i*****g
发帖数: 15 | 1 100 persons, each of them wear a hat which is assigned a number from 1 to
100 (one number can appear several times).
Suppose everyone can see other's hats except his number. Now, everyone guess
the number of his own hat simultaneously. They can cooperate before wearing
the hats, but after that they can not discuss any more. Give a strategy
which guarantees or maximize the probability of at least one person's guess
is right. | o***n
发帖数: 921 | 2 when u say "cooperate", can these 100 people line up according to some order
after the hats are assigned? If yes, I have a strategy to save all 100
people.
guess
wearing
guess
【在 i*****g 的大作中提到】
: 100 persons, each of them wear a hat which is assigned a number from 1 to
: 100 (one number can appear several times).
: Suppose everyone can see other's hats except his number. Now, everyone guess
: the number of his own hat simultaneously. They can cooperate before wearing
: the hats, but after that they can not discuss any more. Give a strategy
: which guarantees or maximize the probability of at least one person's guess
: is right.
| i*****g
发帖数: 15 | 3 Think about 2 person case: A and B are assigned with number 1 and 2. They
can have the following strategy to win the game:
A guess B's number and B guess the opposite of A's number. You can verify
this way guarantees at least one's guess is right. I don't how to generalize
the strategy to 100 people case. | w******r
发帖数: 139 | 4 I think they can sum up all the number each person sees and stand up in a
row from a smaller sum to a higher sum. This means their number on the hat
start from a grater number to a smaller number. They guess their number
starting from 100 for the person with the smallest sum. This strategy will
guarantee a least one person is right. | q**s
发帖数: 10 | 5 I am confused. In your example, A guess his is 2, and B guess his is 2-1=
1? None is right. Thanks.
generalize
【在 i*****g 的大作中提到】
: Think about 2 person case: A and B are assigned with number 1 and 2. They
: can have the following strategy to win the game:
: A guess B's number and B guess the opposite of A's number. You can verify
: this way guarantees at least one's guess is right. I don't how to generalize
: the strategy to 100 people case.
| i*****g
发帖数: 15 | 6 Actual:
A 1 2 1 2
B 1 2 2 1
Guess:
A 1 2 2 1
B 2 1 2 1
win win win win
If A see B's number is 1 then A guess 1. If B see A's number is 1, then B
guess 2 (the opposite of A's
number).
【在 q**s 的大作中提到】
: I am confused. In your example, A guess his is 2, and B guess his is 2-1=
: 1? None is right. Thanks.
:
: generalize
| i*****l
发帖数: 50 | 7 how could they sum up all the numbers each person sees?
Everyone does not know what others see when the game begins.
【在 w******r 的大作中提到】
: I think they can sum up all the number each person sees and stand up in a
: row from a smaller sum to a higher sum. This means their number on the hat
: start from a grater number to a smaller number. They guess their number
: starting from 100 for the person with the smallest sum. This strategy will
: guarantee a least one person is right.
| c******j
发帖数: 149 | 8
guess
wearing
guess
Denote the color number for ith person is Ai, then A = sum(i = 1 to 100)Ai,
take the mode of A with respect to 100, the result could be only 0, 1, ... ,
99, totally 100 possibilities. This tells us the strategy.
Before wearing the hats, each of them will be assigned a number from 0 to 99
. Say the ith person is assigned 7, then after everyone gets his hat, this
ith guy sum up the color number of all other people, say Bi, he tries to
find a color number for himself, say AAi,
【在 i*****g 的大作中提到】
: 100 persons, each of them wear a hat which is assigned a number from 1 to
: 100 (one number can appear several times).
: Suppose everyone can see other's hats except his number. Now, everyone guess
: the number of his own hat simultaneously. They can cooperate before wearing
: the hats, but after that they can not discuss any more. Give a strategy
: which guarantees or maximize the probability of at least one person's guess
: is right.
| n****e
发帖数: 629 | 9 刚才答错了，比我想的难一点，呵呵
考虑100进制，比如32+99=31，以此类推。
设第n个人的帽子号为Y(n)，他的guess为X(n)
那么我们令X(1)=Y(2)+Y(3)+...+Y(100)
如果Y(1)=Y(2)+..+Y(100)那就对了
如果不等，那么有99种情况。令X(2)...X(100)分别对应这99种
比如这样:X(2)=Y(1)-Y(3)-...-Y(100)-1
X(3)=Y(1)-Y(2)-Y(4)-...-Y(100)-2
这样必然有一个人是对的
guess
wearing
guess
【在 i*****g 的大作中提到】
: 100 persons, each of them wear a hat which is assigned a number from 1 to
: 100 (one number can appear several times).
: Suppose everyone can see other's hats except his number. Now, everyone guess
: the number of his own hat simultaneously. They can cooperate before wearing
: the hats, but after that they can not discuss any more. Give a strategy
: which guarantees or maximize the probability of at least one person's guess
: is right.
| l********e
发帖数: 349 | 10 请教一下，为什么mode(Bi+AAi, 100)= 7 代表他猜对了呢？不是应该 AAi=7 才是猜对
了吗？
能否再说的详细些呢？
Denote the color number for ith person is Ai, then A = sum(i = 1 to 100)Ai,
take the mode of A with respect to 100, the result could be only 0, 1, ... ,
99, totally 100 possibilities. This tells us the strategy.
Before wearing the hats, each of them will be assigned a number from 0 to 99
. Say the ith person is assigned 7, then after everyone gets his hat, this
ith guy sum up the color number of all other people, say Bi, he tri
【在 c******j 的大作中提到】
:
: guess
: wearing
: guess
: Denote the color number for ith person is Ai, then A = sum(i = 1 to 100)Ai,
: take the mode of A with respect to 100, the result could be only 0, 1, ... ,
: 99, totally 100 possibilities. This tells us the strategy.
: Before wearing the hats, each of them will be assigned a number from 0 to 99
: . Say the ith person is assigned 7, then after everyone gets his hat, this
: ith guy sum up the color number of all other people, say Bi, he tries to
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