I have some ideas...
1-instead of the 10th man saying BLACK or WHITE based on parity, base it on MAJORITY (duh). That way it has a better chance of helping out the fellow man. THere is an odd # of hats in front of him, so there must be a majority (5/4 at worst). Then things could continue with one of the above algorithms, but in the event that a guess must be made, they could keep track of the colors of hats behind them, and base their guess on probability.
Take it a step further...
10 - starts by making a call on the majority...as stated above.
9 - looks at the 3 guys in front of him, and says "black" if the majority of 8,7,6 is "black", and "white" if the majority is white. #9 himself has a 50/50 chance of survival.
8-looks at 7 and 6 and makes his decision on majority...
EXAMPLE: If 9 says "black", then he is looking at 2 possibilities of the 2 guys in front of him. Both black (which he would know he was white) or mixed (which he knows he's black).
Either way, 8 definitely knows which he is.
7-Sees #6. If 9 said "black", and let's say 8 said "black", and #6 (whe he can see) is white, then he knows he's black. If #6 is black, then he knows he's white.
6-Knows his color based on the answers of 8 and 7.
5-He's gotta be a real man, and take one for the team. He has to have the same task as man #9, and make his call on the majority of the 3 in front of him. Let's say that this time the majority is white, for example. So he says "white". There's also a 50/50 chance of him surviving just based on random chance.
4-same logic as #8
3-same logic as #7
2-same logic as #6
1-Bases his decision on #10's initial majority call, and is definitely saved.
So here's how it comes down...
10=50/50
9=50/50
8=saved
7=saved
6=saved
5=50/50
4=saved
3=saved
2=saved
1=saved (because he kept track of everyone behind him, and knows the majority based on 10's call. Keep in mind, he's not to take #10's color into account, because #10 didn't know his color when he called it).
The above method GUARANTEES us 7/10 survivors. And Chances are that 8 or 9 of them will survive.
:-)
(Edited for spelling error, that ended up not being an error...lol)
1-instead of the 10th man saying BLACK or WHITE based on parity, base it on MAJORITY (duh). That way it has a better chance of helping out the fellow man. THere is an odd # of hats in front of him, so there must be a majority (5/4 at worst). Then things could continue with one of the above algorithms, but in the event that a guess must be made, they could keep track of the colors of hats behind them, and base their guess on probability.
Take it a step further...
10 - starts by making a call on the majority...as stated above.
9 - looks at the 3 guys in front of him, and says "black" if the majority of 8,7,6 is "black", and "white" if the majority is white. #9 himself has a 50/50 chance of survival.
8-looks at 7 and 6 and makes his decision on majority...
EXAMPLE: If 9 says "black", then he is looking at 2 possibilities of the 2 guys in front of him. Both black (which he would know he was white) or mixed (which he knows he's black).
Either way, 8 definitely knows which he is.
7-Sees #6. If 9 said "black", and let's say 8 said "black", and #6 (whe he can see) is white, then he knows he's black. If #6 is black, then he knows he's white.
6-Knows his color based on the answers of 8 and 7.
5-He's gotta be a real man, and take one for the team. He has to have the same task as man #9, and make his call on the majority of the 3 in front of him. Let's say that this time the majority is white, for example. So he says "white". There's also a 50/50 chance of him surviving just based on random chance.
4-same logic as #8
3-same logic as #7
2-same logic as #6
1-Bases his decision on #10's initial majority call, and is definitely saved.
So here's how it comes down...
10=50/50
9=50/50
8=saved
7=saved
6=saved
5=50/50
4=saved
3=saved
2=saved
1=saved (because he kept track of everyone behind him, and knows the majority based on 10's call. Keep in mind, he's not to take #10's color into account, because #10 didn't know his color when he called it).
The above method GUARANTEES us 7/10 survivors. And Chances are that 8 or 9 of them will survive.
:-)
(Edited for spelling error, that ended up not being an error...lol)
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