Teaser game

[Sheremetev]

Following one logic or another, Cuorion got the right answer. So, get creative and give us a new riddle.

P.S: Just to keep you occupied until  and to satisfy my curiousity, did it cross your mind that the answer could be neither linear, nor planar? Imagine it was circular root (or smth), sticking to Tiger's first idea, what would you do with the measure unit part?

[Cuorion]

Imagine it was circular root (or smth), sticking to Tiger's first idea, what would you do with the measure unit part?

Request a nobel prize if we managed to solve it?


Anyway here's my teaser.

You toss a coin three times. First one was heads. What's the chance that it came tails at least once?

[Tiger]

75% or 3/4?

[Cuorion]

Now if that was right, it would be too easy, wouldn't it?

[Tiger]

For most people it wouldnt be to easy. I teach highscool kids same stuff and it needs a lot of explaining to understand it.

[Cuorion]

I don't know, I always thought of conditional probabilities really intriguing.

You grasped the main idea, but there's a catch in there.

[Sheremetev]

Ok, right to the catch- how many sides does the coin have?

I assume that the only possible outcomes of each toss are heads and tails (equally possible- each 50%).
So, equally possible chains of events are:
H H H - 25%
H H T - 25%
H T H - 25%
H T T - 25%

All in all, 75%, as Tiger said.

If we assume this is the "dinosaur problem" from a more generalized point of view, chances are 50%- it will either fall tails atleast once or it will not. Just like the chances of seeing a living T-Rex (velociraptor for all xkcd fans)- you will either see one or not.

[Cuorion]

Hint: What both of you did was "toss the coin twice" (since you both calculate 2² different outcomes).

EDIT: No dinosaurs were harmed during the testing of this teaser.

[Tiger]

If there are 3 tosses, then the solution is 87,5%.
Because if you toss 3 times and you know 1 outcome(first one was head), then it doesnt matter for tail in next 2 tosses. Thats why we calculate with 2 tosses.

[Cuorion]

I never said though, that you make 2 tosses after you learn the first was head.

Let's just say that there's another answer that I would consider more accurate. I think it's a matter of perspective, actually. If you think I caused too much confusion, I can always tell you what's in my mind and post another more... elegant mathematical riddle.

[Cuorion]

After a bit of thought, I came to the conclusion that it's better if we stick to riddles/teasers that can have a definite answer. So let me write my point of view on the last matter and give a more solid riddle to solve.

On the previous problem we had two conditions to fulfill.
1) Get a head on the first toss.
2) Get a tail at least once.

We tossed 3 times, so all the possible outcomes would be 2³, or in detail,

Head Head Head
Head Head Tail
Head Tail Head
Head Tail Tail
Tail Head Head
Tail Head Tail
Tail Tail Head
Tail Tail Tail

Only 3 out of 8 fulfill both conditions. Head Head Tail, Head Tail Head and Head Tail Tail. So the chances of having a head first and then at least one tail is 3/8, or 37,5%.




Now for the more solid one. I want you to find the smallest number with the following properties.

When divided it by 2, the remainder is 1.
When divided it by 3, the remainder is 2.
When divided it by 4, the remainder is 3.
When divided it by 5, the remainder is 4.
When divided it by 6, the remainder is 5.
When divided it by 7, the remainder is 6.
When divided it by 8, the remainder is 7.
When divided it by 9, the remainder is 8.
When divided it by 10, the remainder is 9.

[Tiger]

No problem, its  ridle, but was have meaning that you allready throw it. First should be head would be more clear.
2519 is the answer. If you have positive numbers in mind.

[Cuorion]

2519 is the answer

That's right. Your turn!
Have you found it with the elegant or the brute way?

[Tiger]

Elegant - i wrote program in 4 lines in qbasic to give me answer. i know it exists nice way but i was to lazzy to go on foot.
I leave you the honor to make another question, if you wish.

[Cuorion]

i wrote program in 4 lines in qbasic to give me answer.

Why, I call a basic programme as epicly smart.

So, I am to post another teaser?
It's already here.



P.S.: For those who wonder, as we see on the table, X = a*n + n - 1 for every integral n[2,10]. This gives X + 1 = (a + 1) * n, so (X + 1) is the lesser common multiple (LCM) of the numbers 2, 3, 4, 5, 6, 7, 8, 9 and 10. Using the prime factorization method, we get X + 1 = 2³ * 3² * 5 * 7 = 2520.
Since X + 1 = 2520, X = 2519.