20
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The referee tells Lefty, Righty, and Middly: "I have picked three positive integers $L,M,R$ with $L<M<R$ and $L+M+R=13$, and I have written these numbers onto the three cards lying face down on the table. The left card carries value $L$, the middle card carries value $M$, and the right card carries value $R$."

  • Lefty picks up the left card, looks at it and ponders a while about it. Then he announces to the others: "I am not able to determine the numbers $M$ and $R$."
  • Then Righty picks up the right card, studies the value $R$ and thinks deeply about it. Then Righty announces: "I am not able to deduce the numbers $L$ and $M$."
  • Finally Middly takes the card in the middle and thinks about the value $M$. After some time he says: "I am not able to deduce the numbers $L$ and $R$."

Question: What is the value of $M$? And what can be said about the values of $L$ and $R$?

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23
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M is 4
L and R can be 1 and 8 or 2 and 7

And here's the solution:

Lefty's turn:

L can only be 1, 2 or 3. If L were 4 or more, then M would have to be at least 5 and R at least 6, which sums up to 15, so L is at most 3.
If L were 3, then M and R would be 4 and 6 respectively and then Lefty would be able to guess the value. So L is 1 or 2.
If L were 2, then M+R would be 11, leaving 3/8, 4/7 and 5/6 as possible M/R values.
If L were 1, then M+R would be 12, leaving 2/10, 3/9, 4/8, 5/7 as possible M/R values.


Now let's go to Righty:

If R were 11 or more, then L and M would have to be 0 or negative, so R is at most 10. If R is 10, then L+M must be 3, which only leaves 1 and 2 as possible values, but Righty couldn't deduce it, so R is 9 or less. Also, as a perfect logician, Righty would know that L is 1 or 2, because if it were 3, then Lefty would know the answer.
If R is 9, then L+M must be 4, leaving only 1 and 3 as possible values for L and M. So R must be 8 or lower.
If R is 8, then L+M = 5, with possible values of: 1/4, 2/3
If R is 7, then L+M = 6, possible values: 1/5, 2/4
If R is 6, then L+M = 7, possible values: 2/5
If R is 5, then L+M = 8, no possible values
So L is [1,2] and R is [7,8]


Do the same for Middly:

As a perfect logician he would know than L is 1 or 2 and R is 7 or 8.
So M must be between 2 and 7, L can only be 1 or 2 and R can only be 7 or 8:
If M is 2, then L+R = 11, possible values: none
If M is 3, then L+R = 10, possible values: 2/8
If M is 4, then L+R = 9, possible values: 1/8, 2/7
If M is 5, then L+R = 8, possible values: 1/7
If M is 6, then L+R = 7, possible values: none
If M is 7, then L+R = 6, possible values: none
From here we can deduce that M is not 3 or 5. And because Middly could not deduce it, we can see that M must be 4


Possible solutions:

L M R
1 4 8
2 4 7

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7
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I agree with Novarg's answer, but it's a bit hard to follow. Here's how I solved it:
We can know the possible values are these:

1 2 10
1 3 9
1 4 8
1 5 7
2 3 8
2 4 7
2 5 6
3 4 6

When lefty comes in

If it was a 3, there would be no other options and he would be able to figure it out, but he didn't.
1 2 10
1 3 9
1 4 8
1 5 7
2 3 8
2 4 7
2 5 6
3 4 6

When righty goes in, he knows that lefty couldn't figure it out.

And since he couldnt either, all unique values for R must be eliminated including 6, since he would have figured out given that he knew the other 6 was eliminated!
1 2 10
1 3 9
1 4 8
1 5 7
2 3 8
2 4 7
2 5 6
3 4 6

When M comes in, the same applies for all remaining unique values for M

1 2 10
1 3 9
1 4 8
1 5 7
2 3 8
2 4 7
2 5 6
3 4 6

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  • $\begingroup$ Great answer! I solved it exactly as you did! Interestingly, if the turns of L, M and R would be different, we would always get more possible solutions. $\endgroup$ – craesh Dec 6 '15 at 21:01
3
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Wrote the following python script to help solve this puzzle:

cards = set()

for l in range(1, 13):
    for m in range(l + 1, 13):
        r = 13 - m - l
        if r > m:
            hand = frozenset((l, m, r))
            if len(hand) == 3:
                cards.add(hand)

cards = sorted(tuple(sorted(list(hand))) for hand in cards)

print('L M R ')
for hand in cards:
    print '%d %d %d' % hand

prints out:

L M R
1 2 10
1 3 9
1 4 8
1 5 7
2 3 8
2 4 7
2 5 6
3 4 6

So you can easily see Lefty must have seen a

1 or 2

And then Right had to have seen a

8 or 7

And therefore Middly must have seen a

4

And so the only solutions could be

1 4 8
2 4 7

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  • $\begingroup$ I can't help but think that writing that program must have taken longer than it would have taken to just write them out by hand. :) $\endgroup$ – Chris Dec 7 '15 at 20:17

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