Brain Teasers are Back!

There is no need to wait for me to give a reply, the Brain Teasers are open to everyone. The more the merrier.

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You are right both. It’s 3. I imagine it’s not very difficult, however some mathematics teachers I had during studies gave wrong responses.
It’s funny.

In English, I would say “it’s frightening”

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They tried to apply mathematics formulas instead of simply reasoning.

How many “trys” would you need in order to guarantee at least one red and at least one black ?

Another one:

Pour rentrer à sa ferme, un fermier accompagné de sa chèvre, de son loup et d’un chou doit traverser une rivière. Cependant, il ne possède qu’une minuscule barque lui permettant de transporter qu’un seul de ses compagnons à la fois.
Comment va-t-il alors s’y prendre pour les faire traverser tous, sans qu’aucun ne se fasse manger par son prédateur pendant qu’ils ne seront pas surveillés ?

(le loup mange la chèvre et la chèvre mange le chou)

To go back home, a farmer, walking with a wolf, a goat, and a green cabbage has to cross a river. However the boat is very little, so he can carry only one of the three.
How will he do to cross the river, without any of the passenger being eaten by the predator. He can’t watch at them.

PS: a wolf eats a goat and a goat eats the cabbage.

Take the goat over. Come back leaving the goat, take the cabbage over. Bring the goat back. Take the wolf over. Come back, finally take the goat over.

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I would say also 3, but you have to choose once minimum in the different bag.
I may be wrong, but I am not a mathematician at all. :smile:

51?

L’homme traverse avec la chèvre…et il revient tout seul
Il traverse avec le chou et reviens avec la chèvre
Il traverse avec le loup et revient tout seul
Il traverse avec la chèvre et ils sont tous les quatre de l’autre côté.

EXACT

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With 2 tries in one bag and 1 try in the other, you have 2 red and 1 black, or 2 black and 1 red.

Seriously or you are joking ?

If I went to the same bag 50 times, I would have 50 red or 50 black, so to guarantee at least one red and one black, I need 51 tries.

3 are not enough?

If I went to the same bag 3 times I would have RRR or BBB, so wouldn’t have a RB combination.

No, because I said above that you need once to choose in a different bag.
But apparently we have not the right answer, or Don left us.

I would say that 51 is ok, but not « at least ». It’s the maximum number of tries .

Ah sorry, I misread your statement, I took “you can choose only one bag or both” to mean I could keep going with one bag.

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I tried to seek clarification of your phrase “you can choose only one bag or both” a few posts above.

This was because it was unclear whether you could actually see which bag was which.

I subsequently concluded that you could not see which bag was which, nor could you actually see which colour socket had been selected until you declared “finish!”

If you can “see” the bags, you only need TWO trys to get two sockets of the same colour … you simply pick two sockets out of the same bag !

If you can “see” the bags, you only need TWO trys to get one red plus one black … you simply pick one socket from each bag.

If you can’t see which bag is which, you can guarantee TWO sockets of the SAME colour with a minimum of THREE trys (this is what Eoink and myself described above).

If you can’t see which bag is which, you can guarantee TWO sockets of DIFFERENT colours only after 51 trys … as neatly described by Eoink above.

That’s my thought as well … see above

Well done.