bonanova

Cute idea. But maybe the goats are married and fight with each other. As to your original question, the difference between cases [1] and [2] keep in mind that a goat is certain to appear. The host will under no circumstances open the door that has the car behind it. So, don't start thinking about the "chances" that the open door will show a goat - it always will show a goat. That said, what information do you get when you see a goat? Simply this: your choices have been reduced by one door. Let's take case [2] first: There are 3 doors, and if you choose one, your chances of having picked the car are clearly 1/3. But first, a door opens, showing a goat. Now you know the car is behind one of two doors, not three, so your chances of picking the car have become 1/2. How does that differ from case [1]? There are three doors, and you pick one. You know that your chances of having picked the car are 1/3. That means the chances that the car is behind one of the other two doors is 2/3. But you wouldn't switch your pick, because you'd have to guess, and that reduces the 2/3 into two 1/3 choices. But suppose you didnt have to guess? Suppose the two doors become one door. Then you'd switch and double your chances. That's what happens when the host shows you a goat. Your choices have been reduced by one door, leaving you an opportunity to pick the door with the 2/3 chance.

It goes like this, said Alex, you write down a three-digit number. Davey tried to stifle his yawn, but couldn't. Ian's eyes were glued to the blond at the bar, and Jamie had fallen asleep. It was next to closing time at Morty's, and even Alex wasn't getting much attention. Then I'll tack on three more digits before you can take a gulp of yer ale, and I bet that the resulting six-digit number will be divisible by 37! I don't think so, said Ian, as the blond strolled off to the ladies room. Not even you ... Well, I'll show you, interrupted Alex, and this one is for free. Give it a try. Ian quickly scribbled 412 on his napkin and reached for his mug. But before he could lift it, Alex had written 143 at the end, and quick-witted Davey verified that 412,143 was in fact equal to 11139 x 37! And that's not the half of it, boys, Alex was finding his groove now, I'll do the same thing only this time I'll add on six digits to yer number! Again, Ian obliged, this time writing 341. It took Alex only a second to make that into 341,203,122, and Davey's pocket calculator verified that dividing by 37 gives an even 9,221,706! Well I know now you'll never take my bet, continued Alex, so I'll make you an easier one. Clearly I don't do math that quickly, so we all know there's trick involved. So here's what I'll do. You give me ten numbers, and I'll add three digits, or six digits, before, after, or some of each to all ten numbers. Looking at what I've done ought to lead you to figure out the trick. If you do that, I'll buy one round before closing. Ya don't get an offer like that one every day. That woke Jamie up, and along with the others, they wrote down the following ten numbers, and Alex added the digits in red. Davie verified the results. It took Jamie only a moment. I have it! What did Jamie see? 253,413 = 37x6,849 114,762,012 = 37x3,101,676 620,379 = 37x16,767 247,340,412 = 37x6,684,876 458,541 = 37x 12,393 275,613 = 37x7,449 123,141,513 = 37x3,328,149 201,471,216 = 37x5,445,168 302,364 = 37x8172 121,235,310 =37x3,276,630

That's easy to explain. Your 8 cases do not have equal likelihood. Specifically, cases 1-4 double count the same event.

Pretty quickly I thought of a number that is the square of the sum of its digits. There might be quite a lot of them since it seems an easier problem then finding a cube.

btw topic 892 might be the original link mentioned in the OP.

Quotient of what? Its square and its half?

But 10 octal is just an 8 ...

Proof goes like this. If you stick with your choice until the final losing door is opened, two closed doors remain: the one you initially picked [prob = 1/N] and the only other door that it could be. But you knew that, right?

Posted here. If you didn't write the puzzle, do a search [three +switch +light would have found it] Locked.

I won't delete it, but at the OP's request I will close it.

I have not yet sunk to little markers. Sorry, I quoted the wrong step. I meant to call your attention to this one: GF3 brings GF2 across - sails back. After this step, you have BF2,3,4 with GF3,4 on the near shore. That's ok. And BF1 with GF1,2 on the far shore. That's a no-no. When GF3 was sailed back, GF2 was with BF1 without BF2. And after the next step, GF2,3 are with BF1 without their BF's. ------------- There is a 17-move solution that obeys both the strict and the relaxed versions of the requirement.

You're right ... a 21-move solution. My bad.

taliesin has it. Nice job.

Make 21, using only the four basic mathematical operators [+, -, x, /] on the numbers 1, 5, 6 and 7.

HoustonHokie, k-man and EventHorizon have it. Nice job.

I'm sure you're right that I've missed some. But just to satisfy my curiosity, you say - (1000,0001,0100,0001), will not paint. Doesn't it paint, like this? (1000,0001,0101,0001) (1000,0001,0111,0001) (1000,0011,0111,0011) (1000,0111,0111,0111) (1100,1111,0111,0111) (1110,1111,1111,0111) (1111,1111,1111,1111)

You got me. I meant the more stringent case of being in male company w/o own bf. I've edited the OP.

You are way out of the box on that one! Good job. You too, Mumbles, but by all means try for a red/black solution as well. HoustonHokie has the seed of a more satisfying solution, but yours is no less correct.

No turns or waiting, just guessing. They just guess, and they don't know what each other's guesses are.

Nothing new except the survival criterion. N prisoners, each with a red or black hat. Each can see all the others, but not his own. No one hears the others' guesses. No communication. If all guess correctly or if all guess incorrectly they go free. Otherwise ... well, you know. How do you advise them?

I have not yet sunk to little markers. You say GF1 brings BF1 across - sails back. At this point GF1 is with the other boys, and her BF is no where to be seen. I'll lend you 44 cents if you like.

It shows that girls do most of the work... You say a2 goes to the island and gives the bout to c2. At this point haven't you have left b2 with a1 on the coast?

Use coins.

This is closer, but...

My first attempt leaves it a toss-up.