Assasinative Question...

[HardAttack]

Dear Fellows,
Here i have one more question, i m sure this is not my final question and will go on by the time goes on... I hope i dont bother you with my continious questions...Lets cut the crap and here is my question...
Lets assume we are playing an assasin game, A B C D E F G H are players. I am A, and my target is to assasinate on B. What i want to know if it is possible or not the B have assasin target on me when !!! he is already my target... Ty for your upcoming answers fellows. Good and cute lucks..

[lancehoch]

The way assassin works is the names (player numbers/order) are basically shuffled into a random order, your target is the next number in the list and you are the target of the previous person in the list. So, no your target cannot be the person who has you as a target. (Disclaimer: This is all true unless people deadbeat until there are two players left in the game.)

EDIT: For the mathematically inclined, there is only one disjoint cycle in the permutation.

[HardAttack]

After your explanations, i have one similar proof as follows...
If it was possible my target has me as target, then in some odd number players game, one wud stay without a target, or one wud have 2 hunters. Yes i should have thought it before i asked, shame of me Ty for your answers...

The way assassin works is the names (player numbers/order) are basically shuffled into a random order, your target is the next number in the list and you are the target of the previous person in the list. So, no your target cannot be the person who has you as a target. (Disclaimer: This is all true unless people deadbeat until there are two players left in the game.)

EDIT: For the mathematically inclined, there is only one disjoint cycle in the permutation.

[hecter]

After your explanations, i have one similar proof as follows...
If it was possible my target has me as target, then in some odd number players game, one wud stay without a target, or one wud have 2 hunters. Yes i should have thought it before i asked, shame of me Ty for your answers...

The way assassin works is the names (player numbers/order) are basically shuffled into a random order, your target is the next number in the list and you are the target of the previous person in the list. So, no your target cannot be the person who has you as a target. (Disclaimer: This is all true unless people deadbeat until there are two players left in the game.)

EDIT: For the mathematically inclined, there is only one disjoint cycle in the permutation.

Nope, not possible. Everybody has a target and everybody is targeted.

[Draconian_Intel]

After your explanations, i have one similar proof as follows...
If it was possible my target has me as target, then in some odd number players game, one wud stay without a target, or one wud have 2 hunters. Yes i should have thought it before i asked, shame of me Ty for your answers...

The way assassin works is the names (player numbers/order) are basically shuffled into a random order, your target is the next number in the list and you are the target of the previous person in the list. So, no your target cannot be the person who has you as a target. (Disclaimer: This is all true unless people deadbeat until there are two players left in the game.)

EDIT: For the mathematically inclined, there is only one disjoint cycle in the permutation.

Nope, not possible. Everybody has a target and everybody is targeted.

That's what he was saying. When he learned the answer, he realized it was obvious because otherwise someone could end up without a target.

[hecter]

After your explanations, i have one similar proof as follows...
If it was possible my target has me as target, then in some odd number players game, one wud stay without a target, or one wud have 2 hunters. Yes i should have thought it before i asked, shame of me Ty for your answers...

The way assassin works is the names (player numbers/order) are basically shuffled into a random order, your target is the next number in the list and you are the target of the previous person in the list. So, no your target cannot be the person who has you as a target. (Disclaimer: This is all true unless people deadbeat until there are two players left in the game.)

EDIT: For the mathematically inclined, there is only one disjoint cycle in the permutation.

Nope, not possible. Everybody has a target and everybody is targeted.

That's what he was saying. When he learned the answer, he realized it was obvious because otherwise someone could end up without a target.

Ah, I see, thanks for pointing that out to me