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He points to:

HIDDEN IS >= SHOWN

Because:

#2 says one of the dice is greater than the other. The first two cards can be played in four ways, as it is not specified which die is hidden. For example, a throw of 2,7 could be 2 HIDDEN (card 1) or 7 HIDDEN (card 2). This gives us the two cards.

Then:

Cards 3 implies card 4, but not the other way round. Both the LESSER and GREATER die obey both, one (just card 4) or neither cards together. So #3 implies that the LESSER die must be a $1$ or $4$, and the GREATER die cannot be a $4$ or $7$. Also #3 tells us that the LESSER die is the HIDDEN die.

Now:

Card 2 is false, and card 1 is true.

And:

Removing card 5 violates #3. Removing card 4 disallows card 3, and would only leave 1 true card for the GREATER SHOWN die, namely card 1.

Finally,

So he must have pointed to either card 1 or card 3, and we want the card that leaves the outcome of the dice unknown, i.e. one of the cards fixes the dice. Removing card 1 leaves cards 3 and 4 true, so $12$ or $45$. Removing card 3 means the sum is odd, but the dice are not one apart, and this can only be $45$.

Therefore:

Card 1 is pointed at.

He points to:

HIDDEN IS >= SHOWN

Because:

#2 says one of the dice is greater than the other. The first two cards can be played in four ways, as it is not specified which die is hidden. For example, a throw of 2,7 could be 2 HIDDEN (card 1) or 7 HIDDEN (card 2). This gives us the two cards.

Then:

Cards 3 implies card 4, but not the other way round. Both the LESSER and GREATER die obey both, one (just card 4) or neither cards together. So #3 implies that the LESSER die must be a $1$ or $4$, and the GREATER die cannot be a $4$ or $7$. Also #3 tells us that the LESSER die is the HIDDEN die.

Now:

Card 2 is false, and card 1 is true.

And:

Removing card 5 violates #3. Removing card 4 disallows card 3, and would only leave 1 true card for the GREATER SHOWN die, namely card 1.

Finally,

So he must have pointed to either card 1 or card 3, and we want the card that leaves the outcome of the dice unknown, i.e. one of the cards fixes the dice. Removing card 1 leaves cards 3 and 4 true, so $12$ or $45$. Removing card 3 means the sum is odd, but the dice are not one apart, and this can only be $16$.

Therefore:

Card 1 is pointed at.

HIDDEN IS >= SHOWN

Because:

#2 says one of the dice is greater than the other. The first two cards can be played in four ways, as it is not specified which die is hidden. For example, a throw of 2,7 could be 2 HIDDEN (card 1) or 7 HIDDEN (card 2). This gives us the two cards.

Then:

Cards 3 implies card 4, but not the other way round. Both the LESSER and GREATER die obey both, one (just card 4) or neither cards together. So #3 implies that the LESSER die must be a $1$ or $4$, and the GREATER die cannot be a $4$ or $7$. Also #3 tells us that the LESSER die is the HIDDEN die.

And:

Removing card 5 violates #3. Removing either of cards 3 or 4 doesn't change anything, and so would be a non-clue, and you would have told your friend this. So, either of the first two cards must have been pointed to, and card 2 is now false, and so must have been the card pointed to.

Finally,

As now at least one of cards 3 and 4 is true, the dice must be one of $12, 16, 45$. As you still don't know the dice, card 3 (and therefore also card 4) must be true, so the final options are $12$ or $45$.

He points to:

HIDDEN IS >= SHOWN

Because:

#2 says one of the dice is greater than the other. The first two cards can be played in four ways, as it is not specified which die is hidden. For example, a throw of 2,7 could be 2 HIDDEN (card 1) or 7 HIDDEN (card 2). This gives us the two cards.

Then:

Cards 3 implies card 4, but not the other way round. Both the LESSER and GREATER die obey both, one (just card 4) or neither cards together. So #3 implies that the LESSER die must be a $1$ or $4$, and the GREATER die cannot be a $4$ or $7$. Also #3 tells us that the LESSER die is the HIDDEN die.

Now:

Card 2 is false, and card 1 is true.

And:

Removing card 5 violates #3. Removing card 4 disallows card 3, and would only leave 1 true card for the GREATER SHOWN die, namely card 1.

Finally,

So he must have pointed to either card 1 or card 3, and we want the card that leaves the outcome of the dice unknown, i.e. one of the cards fixes the dice. Removing card 1 leaves cards 3 and 4 true, so $12$ or $45$. Removing card 3 means the sum is odd, but the dice are not one apart, and this can only be $45$.

Therefore:

Card 1 is pointed at.

Either of the $\ge,\le$ cards.

Because:

#2 says one of the dice is greater than the other. The first two cards can be played in four ways, as it is not specified which die is hidden. For example, a throw of 2,7 could be 2 HIDDEN (card 1) or 7 HIDDEN (card 2). This gives us the two cards.

Then:

Cards 3 implies card 4, but not the other way round. Both the LESSER and GREATER die obey both, one (just card 4) or neither cards together. So #3 implies that the LESSER die must be a $1$ or $4$, and the GREATER die cannot be a $4$ or $7$.

And:

Removing card 5 violates #3. Removing either of cards 3 or 4 doesn't change anything, and so would be a non-clue, and you would have told your friend this. So, either of the first two cards must have been pointed to, although there is no way of knowing which one.

Finally,

As now at least one of cards 3 and 4 is true, the dice must be one of $12, 16, 45$.

HIDDEN IS >= SHOWN

Because:

#2 says one of the dice is greater than the other. The first two cards can be played in four ways, as it is not specified which die is hidden. For example, a throw of 2,7 could be 2 HIDDEN (card 1) or 7 HIDDEN (card 2). This gives us the two cards.

Then:

Cards 3 implies card 4, but not the other way round. Both the LESSER and GREATER die obey both, one (just card 4) or neither cards together. So #3 implies that the LESSER die must be a $1$ or $4$, and the GREATER die cannot be a $4$ or $7$. Also #3 tells us that the LESSER die is the HIDDEN die.

And:

Removing card 5 violates #3. Removing either of cards 3 or 4 doesn't change anything, and so would be a non-clue, and you would have told your friend this. So, either of the first two cards must have been pointed to, and card 2 is now false, and so must have been the card pointed to.

Finally,

As now at least one of cards 3 and 4 is true, the dice must be one of $12, 16, 45$. As you still don't know the dice, card 3 (and therefore also card 4) must be true, so the final options are $12$ or $45$.