Talk:Induction puzzles
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In general, if there are n unfaithful husbands, each of their wives will believe there to be n-1 and will expect to hear a gunshot at midnight on the n-1th day. When they don't, they know their own husband was the nth.
Isn't the above statement incorrect? Gunshot will always be heard on the second day for any case other than n<2. For n =>2 the wives will always be aware of n-1 unfaithful husbands on the first day, hence any wife with an unfaithful husband will always know their husband is the nth unfaithful one on the second day.
That's not quite true - the wives of unfaithful husbands will be aware of n-1 unfaithful husbands and the wives of faithful husbands will be aware of n. If a wife from one group discusses numbers with a wife from the other group, they will both immediately know the fidelity of their own husbands, which breaks the rules. So we can assume they don't discuss numbers at all.
This means that no wife knows what n is. By your suggested logic, every wife would kill her husband on the second day, whether faithful or not. The only way they can tell which group they're in is by figuring out the inductive logic and waiting to see if anything happens on day n-1.
Note that the wives of faithful husbands are waiting too, but they're waiting to see if anything happens on day n because they know about one extra unfaithful husband. So when they hear the gunshots on night n, they know they're safe.
Any attempt to shortcut the wait time requires discussion that would give away the answer and break the rules. CupawnTae 09:37, 22 October 2006 (UTC)
[edit] Wording in third solution
I changed the wording in the solution to the third problem because it made heavy use of the word 'repeated' to mean 'seen' or 'appeared.' For instance, seeing a color "repeated once" means it is seen twice because one must have already seen the color in order for it to be repeated. --Ddawn23 12:42, 19 November 2006 (UTC)
