Pure maths, physics, logic (braingames.ru): non-trade-related brain games

[Sceptic Philozoff]

At the request of several respected members of our forums, I'm moving this topic from the "quadruple" forum here. The rules remain the same: if you already know the solution to the problem, don't post it here, and let the others torture themselves. If you really want to prove to me that you really have the right solution, then contact me in person.

The address of the website I took the problem from is braingames.ru. The branch is for those who haven't yet lost their taste for just solving mathematical problems and see the beauty in them.

The first problem (not difficult, weight 3 points):

How to draw a coin toss fairly if you know that this coin has heads slightly more often than tails? A fair toss is defined as equal probabilities of the outcomes.

Explanation: the exact probability of eagle is unknown.

[Avals]

Mathemat:

At the request of several respected members of our forums, I'm moving this thread from the "quadruple" forum here. The rules remain the same: if you already know the solution to a problem, don't post it here and let the others suffer. If you really want to prove to me that you do have the right solution, then contact me in person.

The first problem (uncomplicated, weight of 3 points):

How can a coin toss be used to draw a fair toss if it is known that this coin has heads slightly more often than tails? A fair toss is defined as equal probability of the outcomes.

Explanation: The exact probability of the eagle is unknown.

For example, toss for each (2 times). The one with tails wins. If both have tails or both have heads, another round

[Avals]

Mathemat:
What if they are both eagles?

Another round. Until one has tails.

[Sceptic Philozoff]

Avals: Another round. Until one has tails.

Interesting. I have a slightly different option, but equivalent: make two throws, but assign them to the same person. Success is O-R, failure is R-O, all other options are ignored.

OK, one more, a little more complicated:

N football teams play in the Olympic system. How many TOTAL games must be organised between the teams to determine a winner?

Comment: the Olympic system is when they play a penalty shootout (if there is a draw, a penalty shootout). The winners advance to the next round. If any round has an odd number of teams, one team goes to the next round "for free", the others are divided into pairs and play each other. The game stops when there is 1 winner left.

The answer is obvious, but it has to be justified. And needless to say, the real Olympic system is different. I know. But that's exactly what it is in this problem.

[Sceptic Philozoff]

And one more at once, on the follow-up:

There are 13 yellow, 15 blue and 17 red chameleons living on the island. When two chameleons of different colours meet, they change to a third colour. In other cases, nothing happens. Can it happen that all the chameleons turn out to be the same colour?

[---]

Mathemat:

And one more at once, on the follow-up:

There are 13 yellow, 15 blue and 17 red chameleons living on the island. When two chameleons of different colours meet, they change to a third colour. In other cases, nothing happens. Can it happen that all the chameleons turn out to be the same colour?

Of course. red.

[Sceptic Philozoff]

sergeev: Of course. Red.

Show me how it turns out. The whole sequence.

[михаил потапыч]

sergeev:
Of course. Red.

It is enough to get two families of different colours with the same number of heads

With an initial difference of two heads between the families, it doesn't seem to solve anything.

[Avals]

Mathemat:

Interesting. I have a slightly different option, but equivalent: make two throws, but assign them to the same person. Success is O-R, failure is R-O, all other options are ignored.

OK, one more, a little more complicated:

N football teams play in the Olympic system. How many TOTAL games must be organised between the teams to determine a winner?

Comment: the Olympic system is when they play a penalty shootout (if there is a draw, a penalty shootout). The winners advance to the next round. If any round has an odd number of teams, one team goes to the next round "for free", the others are divided into pairs and play each other. The game stops when there is 1 winner left.

The answer is obvious, but it has to be justified. And needless to say, the real Olympic system is different. I know. But in this task, it is exactly that.

adding one team adds one game:

if there are an even number of teams (N), then the games in the first round would be N/2, and the teams in the next round would be N/2. If the teams were one less (N-1), then the games in the first round would be (N-2)/2=N/2 - 1, and the teams in the next round would be (N-2)/2 + 1=N/2

I.e. the next round will already have the same number of teams and remaining games. Similarly, if N is odd. Therefore adding one team adds only one game. And since with 2 teams there is 1 game, the formula is N-1

[Sceptic Philozoff]

Avals:

adding one team adds one game each:

if there are an even number of teams (N), then the games in the first round would be N/2, and the teams in the next round would be N/2. If you have one less team (N-1), you will have (N-2)/2=N/2 - 1, and the teams in the next round will be (N-2)/2 + 1=N/2

I.e. the next round will already have the same number of teams and remaining games. Similarly, if N is odd. Therefore adding one team adds only one game. And since with 2 teams there is 1 game, the formula is N-1

I did an inductive proof myself, but then I saw a very simple solution - in a few words. I got embarrassed :)

[---]

Mathemat:
Show me how it works. The whole sequence.

I made a mistake, I thought that two turns into one :(

and it can't do that. it's an odd number.