I played some solid poker in just two or three tournaments on the night, leading me to believe that it might actually be possible for me to regain my form from before my vacation after just a few days to shake off the rust. I don't have any huge scores to show for it, due mainly to a gross suckout for almost my entire large stack in the 5050 about two-thirds of the way through the field, but at least I did this again:



I mean I ran through Thursday night's token frenzy like a lawn cutter, mowing down everyone in my path and slowly increasing my stack until I overtook the chip lead with about 25 players remaining (82 runners, 15 would get $75 tier II tokens). From there I only increased my lead, knocking out 14 of the 82 players in the event overall on my way to sittong on double even the 2nd-place-guy's stack by the time the tokens were won. It felt great to make my first big run in a tournament of the week, and hopefully that is something I can build on heading into the weekend and then continuing into the future. I was on an awesome jag there -- for a good 3 or 4 weeks, I was ending up positive almost every single night, just by doing my thing, playing some tournaments and some cash games when the desire piqued. My roll on full tilt probably quintupled during that time, and now it is down just a bit after a week of a few small scores but nothing real big. I think I ran to 50-somethingth in the 50-50 earlier this week, and I cashed in the 28k a few days back as well, but nothing worth being that happy about. But returning to my winning ways in the frenzy seems a bit like a return to my prior form, so hopefully there is more tournament success to come for me in the near term.

I also had some fun on Thursday evening playing low-limit razz with Al, Heather, Drizz and Aposec72, culminating in me bubbling out of a HORSE sng with that crew in Stud Hi when my (TQ)QQ6Tx (Queens Full on 6th street) lost to aposec's (AA)J2JA (2-outered Aces full on 6th street) for a huge stack when we were basically all tied 4-handed for first at around 3k each. For whatever reason that felt like just about the worst Stud Hi beat in the world, but maybe that's just because I don't tend to play so much stud these days, which is true. Anyways as always playing low-stakes with bloggers was a great time and I enjoyed blowing off some steam with some friends. And congrats to Aposec for putting my chips to good use in winning the $5 sng, and to Maigrey for cashing in second place.

Another thing that made all night on Thursday fun was the world of sports. Not only did college football start up, but my Eagles were on tv in New York playing the Jets, and the hated Giants were also on the tube for my viewing pleasure. Eli Manning actually looked pretty good, pretty poised against those scrubs and tryout guys on defense for the one long drive in which he participated. But the best part about Thursday in sports had to be the Phillies**, who completed a 4-game sweep of the Mets in Philadelphia with a dramatic, come-from-behind 9th inning 11-10 win, typical of most of the Phillies' wins this season given their NL-worst team ERA and overall pitching staff (both the rotation and the bullpen). Just four days earlier, on Sunday morning of this week, the Phils stood a seemingly insurmountable 7 games behind the Mets in the NL East. Now as of Thursday late afternoon, the deficit is just 2 games, and suddenly September is gonna be a crazy one for all the closeted Mets fans in this town. And believe you me, there are tons of 'em. Nobody likes to talk about these days anymore, but make no mistake, in the mid 1980's you couldn't give away a ticket to see the loser Yankees, but this city was all Mets, all the time. My how things change. 10 straight years of playoffs, World Series appearances and multiple championships will do that I guess. But for me, getting to watch the Mets collapse at the hands of my beloved Phillies day after day on television in New York was just about the coolest thing I've seen in baseball in a few years.

**Don't think I don't know that now that I've blogged about them, the Phils are about to embark on another losing streak just like they've done every other time they had just about won over the fans that they could really make a playoff run this year. I just had to write about it anyways, because sweeping the division-leading Mets from 2nd place like this does not happen often for us Phillie fans.

God you guys should hear my daughter talk about Barry Bonds. Now whenever she sees me take a pill in the morning (vitamin, aspirin, etc.), she always asks me if those are like the pills that Barry Bonds took to cheat with (rather than explain "the clear" and "the cream" to a 3 year old, I told her that he took some pills that made him cheat). Then she will go off on a 5-minute tirade about how cheating is wrong, Barry Bonds is a cheater, and the real home run champion is Hank Aaron. And then of course how she should always be like Hank Aaron and never like Barry Bonds because cheaters never win, and winners never cheat. That right there is some sickly cute and entertaining stuff coming from the mouth of a 3-year-old, let me tell you. I should post a video sometime. If only I had the first effing clue about how to do that....

Anyways, one other thing about sports this week...how bad did the Yankees thump the Red Sox this week, huh? Now don't get me wrong, I despise the Yankees as an institution because of the ridiculous buy-everyone strategy that they have followed for the past 10 years or so. But I've been to a bunch of games, the stadium is only a few subway stops from where I live (as opposed to Shea, which is like a 50-minute ride away over subway and then a switch to a commuter train), and I know an old-fashioned beating when I see one. After winning the first game of the series on Tuesday, I was at the game on Wednesday night (I heard you guys treated Hammer Wife nicely at the table in the first hour or two), and let me tell you that Roger Clemens laid the beat-down on the Red Sox lineup for 6 straight innings. Sure, Big Papi hit a ball about 10,000 feet deep into the right field seats in the 6th, but Clemens pitched a no-hitter into the 6th before giving up just the one run on the way to a one-run Yankees victory again on Wednesday. Then on Thursday, Yankees ace pitcher Wang took a no-hitter of his own into the 7th inning before the Sox finally got a base hit on their way to a 5-0 shutout loss and a sweep at the hands of the Yankees.

Now to be sure the Yankees are still 5 games behind the Sox with around a month left in the season, so that is a lot of ground to make up, but the Yankees completely took it to the Red Sox in pretty severe fashion over three games. In fact, I was at the game on Wednesday and I have to say, the Red Sox lineup is pretty much not formidable at all like it used to be. You've got Big Papi who is obviously a dominator, even though he too is having a bit of a down year for him, but otherwise you've got Manny Ramirez who is a great hitter but injured, and then...well...that's it. I mean, 5 of the 9 players in Boston's lineup on Wednesday had averages below .270. That is highly unusual for a team with the best record in all of baseball. With Manny back that gets a little better, but considering there is also a pussy DH on that team, it's a pretty poor stat. Meanwhile the Yankees, who admittedly have pitching problems themselves at the back of their rotation, thre 7 of 9 batters in their lineup with averages of .287 or higher. It's not hard to see what happened this week in the Bronx -- with some decent pitching, this Red Sox team is beyond beatable. The Yankees lineup hit two homers off Schilling by the 5th inning on Thursday, crushed Josh Beckett early on Wednesday, and scored two runs in the 1st inning against Dice-K on Tuesday on their way to a sweep. This does not bode well for the Sox over the next month in my view. I bet these two teams find a way to make it interesting at some point during this month.

OK before I sign off today I did want to take a few minutes to address some of the commenters from yesterday who still insist I am totally wrong with my answer to the Monty Hall problem from the other day. I assure you, switching is the correct answer and it does double your chances of picking the right box. As I mentioned yesterday, there are always people (usually most people) who dispute this result vehemently, because it seems so counter to human logic. But it is in fact correct.

In my attempts to further show the correctness of this answer, I posed a similar problem in the comments to yesterday's post that I want to reproduce here, especially because from past experience I know there are probably a great many of you out there who do not agree or at least understand how this conclusion to switch was arrived at:

Instead of having just 3 boxes to choose from as in my original problem, imagine instead that Monty Hall shows you 100 million boxes, and asks you to choose one at random, with one of them containing a million dollars and each of the rest containing just a dead rat. So you're making what is a 1 in 100 million guess to win the million bucks. Then imagine that Monty Hall proceeds to open up every single other box but one -- say box # 98,765,432 -- and shows you a dead rat in each one. You see, he could do this every time to you, because Monty knows where the million actually is, so he can always open up the other 99,999, 998 boxes that you did not choose, leaving unopened only the one box with the million bucks, and your own box. Surely, I posed in the comments yesterday, you would not stick with the one box that you happened to choose with a 1/100,000,000 chance of being right in that situation, rather than switching to the one other box that Monty left over by opening up every other remaining box but that one. It should be more intuitive when you apply this to large numbers that clearly your box has almost zero chance of being the one with the million dollars, given that in every single instance Monty could always use his knowledge of the actual location of the million dollars to open every other box but one, leaving just the box with the million dollars inside. In actuality, just like in the 3-box example I posed the other day, the odds of your box being the money box are still 1 in 100,000,000, while the odds that the money is in the one other box that Monty has left unopened are 99,999,999 / 100,000,000. It is more or less certain in that scenario that you should switch, and in fact you increase your chances of being right by 99,999,999% if you do in fact switch.

The principles and the math in the above example I think show very well why the solution to the simpler, 3-box problem is to switch every time. Yesterday in the comments it was suggested that the law of large numbers somehow changes the math between the 3 boxes example and the 100 million boxes example, but that is flat-out incorrect as I will show here. In the 100 million boxes example, your chances of having picked the right box are 1 in 100 million, while the chances of the last box Monty left unopened being the money box are 99.999999%. Clearly far, far, far more likely to win the money if you switch. In the 3-box example, using the exact same logic and the exact same math, the odds of your box being the money box remain at 33%, while the odds of the other box being the money box are then 66%. Again, clearly it is still far more likely that you get the money if you switch.

And one other point that I will try to rephrase from yesterday's post with the solutions to the problems in it: you people who believe the answer is 50% either way, you would be correct in either one of the following two scenarios:

1. Once he shows the dead rat in box #3, Monty Hall then puts the other two boxes behind a curtain, and randomly jumbles them such that the money could equally be in box #1 or box #2; or

2. Once he opens up box #3 and shows a dead rat, a new contestant is brought in with no knowledge of the prior scenario and simply asked to choose between two boxes, one of which contains money and the other, a dead rat. Although the odds would still be 66% in actuality at this point that the money is in box #2, that new contestant has no way of knowing how the boxes were picked or even that a third box was ever removed from the pool of choices, and thus for that person, the decision is simply a 50-50 guess given what he or she knows about the decision.

So, your answer of 50-50 would be correct if the boxes were somehow re-randomized before the final decision of whether to stay with box #1 or switch to box #2. But that's not what is being done here. In this case, you have the additional very important information that your box was originally picked from a 1-in-3 chance, leaving a 2-in-3 chance that the money was in either box #2 or box #3, and that probabilty factor still holds true after box #3 has been opened and shown to contain a rat. When you come out with an answer that it is 50-50 in either box in this scenario, you are answering the question as if you are simply being asked out-of-the-blue to choose randomly between two boxes, which I think we all can acknowledge is in fact a 50-50 guess situation. But here, you are being asked to choose not between two random 50-50 chances, but rather a 33% box or a 66% box.

As I said yesterday, I myself took several months to accept the veracity of this solution, and it is a truly difficult concept for anyone with a human brain to get his or head around because it is simply so totally counter-intuitive to everything our brains tell us about the situation. I encourage all of you out there if you are interested to explore the various explanations and assessments of this problem available online and elsewhere. There happens to be a pretty decent treatment of several aspects of the problem on Wiki, and more than that, if you're still not convinced, how about some actual testing of your own? This option should be especially attractive to the sciencey types out there, so take a look at the following links and test this theory yourself, as many times as you see fit before you are willing to start accepting the truth of the value of switching boxes:

Decisionhelper.com Monty Hall Problem Simulator -- I just ran it and stopped when switching had worked 10 times and staying had worked only 4 times.

Cut-the-knot.org simulator -- I also just stopped after switching won me 10 out of 14 trials.

University of South Carolina simulator -- here I stopped when switching had won 9 out of 14 trials. This one might be the most satisfying to the nonbelievers out there because it pauses and allows you to choose whether to switch or not, as opposed to the first two which basically just tell you what would have happened had you switched and had you stayed with your original pick.

I urge anyone still interested in the problem but not totally down with the accuracy of the solution I have presented to go to one of the simulator sites above and perform as many trials as you need to feel comfortable that in fact it clearly wins roughly 2/3 of the time if you switch boxes than if you stay with your original pick. The odds are not 50-50 at that point in the problem as posed, but rather it is literally twice as likely (66% vs 33%) that you pick the winning box if you switch when the option is offered in the problem as presented.

I think that's enough heavy stuff for today. I'll be back on Monday when I plan to spend some time discussing Bayes' rule as it applies to poker. Have a great weekend, and you can look for me at some point in that 9:45pm ET token frenzy that I've been killing recently to the tune of 5 shiny new $75 tokens sitting in my full tilt account. As usual because Friday nights can be tough I'm not sure if I can make Kat's DonkeyMint tonight at 9pm ET on full tilt, but as always that tournament is a $1 rebuy and is consistently a fun time, one that turns into an actual fun tournament after an hour of monkeypushing and donkeychasing everything with at least an out or two. Password as always is "donkarama" so please come out and enjoy yourself in this event that is open to anyone and everyone, as long as you've got at least $1 in your full tilt account (Bayne). Ha ha. Good weekend everyone.

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10 Comments:

Blogger pokerpeaker said...

Tune in next week when Hoy FINALLY talks about how all this shit relates to poker!



Posting a video is sickly easy. Post it on Utube and it will post it for you right on your blog. It is super easy and you seem like a smart fellow. Trust me if I can do it....

10:30 PM  
Blogger Astin said...

Ugh, I just posted the statistical proof in the last post, and this one goes up!

So I'll copy it here (and fix a typo), because I find the Wikipedia application of Bayes' theorem to be a bit confusing still.

Oh, and for what it's worth, there's a 3rd time it's 50-50. If Monty picks YOUR door to show it doesn't have the prize. Of course, that doesn't happen.

I will grant that Hoy missed a key point when he originally presented the problem (although it's implied by the fact it's Let's Make A Deal) - Monty will ALWAYS pick the dead rat.

There is a variable that the mind doesn't intuitively take into account here - that your choice of door limits the options Monty has when revealing. If you pick the winning prize, he has two choices. If you pick a rat, he only has one choice.

It's a priori reasoning, so it's a pain to wrap your brain around.

It doesn't matter what door you pick initially, the probility is 1/3. So let's say you pick door A.

Monty can only open a door that doesn't have a prize. He opens C.

Probality he opens C if prize is behind A? 1/2 = P(c|A)

Probability he opens C if prize is behind B? 1 = P(c|B)

Probability he opens C is prize is behind C? 0 = P(c|C)

So, probability he opens C (P(c))?

1/3 * 1/2 + 1/3 * 1 + 1/3 * 0
= 1/6 + 1/3 + 0
= 1/2

And now Bayes theorem:

Let P(x|y)= Probability prize is behind door x if door y is opened.

P(x|y) = P(x)P(y|x)/P(y)

Probability prize is behind A (your pick) if door C is opened:

P(A|c) = P(A)P(c|A)/P(c)
= (1/3)(1/2)/(1/2)
= (1/6)/(1/2)
= 1/3

Probability prize is behind B (not your pick) is door C is opened:

P(B|c) = P(B)P(c|B)/P(c)
= (1/3)(1)/(1/2)
= (1/3)/(1/2)
= 2/3

There's the statistical proof. As long as Monty always picks a door, and always picks a door that doesn't have the prize.

11:00 PM  
Blogger jamyhawk said...

Thanks for these posts as it is a great topic to debate.

Someone brought up the deal or no deal scenario.

You pick 1 box from 25. If YOU are randomly eliminating boxes, and you are down to five, with 1 having a penny and 1 having the million dollars, have your odds changed any? According to the TV show, they will say you have a 1 in 5 chance of having the million dollars. Is this incorrect (you still have a 1 in 25 or 4% chance) or is it actually correct since you have randomly chosen boxes to eliminate?

I would be interested in your thoughts on this.

11:01 PM  
Blogger Astin said...

Also, since Bayes' Theorem is essentially to find the likelihood of something being the case after you have learned more information.

P(A|B) = P(B|A)P(A)/P(B)

Where
P(A|B) = Probability of A given B
P(B|A) = Probability of B given A
P(A) = Initial probability of A
P(B) = Initial probability of B

Hence the Monty Hall problem: You know your chances before, you get more information from Monty, and apply that to your situation.

So ask yourself, "Where in poker would I run into the situation where I receive more information that I can apply to previous odds?"

Then when Hoy finally gets around to applying Bayes' theory to poker, you can see if you were right.

11:13 PM  
Blogger Astin said...

I should just post this in my blog dammit.

Deal or no Deal. Oddly, I mentioned it. And now I'm giving it some thought. This will be real-time :).

You have a 1 in 25 chance initially. Let's simplify it to Monty again, and make it 1 in 3. Except now instead of Monty, YOU pick the door, and have no idea what's behind.

You pick A (1/3) for the prize.

You pick C for the elimination, and it doesn't have the prize.

This is now essentially the Deal Or No Deal situation I think.

Your chance of picking C is always 1/2 regardless of which door has the prize.

So:

P(c|A)=P(c|B)=P(c|C)=1/2

so P(c)=(1/3*1/2)*3=1/2

P(A|c)=P(A)P(c|A)/P(c)
=(1/3)(1/2)/(1/2)
=1/3

P(B|c)=P(B)P(c|B)/P(c)
=(/3)(1/2)/(1/2)
=1/3

So now the odds are even, or 50-50.

In the Deal case, you have 5 cases left (including yours), and I think they all have a 1/25 chance of having the million, because it's been random chance. So 4:1 odds against you, or 1 in 5.

Deal skews the game though by it not being a binary problem. It's not that one case has $1 million, and the other 24 have 0. So the above solution assumes you only care about the million, and everything else is chump change to you. Once you toss EV into the equation, it alters your decision-making process (but not the odds of a case have $1 Mil).

11:29 PM  
Blogger Hammer Player a.k.a Hoyazo said...

Jamy, it's a good question about Deal or No Deal. I reproduce here the treatment of that very issue way down near the bottom of the Wiki entry I linked to:

"This problem appears similar to the television show Deal or No Deal, which typically begins with 26 boxes. The player selects one to keep, and then randomly picks boxes to open from amongst the rest. In this game, even until the end, the box the player initially selects and all boxes left unrevealed are equally likely to be the winner. The distinction is that any box the player picks to open might reveal the grand prize, thereby eliminating it from contention. Monty on the other hand, knows the contents and is forbidden from revealing the winner. Because the Deal or No Deal player is just as likely to open the winning box as a losing one, the Monty Hall advantage is lost. Assuming the grand prize is still left with two boxes remaining, the player has a 50/50 chance that the initially selected box contains the grand prize."

So that's the difference. In Deal or No Deal, the contestant could pick to open the suitcase with the million at any time along the way in running through all 25 other boxes. In Let's Make a Deal, however, Monty is always going to open one of the doors with the gag prize, and he always knows where there is one. That difference preserves the 1/26% chance of your suitcase being the million from beginning to end in the Deal or No Deal situation.

Good question though, I've thought about the same thing myself since I first watched Deal or No Deal sometime a year or two ago.

11:31 PM  
Blogger B said...

The key is in stating that Montey *always* picks the dead rat. If I missed that, then my apologies.

The problem is easier now, although nice work from Astin in showing it explicitly.

Here is the easiest way (for me) to explain it, when you say he *always* picks the dead rat.

You pick a box, there is a 1/3rd chance you pick the money. You see the dead rat and pursue the always switch strategy. Thus 1/3rd of the time you will have chosen the money and will switch to a dead rat and lose.

Probabilities must sum to 100%, 1-(1/3) = 2/3 = success rate in switching (again, with the stipulation that Montey always shows the dead rat).

Probably silly to have to stipulate that anyway, because what kind of an ahole game show host would show you the money and then ask you if you wanted to switch???????

So I change my stance. It is in my nature to consider all possible outcomes (ie Montey showing the money) when attacking a problem!

Biggestron

11:37 PM  
Blogger smokkee said...

the Angels have been b-slappin' AL division leaders and wannabe wild card teams for the last month or so.

i can't wait for the post season.

12:14 AM  
Blogger Francase said...

that was a sick beat there......I was thinking you had another pair in the hole and so you had queens up (not trips) and I was way ahead with aces up.

it's probably the worst beat i've laid out in a long time.

sory about the 50/50 doing the same...

2:02 AM  
Blogger Dr Zen said...

The one mistake you make in this post is "Although the odds would still be 66% in actuality". They would not. They would be 2 to 1 for you, but not for the new contestant. Different players can have different odds (obviously, Monty's odds of picking the right box would be different from yours!) because they are actually offered different bets (and Monty would be taking a whole different bet too). The new player faces exactly the same position as you would if Monty randomised the boxes. You are mistaking what the bet is that you are being offered. It is that you chose the right box in the first place, not that you will be right to switch. (This is why people cannot grasp the right answer conceptually. They are convinced that they are being offered a bet on which box will now be correct, when they are actually being offered a bet on their initial choice.)

If you don't agree, think about the 100 million boxes again. You pick a box and it's nearly 100 million to one that you picked the right one.

Then you sit back down. Monty eliminates all bar one other box and then your wife is brought from backstage and asked to choose a box from the two remaining.

Clearly, it's not nearly 100 million to one that she picks the right one. It is even money. But if Monty offered you the bet, not her, he is not offering you odds on the choice, but on whether you picked right in the first place. That is not even money. It's nearly 100 million to one!

4:27 PM  

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