Best Answer Prime, 07 February 2013 - 09:10 AM

I’ll try again.

Spoiler for not that long

Just the case of 6 red and 6 blue labels, disregarding the round before the host revealed two of the stamps.

Let’s use the following terminology:

* = AND;

There are only 6 significant cases (disregarding the order) based on the number of mixed colors in the lineup -- from zero to 5:

1)

Example 1: r

2)

Ex.1: r

3)

Ex.1:

4)

Example: x

5)

Ex.1:

6)

Example: x1''

I derive deduction strings for each case based on perspectives that the men in the lineup may have and deduction strings from those perspectives. Take for example case (3). An

It appears,

x

x

r''

The first round, where 14 stamps were in play, does not seem to offer any useful information for those longest deduction variations above.

Let’s use the following terminology:

**r**= a man with two red stamps;**b**= man with two blue stamps;**x**= man with a mix.**x'**= NOT (x was unable to deduce his stamps.)* = AND;

**+**= OR;**x(condition)**= x can deduce his stamps based upon the condition. E.g., x(r’) means x can deduce his stamps after r has failed.There are only 6 significant cases (disregarding the order) based on the number of mixed colors in the lineup -- from zero to 5:

**rrrbb, xrrbb, xxrrb, xxxrb, xxxxr, xxxxx**. “r” and “b” may be swapped for one another making logically identical cases.__The 6 cases may be resolved as following:__1)

**r**==> b() + r_{1}r_{2}r_{3}b_{1}b_{2}_{3}(r_{2}’(r_{1}’)).Example 1: r

_{1}’__r__**b**_{1}_{2}r_{3}b_{2}, where b_{1}solves on his turn; OR Ex.2: r_{1}’ r_{2}’__b__**r**_{3}_{1}b_{2}, where r_{3}solves after r_{1}and r_{2}failed.2)

**x r**==> x(r’ * b’) + r_{1}r_{2}b_{1}b_{2}_{2}(r_{1}’) + b_{2}(b_{1}’)Ex.1: r

_{1}’ b_{1}’__r__**x**_{2}b_{2}. Ex.2: r_{1}’__b__**r**_{2}_{1}b_{2}x. Ex.3: b_{1}’ x’ r_{1}’__r__**b**_{2}_{2}.3)

**x**==> b() + x(r_{1}x_{2}r_{1}r_{2}b_{2}’(r_{1}’)).Ex.1:

__r__**b**_{1}r_{2}x_{1}x_{2}. Ex.2: r_{1}’ r_{2}’__b x__**x**_{1}_{2}.4)

**x**==> x(r’ * b’)._{1}x_{2}x_{3}r bExample: x

_{1}’ r’ b’__x__**x**_{2}_{3}.5)

**x**==> x_{1}x_{2}x_{3}x_{4}r_{1}(x_{2}’(x_{1}’ * r’)).Ex.1:

__r' x__**x**_{1}'_{2}' x_{3}' x_{4}' (x_{1}solves on his second turn). Ex.2: x_{1}''__x__**x**_{2}'_{3}' x_{4}' r' (x_{2}solves on his second turn.)6)

**x**==> x_{1}x_{2}x_{3}x_{4}x_{5}_{2}(x_{1}’(x’(x_{1}’ * x_{2}’))).Example: x1''

__x__**x**_{2}'_{3}' x_{4}' x_{5}', where x_{2}can deduce that he is a mix on his second turn after x_{1}fails for the second time.I derive deduction strings for each case based on perspectives that the men in the lineup may have and deduction strings from those perspectives. Take for example case (3). An

**x****in that line up may suppose he is either part of the case (2), or (3). Without any additional data,****r**may think he is in either (3) or (4). Whereas**b**in lineup (3) simply knows he is b.It appears,

**the longest deduction string here is one where the men made one full round of non-guessing whereafter in the second round the second man in the line up must deduce his stamps**. There are several possibilities for that:x

_{1}''__x__**x**_{2}'_{3}' x_{4}' x_{5}';x

_{1}''__x__**x**_{2}'_{3}' x_{4}' r';r''

__x__**x**_{1}'_{2}' x_{3}' b'.The first round, where 14 stamps were in play, does not seem to offer any useful information for those longest deduction variations above.

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