Off-Topic

[Caulfield]

re: Off-Topic

Lol. Are we gonna do Chevy vs Ford next?

automobiles are for men with weak legs

[WaitTilNextYear]

re: Off-Topic

Lol. Are we gonna do Chevy vs Ford next?

We could, but I've owned both and both have been reliable vehicles. Come to think of it, I've owned 1 Ford, 1 Dodge, and 1 Chevy in my lifetime.

Perhaps foreign vs domestic would be a more spirited debate? I am firmly in the camp of "made in the USA." Buy local as much as you can and regional or national where you can't.

[TheMatrix31]

re: Off-Topic

https://www.thedailybeast.com/how-an-ex-cop-rigged-mcdonalds-monopoly-game-and-stole-millions

Fascinating read here.

[Blzer]

re: Off-Topic

@Blzer

Do you teach probability theorms?

Sent from my SM-G950U using Operation Sports mobile app

Hmm, we've covered probability in Algebra II, but what do you specifically mean when you say probability theorems?

[DamnYanks2]

re: Off-Topic

Trying to remember a YouTube show that came out a while back. I can't remember much. Except a guy named Terry said Heyyy ooo alot and there was a character named Greg. It was hilarious and I wish I could remember it. It had an Office vibe to it.

Never mind figured it out. Man in the box!

Anybody remember this ?

https://youtu.be/suQgpqvYw1Q

Sent from my SM-J327VPP using Tapatalk

[Speedy]

re: Off-Topic

Hmm, we've covered probability in Algebra II, but what do you specifically mean when you say probability theorems?

I was watching the movie 21 (blackjack, counting cards) and they tried to explain the Monty Hall problem.

https://youtu.be/Zr_xWfThjJ0

So it proposes an old game show that has 3 doors, where one has a car and the other 2 a goat:

x x x

If I choose door one, the host opens door 3 that has a goat behind it.

x x O

The host then gives me the choice whether to stick with door 1 or change to door 2.

The theory is that my win probability shifts to 2/3 if I choose to switch my door choice versus 1/3 if I stay the same.

I don't understand how it's not 50/50 after the host opens the first door. I've looked at a bunch of explanations and it's just not coming to me.

[Blzer]

re: Off-Topic

I was watching the movie 21 (blackjack, counting cards) and they tried to explain the Monty Hall problem.

https://youtu.be/Zr_xWfThjJ0

So it proposes an old game show that has 3 doors, where one has a car and the other 2 a goat:

x x x

If I choose door one, the host opens door 3 that has a goat behind it.

x x O

The host then gives me the choice whether to stick with door 1 or change to door 2.

The theory is that my win probability shifts to 2/3 if I choose to switch my door choice versus 1/3 if I stay the same.

I don't understand how it's not 50/50 after the host opens the first door. I've looked at a bunch of explanations and it's just not coming to me.

We don't specifically teach this, but I have presented this to my students many times before. I know this problem all to well, and have explained it to my little sister once. We then saw 21 in theaters and she nudged my shoulder and whispered: "I know this, Brub. Switch, switch!" Haha I'll try and explain it in full, so read carefully.

The idea of the problem is that the game show host knows which door has the car behind it. So whether your initial choice has the car behind it, he will always reveal another door that always has a goat. He will never reveal what is behind the door that you chose. He is asking whether you want to stick with your original choice (which you had a 1/3 chance of getting right) or switch to the other door that has yet to be mentioned at all.

Allow me to revise the scenario to 100 doors instead, 99 are goats and one is a car. You have 100 doors to choose from, and you pick one of them... those do not sound like incredibly good odds that the first door you chose also just so happened to be the car, now does it? Anyway, the game show host will now reveal ninety-eight other doors, all which have goats. He knows which door has the car: it's either the initial door that you chose (fat chance), or it's the one other door that he didn't reveal anything behind. He is asking if you would like to switch or not. Of course you would make that switch, because you had a 99% chance of being wrong with your initial pick.

If you want the problem to be thought of correctly, think less about how many doors you are down to and think more about what your odds are of your initial choice being wrong. The real question at hand is: "Would you rather stick with your choice, or not?" So in the 100 doors example, it really is just asking: "Would you like to stick with the first door you chose, or would you instead like to switch to the 99 other doors that are there? Remember, if the car is behind any one of those 99 doors, I will give you that car." Now to the lesser extreme example with three doors, they're really asking: "Would you rather stick to your first door, or guess that the car may be in the other two doors?" It is 1/3 versus 2/3.

Now, let's say some gal entered the room late and went: "Hey guys, what'd I miss?" All she sees are two closed doors and a door open with a goat. If the game show host asks her which door has the goat, she has a 50% chance of getting it correct. She has no prior knowledge of the events that went down beforehand, and that is a completely different scenario. You, however, know which door you chose to begin with and the game show host intentionally revealed another door (not your choice) that he knows has a goat behind it, leaving your door sealed and some other door sealed as well, which may or may not have the car... but because your initial chance of getting the car.

And remember, this is all theoretical probability and not experimental. You may switch doors and find out that your first choice was correct; after all, 1/3 is still a pretty good chance that you were correct the first time. It's just that 2/3 is twice as likely, so statistically it is in your best interest to switch doors. Truthfully, there are no probabilities about which door has the car, because there really is only a 100% chance that one door has the car and a 0% chance for the other doors if you really think about it haha. It's asking about whether your initial choice has a higher probability of success than not, and the answer is that it does not.

I'm not sure that I can explain it any better outside of us physically doing it together.

[pietasterp]

re: Off-Topic

I was watching the movie 21 (blackjack, counting cards) and they tried to explain the Monty Hall problem.

https://youtu.be/Zr_xWfThjJ0

So it proposes an old game show that has 3 doors, where one has a car and the other 2 a goat:

x x x

If I choose door one, the host opens door 3 that has a goat behind it.

x x O

The host then gives me the choice whether to stick with door 1 or change to door 2.

The theory is that my win probability shifts to 2/3 if I choose to switch my door choice versus 1/3 if I stay the same.

I don't understand how it's not 50/50 after the host opens the first door. I've looked at a bunch of explanations and it's just not coming to me.

It's not just you who doesn't understand this - this was debated for years by some of the smartest people on earth, and it took an extreme amount of work and convincing for people to come around (allegedly even Paul Erdos required a computer simulation to be convinced). It's extremely counter-intuitive, but the math is correct.

The key to the problem lies in the fact that the particular set of rules in the game are very specific. The part that trips people up and makes it hard to conceptualize is the fact that after you make your initial choice, the host (Monte Hall) MUST open a door that does NOT contain a car. This is the confusing part, because it naturally leads one to believe that the odds of choosing the right door initially have changed, when in fact they have NOT changed at all.

I'm not sure I'm going to do a better job of explaining this than anyone else, but one way to think about it is: there is a 1/3 chance of the car being behind any of the 3 doors. After you make your choice, you have a 1/3 chance of having chosen correctly and 2/3 chance of having chosen INcorrectly (e.g. there is a 2/3 chance the car is behind one of the other 2 doors that you did NOT pick). Once Monte Hall reveals the goat in one of the unchosen doors, you STILL have a 1/3 chance of having chosen the car on your first guess. That didn't change to 1/2 just because you now know the goat is behind one of the other doors. There is STILL a 2/3 chance of the car being behind one of the other 2 doors that you did NOT pick, nothing has changed at all with regard to the odds. Therefore, you are better off changing because you have a 2/3 chance of being right vs. 1/3 chance if you stick with your original guess.

Put another way, you are effectively being given the chance to choose 2 doors if you switch, vs. only choosing 1 door if you stand pat. (it helps to be able to diagram this out but I can't draw on this post obviously...)

I completely understand why people think your odds change to 50-50 after the goat is revealed in one of the other doors, but that would only be true if you did not make a choice originally, when there were 3 doors each as likely to contain a car.

TL;DR: You have a 1/3 chance of choosing the correct door and a 2/3 chance of choosing the incorrect door when the game begins, which does not change after the goat is revealed behind one of the 2 doors you did not choose.

[Speedy]

re: Off-Topic

After you make your choice, you have a 1/3 chance of having chosen correctly and 2/3 chance of having chosen INcorrectly (e.g. there is a 2/3 chance the car is behind one of the other 2 doors that you did NOT pick).

This makes sense...no issues here.

So let's now move forward with a door being revealed and 2 doors closed.

Once Monte Hall reveals the goat in one of the unchosen doors, you STILL have a 1/3 chance of having chosen the car on your first guess.

This is barring I cannot change my selection though...if the host reveals door #3, I still have a 1/3 chance that (PAST TENSE) I selected correctly if he then goes to reveal door #2.

That didn't change to 1/2 just because you now know the goat is behind one of the other doors. There is STILL a 2/3 chance of the car being behind one of the other 2 doors that you did NOT pick, nothing has changed at all with regard to the odds.

The bolded here I understand...so how is it a 2/3 probability of CORRECTLY choosing the car if I change my door choice?

Now, let's say some gal entered the room late and went: "Hey guys, what'd I miss?" All she sees are two closed doors and a door open with a goat. If the game show host asks her which door has the goat, she has a 50% chance of getting it correct. She has no prior knowledge of the events that went down beforehand, and that is a completely different scenario. You, however, know which door you chose to begin with and the game show host intentionally revealed another door (not your choice) that he knows has a goat behind it..

I'm not understanding how my prior knowledge of past events suddenly changes my probability from 50%. Whether my choice was a goat or a car, the host would've opened a door with a goat behind it regardless.

I know that the statistics bear out that I would win roughly 2/3 of the time by switching, it just really is counter-intuitive to think that my initial odds of LOSING 2/3 can be switching to having odds of WINNING 2/3 if I change my selection.

[l3ulvl]

re: Off-Topic

I was reading all about that Monty Hall thing after a joke from Brooklyn 99 a few months ago lol

https://lgbtlaughs.tumblr.com/video_file/t:7GuO_LXqJHX2Q8I-oOfvAw/162418817442/tumblr_ohx250vY9x1s5qteu

I went about a week straight reading about it and other examples like the Sleeping Beauty problem, idunno why it was all fascinating to me, but I was sucked in.

[Blzer]

re: Off-Topic

The bolded here I understand...so how is it a 2/3 probability of CORRECTLY choosing the car if I change my door choice?

Because in a binomial probability scenario, there is a 100% chance that the car is behind one of the doors. Your probability sum needs to be 100% for this idea to take fruition in the first place.

If you have a 1/3 chance of being correct on your first pick, that means you have a 2/3 chance of being incorrect on that first pick. That's really all there is to it.

I'm not understanding how my prior knowledge of past events suddenly changes my probability from 50%.

To an unsuspecting individual, a coin has a 50% chance of landing on heads or tails. Somebody else may be in the know that it's a loaded coin and actually has an 80% chance of landing on heads. You as the Joe Nobody is not privy to such information, though. It's all about perception and awareness.

Whether my choice was a goat or a car, the host would've opened a door with a goat behind it regardless.

I know that the statistics bear out that I would win roughly 2/3 of the time by switching, it just really is counter-intuitive to think that my initial odds of LOSING 2/3 can be switching to having odds of WINNING 2/3 if I change my selection.

Because you go from a door that has a less likely chance of having the car to a door that has a more likely chance of having the car. There really is nothing else to it.

Try and read over my 100 doors scenario again and see if that makes more sense.


EDIT: Better yet, I'm going to run through the 100 doors scenario with you, providing you full transparency and making every choice.

I am the game show host standing in front of 100 closed doors, and I have knowledge that a brand new car is sitting behind door #62. I then ask a contestant to pick a door at random for a chance to win that car. To make this easy to write out, this contestant selects door #1. Clearly, not only is the contestant incorrect, but that contestant had a snowball's chance of being correct here.

At any rate, I try and uncomplicate it for the individual by tipping my hand slightly. I let the contestant know that I am going to narrow it down to two doors: his choice (door #1) and another choice (in this case, door #62). So I am going to reveal all 98 other doors and show that there are goats (that is, doors #2-61 and #63-100).

The contestant is now down to two doors: the door that the contestant chose (which we know has a goat behind it) and the door that I have not revealed in any which way (the door which I know has the car behind it).

So, which would you choose?

And let it be known that if the contestant luckily selected door #62 to begin with, that I would still choose another door to keep closed, and should the contestant unluckily switch, then it's just exactly that... bad luck. The contestant mathematically had a 1% chance of being right the first time, and in that particular instance the contestant would have not won the car, but you do not ever take those odds.

In the much less extreme example, everything in place holds true for the 1/3 vs. 2/3 debate, but it is probably best to choose the 2/3 case if you can. No matter either case though, if an individual came into the 100 doors scenario blind only seeing two closed doors, they are quite literally selecting their door by a coin toss, and they have an even chance of being correct on either guess (though technically there is a 0% chance of being correct on door #1 and a 100% chance of being correct on door #62, but speaking in theoretical terms we are looking at probability of result by choice and not by actuality).

This is why the problem is what it is. It is to make you believe that your probability changed simply because I have changed the number of closed doors. But you having known that you selected one door out of 100, or three, or any number more than two, means that you have less a chance of being correct then you do at being incorrect.

Screw doors for a second, let's go to a deck of cards. I lay out 52 cards face-down and ask you to point out where you think the ace of clubs is (and I know which particular card it is). You point at a random card, then I reveal 50 of the cards (none of which of course are the ace of clubs), leaving your card you first pointed at and another "seemingly random" card faced down. Do you really think that it is now a 50% chance that the card you pointed at was the ace of clubs? Your card is still face-down for one of two reasons: 1) it is the ace of clubs; 2) it is not the ace of clubs, but you chose it and I don't want you to know what it is. I will never reveal your card in that scenario. What is the other card that is face-down? One of two things: 1) it is not the ace of clubs because yours is and I just picked a random card for show; 2) it is the ace of clubs because I know exactly where it is in the pile, which is why I kept it face-down.

Your odds of being correct when you began do not change, unless I shuffle the cards and you are now unknowledgeable as to which card was your initial card. I know I'm doing bigger examples than three doors right now, but I'm using the extreme scenarios to enhance the point... and the only point here that matters is that more than 50% is more than less than 50%, and unless you have psychological doubts stricken by fear or paranoia (as Kevin Spacey puts it in the film), you should let statistical probability take over and switch... even if you're wrong every so often.

On an unrelated note to that, let's say we did have a loaded coin that landed on heads 80% of the time, theoretically (that's how it was weighed to be done). Even if you did 100 coin flips and they all landed on tails, if someone asks you what you think the next one would be, you choose heads. Despite the fact that all events are independent of one another, you always choose a higher probability of success when you are able to. This is what baseball managers do when they choose certain pitching and hitting matchups. It doesn't always pan out in hindsight, but we aren't looking that way when these questions are asked.

[Speedy]

re: Off-Topic

One of my colleagues at work created this graphic.

It's basically showing that the decision of switching is NOT an independent choice but rather, the probability takes into account the initial choice as well.

[Blzer]

re: Off-Topic

^ Umm... I guess?

Didn't think I needed to make a pie chart to show the difference between 1/3 and 2/3, but if that gets you to understand it then I guess that's all that really matters.

If it makes you feel any better, the initial choice is the only aspect of the probability that is relevant.

[pietasterp]

re: Off-Topic

One of my colleagues at work created this graphic.

It's basically showing that the decision of switching is NOT an independent choice but rather, the probability takes into account the initial choice as well.

I'm glad this helped you conceptualize it. This is not an easy concept, and it's counter-intuitive. As I mentioned before, some of the smartest people in the world did not believe this evaluation until a lot of discussion/analysis. If it were straightforward, they wouldn't have bothered to name it the "Monty Hall Paradox". Even though it isn't paradoxical at all, it sure seems like it on first blush....

[Blzer]

re: Off-Topic

I have another one that some of the greater minds (like Einstein) had trouble solving the first time. It's not probability, just math (and some logic).

Say you're riding a bike on a two-mile path, the first mile being uphill and the second mile downhill. At best, you are only able to ride the first mile uphill at an average speed of 15 MPH. How fast would you have to bike on the downhill mile to achieve an overall average speed of 30 MPH for that two-mile path?