On 10/10/06, maru dubshinki <[EMAIL PROTECTED]> wrote:
Now, the stranger appears to be absolutely useless, but nevertheless, removed from the picture the whole thing breaks down in the case where N = 2. What is the use of the useless stranger?
The key here as I see it is that prior to the stranger's announcement, each of the blue-dot natives thinks that either: 1) he is red-dot and there is only the one blue-dot native, who in turn sees all red-dot natives 2) he is a blue-dot, and the other blue-dot also sees one blue-dot native. Similarly, the red dot natives see two blues, but can't be sure about themselves: they each think they could be blue or red and don't know for sure either way. So, for all the natives, seeing everyone else's color doesn't tell them anything about their own, without any extra information becoming available. So in the initial state, both the blue-dot natives cannot distinguish between cases 1 & 2 and do not act, and each red-dotters doesn't know for sure if there are two or three blue dotters (the two he sees, plus potentiall himself). But, once the stranger blabs, all the natives, particularly the blue-dot ones, knows that the blue-dot native he sees now has enough information to act, if he sees all red-dots as in case 1 above. If there was only 1 blue-dot, he would have seen every one else with reds and known he must be the blue and killed himself that first night. When everyone is still alive on the second day, both blue dotters know that case 1 above cannot be true, so case 2 must be correct, and thue they kill themselves that night. That is, assuming they all took the time to work out the logic and didn't just say "yeah, we know" and blow it off. Funny, if both blue dotters cheat on night 2 and didn't kill themselves, all the honest red dotters would assume they were a third blue-dotter, and kill themselves on night 3. -Bryon _______________________________________________ http://www.mccmedia.com/mailman/listinfo/brin-l
