Yes, but every prisoner can only see the hat on the one person directly in front of him. So person number three can only see the hat on person number two. Not the one on person number one, not the one on person number four, not his own, only that of number two.
Plus, you don't know how the hats are divided. For all we know there's one black hat on person 1 and all the other 99 have whites. Or, it could be the other way around. Or, more likely, somehting inbetween.
I still like my cricket solution best

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