Swinging Swede

Or Clark Terry. Actually this question came up recently, and this was the answer then: Yep. Not sure that it's applicable now!

Personally I have everything on it on CD already except Hipnosis, which I would like to have. But should I buy the whole Sele(c)t for it, or wait for a possible Conn?

Well, Booby and Handcock are both playing on it, so why not a Selet?

Some misspellings are more common than others. I don't know how many times I've seen McClean and Adderly. And who can forget Blue Note gems like Handcock and Booby?

I think he plays bass guitar.

Yes, tracks 2-8 match the Movin' Along album, albeit in different order, and I would guess that 'Round Midnight is the Paris 1965 version. It is also about 9½ minutes long. Lol at hinting at both RCA and Columbia for a Riverside album.

Your question is interesting, but just one remark: Kenny Clarke moved to Europe in 1956 AFAIK, so I don't think he was a factor.

OK, that explains a thing or two!

Actually you have a 100% chance of getting that 2 out of 3 chance, and you realize it by switching door. So you're saying that you do not have the option of choosing not to switch? Even when you have a 100% chance, you can choose not to take it. Monty himself, if he were to step in, could choose not to pick the door he knows that the car is behind, but pick the other door instead. I'm not sure why he would do it, but he certainly has the option.

Once again it doesn't matter whether it is about one time or an endless stream of times, the odds are still the same. The outcome in a one shot game is always 100% or 0%. That doesn't prevent the odds from being 67% and 33% respectively. OK, then I would say that the odds are 67-33, and the odds of the odds 100-0!

Of course they are. That has never been a point of contention. Well, you did actually say the following (see question in red): So I answered that the other possible odds than 50/50 that the smart choice will be successful this one time are 67/33. Actually it is the only correct odds, 50/50 is not.

Ok, one...more...time... 50-50 as I'm using it does not refer to the probability that switching will be successful. That is clearly 2/3. It (50-50) refers to what you can guarantee as far as your choice being the right one for any given time, before the results are revealed. You can never guarantee that your choice is the right one, unless you're Monty. Whichever door you pick you might lose, so the guarantee factor is 0.

Then that means you have three choices to make - switch, don't switch, and....? A 2 out of 3 chance does NOT mean that you have three choices. In the example I mentioned with 100 initial doors, you have a 99 out of 100 chance, but that doesn't mean that you have one hundred choices. You still have only 2 doors to choose between, but by switching you have a 99% chance of winning the car.

Actually you have a 100% chance of getting that 2 out of 3 chance, and you realize it by switching door.

But if you would have a 50% chance of getting a 67% chance, and a 50% chance of getting a 33% chance, then you would on average only have a 50% chance of winning the car, which just isn't the case. With the knowledge you have, you have a 67% chance of winning the car. Just as Monty has a 100% chance by the way. Otherwise, with your way of reasoning, Monty would have a 50-50 chance of getting a 3 out of 3 chance, whatever that means. By the way, remember the alternative case I mentioned earlier with 100 initial doors, and Monty opening 98 of them, with 2 doors remaining. Would you then say that you have a 50-50 chance at getting a 99 out of 100 chance?

Sure. In 33% of the cases this scenario is repeated the person who does this chocie once will realize, after Monty opens the door, that it didn't pan out, and in 67% of the cases the person who does the choice once will realize that it did pan out. How many times you get doesn't matter for the statistical probability. If it's a 67% chance if you do it 1000 times, then there's also a 67% chance if you do it once. Of course 67% does not guarantee anything, but it is still a lot better than 50%. What will happen is that you will 100% win in 67% of the cases, and 100% lose in 33% of the cases, if you want to express it that way. Well, a 33% chance is still a decent chance. It's just not as good as a 50% chance or a 67% chance. But people who keep their originally chosen doors will fairly often win a car. To be more specific, it will happen in one third of the cases! They do apply every time. 67% is not 100% and 33% is not 0%, so of course it will happen that people who switch doors will lose and people who keep the door will win. That doesn't mean that the odds don't apply. The only correct odds that the smart choice will be successful (i.e. winning the car) are 67/33, not 50/50. If the odds somehow were 50/50 for one incident, then it would be 50/50 for 1000 incidents too, since they only consist of 1000 single incidents counted together. I can't understand how you can think that it's a 50% chance for one single incident, but a 67 % chance for 100 or 1000 of them. That's just illogical.

According to the two quotes above you seem to be under the common misperception that if there are two choices, then it must be a 50/50 chance. That's just not so. If that were the case, then Monty would also have a 50/50 chance, even if he actually knows which door the car is behind. Switching doors does automatically double your chances. Keeping the door gives you a 33 % chance, switching to the other door gives you a 67 % chance. That's doubling your chances. You also seem to make a distinction between playing the game once and multiple times, in that you accept that over time it would be a 67 % chance. But playing the game 100 times and winning 67 times is the same as having a 67 % chance if you do it once. Those 100 times are just 100 single shots counted together. Apologies if I misunderstood your argument.

The chances are always evaluated in relation to prior knowledge. If Monty himself were to step in at this point, he would have a 100 % chance of winning the car if he chooses the door he already knows that the car is behind, and a 0 % chance if he chooses the other. You would have a 67/33 chance because of your prior knowledge. The third person who has no prior knowledge and just knows that there are two doors with a car behind one would correctly evaluate his chances as 50/50. In the other example with 100 initial doors, you would have a 99 % chance if you switched, but the third guy would once again have a 50/50 chance.

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? No, in this example there is a 1 % chance that the car is behind your door, and a 99 % chance that it is behind the other door. Just one question at this point: Are you aware that Monty isn't randomly opening another door, but is purposely avoiding opening the door with the car behind it? In other words, if the car is behind 1 of the other 99 doors (which it has a 99 % chance of being), Monty will purposely avoid the door with the car behind it and open the other 98 with goats behind them. If you are aware of that, and still think there is a 50/50 chance in the end, then I/we have to think hard about some other way of convincing you.

Picking one of two is not the same as having a 50/50 chance, if you have additional information about the two choices. Suppose that there were 100 doors with 99 goats and 1 car behind them. You pick one door randomly. You should now have a 1 % chance of having picked the door with the car behind it. In other words it's very unlikely that you have picked the right one. Then Monty opens 98 of the other 99 doors and reveals 98 goats behind them. Once again you are left with two doors. Would you still think that you have a 50 % chance if you keep the initially chosen door?

The thing is that it isn't a new ball game. You already knew that there would be a goat behind at least one of the two other doors, so Monty opening one and showing a goat doesn't actually reveal any new information and therefore you still have a 1 in 3 chance if you keep the door you first picked. Here's another way of looking at it. Suppose that you always pick door 1 (in bold). Now there are three possibilities. door1 door2 door3 car goat goat goat car goat goat goat car Now Monty will open one of the other doors and show a goat (which is crossed over below). We will then have: door1 door2 door3 car goat goat [Monty could also open door 3; it doesn't matter.] goat car goat goat goat car As you clearly see here, in 2 of 3 cases you will win the car if you switch to the other remaining door, and in 1 case of 3 you will win the car by keeping your originally chosen door.

That was my tentative conclusion too. But if Lee's chops tired at the end of the session, I guess the solo on Judiful is something he might have played. Otherwise I agree that the trumpeter on Judiful sounds more like the one on New Date than the one on the three other tracks, where it's obviously Morgan. By the way, the original liner notes mention Morgan's death. Was this an album that was given a catalogue number (MG 12164) at the time, but like several Blue Note albums wasn't actually released until much later? The liner notes only mention Morgan as the trumpeter on the entire album, but if those notes were written in the 70s, I guess noone remembered any longer what really happened on the sessions.

This was not published on 1 April, was it? Thanks for the good news. A low dollar and no added charges. Think of the possibilities!

"Join in the community or get the fuck out."

What's lamentable is that many of the 1965 Prestige tracks only are in mono on the CD reissues, despite being in stereo on the original LPs.