Can anyone give me some puzzle questions

Type 1: The one at the front must be the yellow hat. Because there are nine people in the back who are not sure what kind of hat they are, it means that it may be a yellow hat or a blue hat.

This is not enough to convince, this has to be reversed.

For example, in the case of 2 people, there are 2 yellow hats and 1 blue hat, if the first person is a blue hat, because the second person can see the front one, and there is only one blue hat. In this case, the second can be deduced to be a yellow hat. If the first one is a yellow hat, it is uncertain whether the second person is wearing a yellow hat or a blue hat.

For example, in the case of 3 people, there are 3 yellow hats and 2 blue hats. There are several possibilities in this case:

A: 1 blue, 2 blue case, 3 know what color you are, because there are only two blue hats.

b: 1 blue, 2 yellow, 3 don't know what color they are, 2 know what color they are. Because 2 thinks like this, 1 is blue, if he is blue, then 3 should know what color he is, and 3 does not know, then he must be yellow.

C: 1 yellow, 2 blue, 3 don't know what color they are, 2 can't be sure what color they are.

D: 1 yellow, 2 yellow, 2 and 3 are not sure what color they are.

Excluding the cases of A and B, only C and D remain, and in both cases, 1 is yellow.

For example, in the case of 4 people.

5 people. And so on.

So in the case of 10 people, only the first person is a yellow hat, and the others can't be sure. If the first person is a blue hat, there will always be one of the remaining nine people who can be sure.

I also asked this question in today's interview, I thought about this answer for a long time and felt that this answer is not right, it is only suitable for the situation of 3 people, if it is the case of four people: 1 blue 2 yellow 3 blue or 1 blue 2 yellow 3 yellow 4 no matter what it is, I can only not know and then to 3, 3 think that I can not give 4 a clear answer whether it is yellow or blue, and then 3 can only not know not to mention so there is no answer to the question.

Shoot A-B-C Because B is a sharpshooter, and C's hit rate is only 50%, A must hit B first, B is dead, and C's is left, C shoots A, and if A is dead, C is alive; If it misses, A shoots and A survives, because it says the rest of the people shoot one after another. That is, the survival rate of A at this time can be said to be 1 2, B is 0, and C is 1 2

A-C-B is the same as above, A hits B first, then C shoots, A is 1 2, B is 0, C is 1 2

B-A-C B hits A first, A dies, and then C hits B, at this time B's survival rate is 1 2, A is 0, and C is 1 2

B-C-AIbid.

C-A-B At this time, C shoots A or B first, and the survival rate of A and B is 1 4, assuming that he hits A, and A dies, then B hits C; If you don't die, A hits B, and C has a survival rate of 1 2 The same goes for B, where A and B have a survival rate of 1 4, and C is also 1 4

C-B-A.

a: 1 4, = 25% b: 1 4, = 25% c: approximately equal to So C has the highest survival rate.

A or B is 1 4 each, and C is 1 2 but C has to deliberately miss.