No. of Recommendations: 38

This is a thought-provoking article to take to heart. It presents 5 real-world problems and the counter-intuitive answers. Read each one nd think the answer through. They make perfect sense, but it's not what "common sense" would say was the right answer. It begins with the "Monty Hall Problem".

Pick one door of three possible. Monty opens a non-picked door that shows the coveted prize isn't there. Do you change your pick?

http://www.theatlantic.com/business/print/2012/11/5-statisti...

No. of Recommendations: 12

Hmmm. Where to begin? First, the Monty Hall scenario: They have it wrong. You have a one in three chance of being right. You pick a door. One you did not pick is opened, revealing no big prize. The game just changed. It is now a new game, 2 doors, pick one, the odds are 50/50. Change, don't change...it does not change the odds. The only way this is not true is if Monty Hall knows which is the right door and you can game his psychology in which door he shows you. Otherwise, you are in a new game, 2 doors, equal odds.

The next three I do not have issues with. However, the last one.... I think I would choose treatment A, not B. And, if I were the manufacturer of treatment A, I think I would run more tests on small stones. Clearly small stones are easier to treat with EITHER medicine. The better overall results to treatment B is a factor of the significantly larger number of small stones vs. large stones in their test, not a superiority of the treatment. I am reminded of the quote about three types of lies: "lies, damn lies, and statistics". Here the statistics do not tell the whole truth as the variation in stone size seems to not be adequately considered when evaluating over-all results.

I guess editors and journalists, even at a prestigious magazine such as the Atlantic, do not understand even basic statistics.

Cheers,

Doug

No. of Recommendations: 7

PM

ABRAHAM WALD'S MEMO is an old story I've heard before and a great description of the myths of survival in war. While survivors attribute it to an assortment of factors such as skill, training, bravery, heavenly intervention and even the lucky dice it really was a case of survivor bias and really mostly just chance.

A more interesting example of statistics used to good purpose would have been the RAF reduction of pilot losses in bomber command by 50%. They accomplished this by reducing the number of pilots in the bombers from two to one. Bomber losses were for the most part caused by night fighters and flak. Once the basics (fly as high as possible and stay in the bomber stream (an interesting statistical story in itself***)) there was almost nothing the crew could do to improve the odds. There was about a 10% chance of aircrew surviving by bailing out of an RAF bomber that had been shot down.

*** - The bomber stream was based on the difficulty the German night fighter controllers had when many were concentrated in a small area with the basic RADAR of the day. They were therefore assigned to geographical boxes. The concentrated bomber stream would provide a wealth of targets for those few night fighters in the boxes the "stream" went through but those few fighters would only be able to engage a limited number of bombers and would also soon run out of ammo.

Tim

http://en.wikipedia.org/wiki/Bomber_stream

**British analysis of the loss of bombers to night fighters of the Kammhuber Line was one of the first applications of statistical analysis which would become known as operational research.** The introduction of GEE allowed the RAF bombers to fly by a common route and at the same speed to and from the target, each aircraft being allotted a height band and a time slot in a bomber stream to minimize the risk of collision. Data provided to the British scientists allowed them to calculate that the bomber stream would overwhelm the six potential interceptions per hour that the German night fighters could manage in any one Himmelbett zone. It was, then, a matter of calculating the statistical loss from collisions against the statistical loss from night fighters to calculate how close the bombers should fly to minimise RAF losses.

No. of Recommendations: 0

*Thank you for recommending this post to our Best of feature.*

Hmmm. Where to begin? First, the Monty Hall scenario: They have it wrong. You have a one in three chance of being right. You pick a door. One you did not pick is opened, revealing no big prize. The game just changed. It is now a new game, 2 doors, pick one, **the odds are 50/50**. Change, don't change...it does not change the odds.

Yup!!!

The only possibility I see would be if:

`The host says that once you pick a door, he'll open one of the doors you didn't pick to reveal a goat.`

....

This is actually based on a real game show, and the result has been the source of controversy for years.

the host was somehow hinting with his actions... which would not be impossible?

Tim

No. of Recommendations: 2

*the host was somehow hinting with his actions... which would not be impossible? *

ummm, more likely, the show was paid by the prize vendors for exposure of their products, so showing a prize not won was in fact fulfilling the terms of an advertising contract, not a hint or attempt to manipulate the contestant.

Steve

No. of Recommendations: 14

Lobodoug:

*The game just changed. It is now a new game, 2 doors, pick one, the odds are 50/50. Change, don't change...it does not change the odds.*

tim:* *

Yup!!!

+++++++++++++++++++++++++++++++++++++

uuuhhhmmmm, **NOPE!!! **

PolymerMom and Atlantic Mag have it correct. The crux is correctly identified by Lobodoug, but not correctly understood. IT IS NOT A NEW GAME!!!

Two different ways you can readily chew on this

1) read, for example, this wikipedia article with its multiple presentations using Bayesian formulations, Monte Carlos, etc...: http://en.wikipedia.org/wiki/Monty_Hall_problem

OR

I had an obsessive friend who did not believe me when I told him this, and on a bet from me he simply went through 100 of the Monty Hall shows and found that the show's results (REALITY indeed) corresponded with the 2/3 chance predicted by statistical theory!

PolymerMom's excellent point was that "Common Sense" is alarmingly wrong, and I will add that our brains are wired to be excellent hunter/gatherer/lovers rather than excellent denizens of modern times with curtains and game shows -- or for that matter investment funds and sexting.

The phrase "Common Sense" has lost its original extremely useful meaning. It now means (as PM used it) *understanding or reasoning that is basic or common*, a human version of "horse sense". However, the original meaning is from Aristotle and meant a kind of 6th sense that combines and amplifies everything that can be brought in by the other senses -- the reasoning that occurs with the senses themselves *comm*unicate. This was taken a step further in the Scottish Enlightenment, which first popularized the English phrase common sense, but they meant by that a form of reasoning based on skeptically combining (communing) received traditional understanding, the input of the 5 senses structured through objective tests, and intuition -- in other words Scientific Testing.

It is useful to remember that original meaning of "horse sense" was ironic, that a horse will often have better understanding than found in "common sense" (in the modern sense of off-hand and off-the-cuff understanding). If your horse does not want to ford a river, despite your common sense telling you all is well, better think again, or you may drown. Or open the door with the goat!!

david fb

No. of Recommendations: 7

*The game just changed. It is now a new game, 2 doors, pick one, the odds are 50/50. Change, don't change...it does not change the odds. The only way this is not true is if Monty Hall knows which is the right door and you can game his psychology in which door he shows you. Otherwise, you are in a new game, 2 doors, equal odds.*

You're wrong.

It's NOT a new game.

Think of it this way:

You initially pick A. The chance that the car is behind A is 1/3. The chance that it is behind B or C is 2/3.

So if you could pick BOTH B and C instead of A, would you do it? Of course.

And that is, essentially, what you do if you switch after being shown that B or C don't have the prize.

Because we knew already that one of them didn't have the prize, but that collectively they have a 2/3 chance.

Looking from the beginning of the bet, if you switch, you will get the prize in case EITHER B or C had it.

I think you're right about the stones, though. It's a bad example for Simpson's paradox. A really bad one.

No. of Recommendations: 1

This explanation made the most sense to me:

http://en.wikipedia.org/wiki/Monty_Hall_problem

*That switching has a probability of 2/3 of winning the car runs counter to many people's intuition. If there are two doors left, then why isn't each door 1/2? It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. His initial probability of winning is 1 out of 1,000,000. The game host goes down the line of doors, opening each one to show 999,998 goats in total, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door...*

...

Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. **It's as if Monty gives you the chance to keep your one door, or open all 999,999 of the other doors, of which he kindly opens 999,998 for you, leaving, deliberately, the one with the prize.** Clearly, one would choose to open the other 999,999 doors rather than keep the one.

Also, this amusing note:

*Interestingly, pigeons make mistakes and learn from mistakes, and experiments show that ***they rapidly learn to always switch**, unlike humans (Herbranson and Schroeder, 2010).

Of course, Herbranson and Schroeder should note that humans are not regularly stuck in a cage and asked to peck for food, or I imagine we'd come to the same conclusion rather quickly. Practical experience modifies "common sense".

No. of Recommendations: 3

* The only way this is not true is if Monty Hall knows which is the right door and you can game his psychology in which door he shows you.*

One of the reasons that the Monty Hall problem seems so counter intuitive is that the problem is stated badly and does not match the "game" the way that is was really played. He did know the right door and was not required to show you a door.

If someone was playing with his own money then they would know that they had already won two thirds of the time and they would only offer to let you swap doors when they knew that you had already won (or to entice you to swap doors more often in the future).

If anyone wants to play that game with me for real money at even odds I'll be glad to play Monty Hall, if I get to know the right door and if I do not have to offer to let you trade doors each time. If we play long enough I will guarantee that I will win at least two thirds of the time.

No. of Recommendations: 3

*I think you're right about the stones, though. It's a bad example for Simpson's paradox. A really bad one.*

Nate Silver will be aghast.

And I don't care what the math says, I think it's 50-50 for Monte Hall. How does the door know if this is "the same game" or a new one with only two doors to chose from? Monte may know, but he's not doing the choosing. Door=roulette wheel. No institutional memory. Every spin is independent of whatever came before.

No. of Recommendations: 0

No. of Recommendations: 0

*How does the door know if this is "the same game" or a new one with only two doors to chose from? *

You don't even NEED him to show you a door without the grand prize. Basically, your choice is between your original door and the other two doors. You could make your original pick FIRST, THEN switch, and THEN let Monty show you a door that doesn't have the grand prize. His action is predetermined if you haven't picked the grand prize, which was one chance in three.

No. of Recommendations: 1

[i]And I don't care what the math says, I think it's 50-50 for Monte Hall. How does the door know if this is "the same game" or a new one with only two doors to chose from? Monte may know, but he's not doing the choosing. Door=roulette wheel. No institutional memory. Every spin is independent of whatever came before. [/i]

The answer is definitely 2/3 to switch. The problem is with the article in that it left out what I think is the most important piece of information when describing the Monty Hall problem:

The position of the prize does not change after the empty door is revealed.

Once you make a choice, the prize is either behind the door you choose (1/3 chance) or one of the doors you didn't choose (2/3 chance). One of the doors you didn't choose now gets opened revealing nothing, but if the prize stays behind the same door then the chance you picked the door with the prize the first time is still only 1/3 and the chance that you didn't pick door with the prize (one of which is now open) is still 2/3, so you should switch to the unopened door that you did not pick the first time.

If the prize got randomly assigned again once the wrong door was revealed then the odds would change to 1/2-1/2. If it stays in the same place, the original probabilities of 1/3-2/3 remain.

No. of Recommendations: 2

* I think it's 50-50 for Monte Hall. How does the door know if this is "the same game" or a new one with only two doors to chose from? Monte may know, but he's not doing the choosing. Door=roulette wheel. No institutional memory. Every spin is independent of whatever came before. *

Nope. The key is that the host knows, opens a booby prize door after you pick, and thereby gives you information that you didn't have upon your initial choice. If your initial choice was wrong (2/3 probability) then switching will ensure you win at 100%.

No. of Recommendations: 8

PM: *This is a thought-provoking article to take to heart...*

Yes, indeed. Having been a consulting statistician for, um, the past forty years, I had heard these old chestnuts many times. Except for #3 (Abraham Wald's Memo), that is, which was new to me -- and I loved it.

The bedrock foundation of modern probability and statistics is something called "measure theory", which has its own collection of counter-intuitive puzzlers. Your readers may enjoy testing their mettle against these as well. It helps to know that a probability is just a special case of a measure.

These head-scratchers are arranged in order of ascending difficulty. The answer to the third could make your fortune!

1. (Lebesgue measure) In measure theory, and hence in probability and statistics, the "Lebesgue measure" of the interval from zero to one is just its length, i.e. 1. Similarly, the Lebesgue measure of a unit square is just its area, also 1. The Lebesgue measure of a unit cube is just its volume, again 1. The Lebesgue measure of a unit hypercube of, say, N dimensions is also 1, for any finite N. What is the Lebesgue measure of a hypercube of *infinite* dimensions?

2. (Vitali set) We can all agree, I think, that if we throw a dart at a dart board in a completely fair way -- so that all zones of the same size are equally likely to be hit -- then the probability of hitting any particular zone is proportional to the size of the zone. Okay, let's construct a paradoxical dart board:

Start with a dart board that is just a straight line of length one. Paint it red. Now divide the dart board into zones so that two points are in the same zone if they differ by a rational amount (an amount expressible as a ratio of two integers). From each zone, pick a single point and paint it black.

You grab a dart whose point is sharpened so well that it is truly just a point, not a surface. You throw this dart at the board, in a completely fair way, and it hits the board. What is the probability that the point of your dart has hit a black point?

3. (The Banach-Tarski paradox) Consider a solid sphere in three dimensions. For motivation, it may help to imagine that it is solid gold. Is there a way to dissect the sphere into a finite number of non-overlapping pieces, and then to reassemble them into **TWO** spheres of exactly the same size and volume as the first? The reassembly process must use only moving the pieces around and rotating them, without changing their shape. [Hint: in this puzzle, size = volume = measure]

Answers will appear here on Monday night.

Loren

No. of Recommendations: 2

*Hmmm. Where to begin? First, the Monty Hall scenario: They have it wrong. You have a one in three chance of being right. You pick a door. One you did not pick is opened, revealing no big prize. The game just changed. It is now a new game, 2 doors, pick one, the odds are 50/50. Change, don't change...it does not change the odds. The only way this is not true is if Monty Hall knows which is the right door and you can game his psychology in which door he shows you. Otherwise, you are in a new game, 2 doors, equal odds.*

Switching is the right thing to do. Mythbusters even demonstrated it and had a nice colored chart showing how many times the contestant would win by switching.

http://dsc.discovery.com/tv-shows/mythbusters/videos/wheel-o...

http://mythbustersresults.com/wheel-of-mythfortune

When presented with the Monty Hall Problem, people would be more likely to win if they changed their decision.

confirmed

They built a small-scale simulator to do 50 trials each, with Adam always switching his choice and Jamie never switching. Adam won far more often than Jamie did, and Jamie explained the reason: because the player has a 2/3 probability of choosing a losing door at first, switching turns the odds in his favor.

No. of Recommendations: 1

lobodoug,

I totally agree with you on the Monty Hall problem.

PM

No. of Recommendations: 2

Never mind... I just read SB's missive.

No. of Recommendations: 2

* (Abraham Wald's Memo), that is, which was new to me -- and I loved it.*

While certainly interesting, all military combat equipment is a compromise. You cannot just start loading heavy armour on fighter planes without giving up something that may be even more critical.

The key is always the three legged stool of mobility, fire power and defensive armour. Take away one and the stool falls over.

Battleships for instance were heavily armoured with big powerful guns and engines to drive them. Battlecruisers had the same sized big guns, powerful engines but reduced deck armour in order to improve speed. The most famous result was when the battle cruiser HMS Hood blew up because a couple of Bismarck's shells penetrated the decks and blew up in the magazine killing all but three of the over 1400 man crew.

http://en.wikipedia.org/wiki/Battlecruiser

*A battlecruiser, or battle cruiser, was a large capital ship built in the first half of the 20th century. They were similar in size and cost to a battleship, and typically carried the same kind of heavy guns, but battlecruisers generally carried less armour and were faster.*

The US Sherman tank had great mobility, could literally run circles around the German tanks, unfortunately the armour was less than adequate to stop anything but the smallest caliber anti-tank rounds and the infantry support 75 mm gun needed to be used at very short range to have any hope against the better armoured German tanks. Later in the war the Brits stuck a 17 pounder AT gun in the turret and turned it into a tiger killer.

http://en.wikipedia.org/wiki/Sherman_Firefly

The Japanese Zero started the Pacific war as the preeminent fighter with range, speed and manoeuvrability unmatched. It didn't take long for the US pilots to figure out that these benefits were due to the unprotected pilot and fuel tanks and made them an easy torch if you could get a burst in.

http://en.wikipedia.org/wiki/Mitsubishi_A6M_Zero

Tim <desperately forced to watch a football game> 443

No. of Recommendations: 1

**The only way this is not true is if Monty Hall knows which is the right door and you can game his psychology in which door he shows you. Otherwise, you are in a new game, 2 doors, equal odds.**

Monty Hall knows which is the right door and will not open it. Because if he opens the door with the car the game is over. How do we know he knows? Because if he didn't know, then some of the time when he opened the door the car would be behind it. And since that doesn't happen, he knows.

Monty Hall not only knows, he opens the doors behind your back.

No. of Recommendations: 0

*If your horse does not want to ford a river, despite your common sense telling you all is well, better think again, or you may drown. Or open the door with the goat!!*

I suppose you could use the door as a raft to get across the river????

No. of Recommendations: 0

*Every spin is independent of whatever came before.*

The key to the problem is to understand that it's NOT a new spin.

It's still the same spin. You're just given some additional information.

Perhaps the easiest to understand example:

There a 1000 doors. You pick one. Monte opens of the remaining doors to show you the goats, except one (998).

Do you think the last unpicked door he hasn't opened has a 50% chance?

No. of Recommendations: 0

*2. (Vitali set) We can all agree, I think, that if we throw a dart at a dart board in a completely fair way -- so that all zones of the same size are equally likely to be hit -- then the probability of hitting any particular zone is proportional to the size of the zone.*

Let us make a dartboard which is a uniform tesselation of two shapes, one of them being one square inch and the other being three square inches.

Now let us put a powerful magnet behind each of the smaller shapes.

And let us throw at it a dart with an iron tip.

We have a dartboard where zones of the same size are equally likely to be hit, but zones of different sizes are hit in proportions which do not match their area.

No. of Recommendations: 10

Can't believe after lurking here for so long that I am compelled...COMPELLED...to jump in on this issue.

Here's my problem. All those decision trees show three door paths. However, the real decision comes after Monte has removed one of the doors. And not a random door, I might add, which puts the kink in the logic. He's going to show you a goat irregardless of your pick. You have already "own" a door which either has a goat or a car, and the door he has shown you is now irrelevant. I believe the decision trees should begin here, when one door has a goat (or worse), and the other has a car. To "not switch" is to choose the door you have. To "switch" is to choose the door you do not have.

Or so it seems to me.

No. of Recommendations: 0

*However, the real decision comes after Monte has removed one of the doors.*

Doesn't matter when the decision is made. If either of the two doors has the grand prize, the door he will show you is predetermined. And there's only one chance in 3 that neither of the two doors has the grand prize.

Once you pick a door, Monty ALWAYS has at least one goat to show you.

If there are 100 doors, he will ALWAYS have at least 98 goats to show you after you pick a door, until it comes down to the two doors. ALWAYS.

Here's a Java simulation:

http://mste.illinois.edu/reese/monty/MontyGame5.html

No. of Recommendations: 0

I played ten and won five which totally matches my 50% expectation.

No. of Recommendations: 1

*I played ten and won five which totally matches my 50% expectation.*

I ran 5000 and got 68.7% when switching...

No. of Recommendations: 1

*Hmmm. Where to begin? First, the Monty Hall scenario: They have it wrong. You have a one in three chance of being right. You pick a door. One you did not pick is opened, revealing no big prize. The game just changed. It is now a new game, 2 doors, pick one, the odds are 50/50. Change, don't change...it does not change the odds. The only way this is not true is if Monty Hall knows which is the right door and you can game his psychology in which door he shows you. Otherwise, you are in a new game, 2 doors, equal odds.*

They do not have it wrong. What you are missing is the information asymmetry between you (the contestant) and Monty. Monty Hall DOES know which door is the right door, and he always opens one without a prize. That tips the odds in your favor to switch. The problem often isn't set up very well and this point is missed.

I fought this one tooth and nail too. Most people do! That's why it's a great thought experiment.

No. of Recommendations: 0

1, 0, and I don't know.

Sometimes I hazard wild guesses and make a fool of myself.

No. of Recommendations: 3

A couple of days ago PolymerMom posted an excellent set of five classical problems "that will change the way you see the world". Pondering these kinds of paradoxes and counter-intuitive situations is great for the mind -- one can almost always learn something. In reply, I posted another set of three paradoxes, also from probability and statistics, which have in recent years achieved a considerable degree of notoriety in mathematics. From the meagre responses it is clear that I employed language and concepts that are foreign to most people, and so I owe an apology to the board. It was not my intention to mystify, only to show how easily paradoxes arise in the realm of the infinite as well as finite probabilities. These too can "change the way you see the world", but only if you have some experience with the Axiom of Choice.

FWIW, here are some answers, with links for the curious.

Q1. (Lebesgue measure) In measure theory, and hence in probability and statistics, the "Lebesgue measure" of the interval from zero to one is just its length, i.e. 1. Similarly, the Lebesgue measure of a unit square is just its area, also 1. The Lebesgue measure of a unit cube is just its volume, again 1. The Lebesgue measure of a unit hypercube of, say, N dimensions is also 1, for any finite N. What is the Lebesgue measure of a hypercube of infinite dimensions?

A1: ZERO.

Why zero? Here is a cute but informal argument: Imagine for a moment that the Lebesgue measure of a hypercube of (countably) infinite dimensions has a positive measure, say one. Since the hypercube is infinite-dimensional, it can be carved a (countably) infinite set of pairwise disjoint identical hypercubes, each one with side length one-half that of the original. Since the sum of the volumes of this infinite set of smaller hypercubes is finite, each of the smaller hypercubes must have measure exactly zero. But the volume of the original hypercube is the sum of its parts, so it too must have volume zero. This is a contradition, therefore the original hypercube cannot have a positive measure.

Wikipedia article: en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebes...

Q2. (Vitali set) We can all agree, I think, that if we throw a dart at a dart board in a completely fair way -- so that all zones of the same size are equally likely to be hit -- then the probability of hitting any particular zone is proportional to the size of the zone. Okay, let's construct a paradoxical dart board:

Start with a dart board that is just a straight line of length one. Paint it red. Now divide the dart board into zones so that two points are in the same zone if they differ by a rational amount (an amount expressible as a ratio of two integers). From each zone, pick a single point and paint it black.

You grab a dart whose point is sharpened so well that it is truly just a point, not a surface. You throw this dart at the board, in a completely fair way, and it hits the board. What is the probability that the point of your dart has hit a black point?

A2: The probability cannot be determined. It does not exist.

Why does this probability not exist? Because the set of black points does not have a measure. The "Vitali Set", which this puzzle describes, is the first-discovered (1905) and best-known example of a non-measurable set. The dart board is also non-constructive -- meaning you cannot go out and make one -- because it relies on the infamous Axiom of Choice (picking one each from an uncountable set of sets [cue evil music here]). We can conclude from this exercise that probabilities, even simple ones based on throwing darts, are not as straightforward as they appear.

Wikipedia article: en.wikipedia.org/wiki/Vitali_set

Q3. (The Banach-Tarski paradox) Consider a solid sphere in three dimensions. For motivation, it may help to imagine that it is solid gold. Is there a way to dissect the sphere into a finite number of non-overlapping pieces, and then to reassemble them into TWO spheres of exactly the same size and volume as the first? The reassembly process must use only moving the pieces around and rotating them, without changing their shape.

A3: YES, there is a way to do it.

Alas NO, you cannot write an algorithm to achieve it, or design a machine that will do it, because the manufacture relies on the Axiom of Choice [more evil music], which is non-constructive. The Banach-Tarski paradox remains an intriguing and active research area in mathematics.

Layman's guide to Banach-Tarski: www.kuro5hin.org/story/2003/5/23/134430/275

Wikipedia article: en.wikipedia.org/wiki/Banach-Tarski_paradox

Loren

No. of Recommendations: 1

First, I was gone for 5 days to Singapore, then was securing boats and battening down for typhoon Bopha (I have now been thru the eye of 4 hurricanes and 1 typhoon! Figure the odds on that!), so am only now getting back to this thread.

I concede that I am wrong if EVERY TIME you play you are given a second chance. If, however, it is random, or at the whim of Monty Hall, then I would say perhaps.... For me, it took viewing the problem to understand (I learn best visually).

Of course, I ran the simulator a half dozen times and LOST EVERY TIME. I have the same ability to bend statistics at black jack as well.

Cheers,

Doug

No. of Recommendations: 0

*...was securing boats and battening down for typhoon Bopha (I have now been thru the eye of 4 hurricanes and 1 typhoon! *

I'm glad you're safe.

As to the Monty Hall statistics problem &/or Blackjack, your track record is the same as mine on both.

However, I have a bit better luck at Roulette.

I read instructions one time on how you can improve your odds at Roulette, but I never even followed them - nor could I even tell you what they were.

But I generally do pretty well at Roulette so long as I quit while I'm ahead - or keep my bets limited.

;-)