Aggie87

I tend to agree with Seeline. Threads like this (and the other sexy women thread) don't promote equal dialog with women, but rather objectify women. I've seen the arguments that have come up previously - that it's just the way guys are wired, it's natural to be attracted to feminine beauty, etc. But I wouldn't think that posting pictures and videos of women that we think are sexy, is going to inspire women to want to talk with the guys on this board about music and other topics, if they're looked at like they're women first, and then their thoughts and ideas come second.

Knee ligament. It's a pretty well known sports injury in humans, hadn't heard of it happening to a dog before. ACL Hope Coltrane gets well quickly!

Maybe "sexist" is misspelled in the title...

I bet it just recently came up on his word-a-day calendar.

Jim's 50/50 thing isn't related to any odds. He's just saying he has two choices - stay or change.

How do you guarantee that you've chosen correctly? I'd like to have that talent!

Switching or not switching ARE the only choices you have, so it IS a binary decision (either/or), which I think is what you're trying to say. But it's still a 67/33 split of likelihoods, not 50/50.

Then that means you have three choices to make - switch, don't switch, and....? No it doesn't. the 66-33 is still dealing with only two alternatives - switching or staying. There are four separate scenarios though, one of which WILL play out: 1 - stay with door A and be correct (33% likelihood of this happening) 2 - stay with door A and be wrong (67% likelihood of this happening) 3 - switch to door B+C and be correct (67% likelihood of this happening) 4 - switch to door B+C and be wrong (33% likelihood of this happening)

I don't think that's true. Well, here we go back to what everybody's been saying all along - you got two choices, right or wrong. 50-50. I still think it's "two choices, right (switch) or wrong (stay), 66-33." And I was one of the few that started out kinda thinking like you are, I think. Making the right choice is switching, 66% of the time. Making the right choice is staying with your original door, 33% of the time. I still say you have a 66-33 chance at getting the 2 out of 3 chance.

Indeed, the will apply every time. But they will not come through every time. Remind me to play poker with you some time... That's very true. Odds can be in your favor and you can still lose, or odds can be against you and you can still win. That's why so many people play the lottery. Not worth throwing away my money.

Nice review. I've enjoyed the Sco shows I've been to quite a bit. Also, Phil G's a board member here, and I know he was looking forward to this tour.

I don't think that's true.

Outcome A = winning the car Outcome B= losing the car You can still win the car with the 33% likelihood (by keeping with Door A), or lose the car even with the 67% likelihood of winning (by swithing to Door B+C). It's just a probability. You still had two choices. But it wasn't a 50/50 thing. There's no way to guarantee the win (unless you're Monty and know which door has the car).

I think basically what Jim's saying is that you have two choices - switch or don't switch. He's saying that you have an either/or decision, which isn't the same thing as a 50/50 chance of winning the car.

That's the odds of winning the car. What are the odds that you made the right choice to do so? The odds of having made the right choice by picking Door A would be 33%. And the odds of having made the right choice by picking Door B+C would be 67%. Even though there are only two choices.

If a car is placed behind one of three doors, randomly, and if you were offered a choice of what's behind (Door A) or what's behind (Door B+Door C), you'd pick (Door B+DoorC) wouldn't you? Just about every time? It's a 50/50 choice (i.e. either the one door or the set of two doors), but the odds of winning the car by picking the two door set is greater. By a 2 to 1 margin.

I'm not sure usuality is a word (it's a fuquitous one if it is! ), but the measurement would still be 66/33 in favor of switching for that one isolated decision.

As far as I can tell, your logic is still the same as it would be if you were the person who walked up to the game show with only two doors remaining, and got to pick one of them. Then it's a 50/50 chance. If you were the person standing there the whole time, there's STILL a 66% chance that the car is behind the other door. The other folks here have done alot better job of explaining it already, than I could.

Happy Birthday!

That's not correct, but it's what I was trying to do too, LOL. It's only 50/50 for the guy who walks up at the point where there are two doors left, and he is allowed to pick one of the two doors. For the person who was there when there were 3 doors, his likelihood of winning the car is still better if he switches - by a 2 to 1 margin. It's counterintuitive for sure.

testing the server clock... edit - and it's off by 5 hours or so.

I got it. Thanks. Door A (my initial pick) = 33% chance of car Doors B + C (the ones I didn't pick) = 67% chance of car When Monty opens one of the other doors, it still means my door has a 33% chance, and the remaining door has the 67%. I understood that logic from the beginning, I just wasn't sure I was accepting of it. Thanks Swede & Dan (& Sen. Rockefeller).

Thanks for posting that link, Rod! LOL I think this quote is where I'm tripping up: I keep putting myself into that point of time where the player is asked whether to switch, which isn't the correct logic to use, I guess. But try this: Initially I pick Door A. Monty opens Door C with a goat behind it, leaving Door A or Door B. He lets me choose which one I want. I should switch to Door B, because it gives me a 67% chance of winning. Another person is brought into the studio at that point (without any prior knowledge of this game), and given a choice of Door A or Door B as well. Are his odds different than mine? Why?

I haven't run the simulation, Dan - can't access it from this computer. I also understand the arguments that have been shared here, and they make sense. I'm still saying that Monty's opening of a door with a goat behind it leaves 2 closed doors, one with a car behind it. It seems immaterial which door you chose first, KNOWING that you will again have to choose one of the two doors that are left. The data at that point in time is that you have two closed doors, and a car behind one of them - and that's it. That's what you have to go by AT THAT POINT IN TIME. I'm not arguing that you're wrong. I just am not convinced by the logic, whether it's true or not. Well, if the result is that there are two remaining closed doors, one of which you've picked and one you didn't, there's one with a car behind it and one with a goat. At that point in time there's a 50/50 chance the car is behind your door, is there not?

To me it almost sounds like the paradigm has shifted for this guy - from the "old" days before cell phones, to modern times - and he can't accept it. It's not his place to be a vigilante about it, whether he likes it or not. If you're in a public place (like a train car), you have to accept what society says is acceptable public behavior. And using cell phones in public has become part of that.