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The caretakers in a Home for the Golden Agers decided to have a guessing game for the old fellows. Two partner teams shall play the game and the winners shall have free access to the fridge for 1 week. One partner of the participating team shall roll a die and draw a letter tile from the Scrabble set. If blank tile is drawn the team loses the game. There is also a box of a complete set of playing cards on the table. This can be used for a clue so that the other partner on other table could guess what number was rolled and what letter was randomly picked. But the guessing partner is only allowed to see just 1 card inside the card box. In what way can any team win this parlor game?

NOTE: I've attempted this problem twice before. If you'd like to view those tries, check out the edit history.

Hopefully the third time's a charm.

I'm going to try a more mathematical approach this time. There are exactly

$26*6=156$ possible variations (ignoring the blank tiles) of tiles and numbers that can be chosen. Some tiles are more common than others. Perhaps we can use this to our advantage later - and perhaps not. This solution is still in progress.

So, there are 52 cards in a deck. EDIT: Er nope, we're using a 54 card deck. I'll add the jokers in to my solution at the end.

That falls short of the required 156 variations. We can improve the number of available variations from 52 to 104 by changing the card's orientation within the card box - front facing and back facing.

That still falls short. However, there is something that I failed to consider in my previous attempts:

Certain cards also have different upside-down and right-side up orientations. These cards are the Ace, three, five, seven, and nine of clubs, the same values of hearts, the same values of spades, and the seven of diamonds. That's a total of sixteen cards. These cards can not only be placed in the box front-facing and forward facing, but also upside-down and right-side up. Thus, we can add another thirty-two variations to our communication, for a grand total of 136 - 20 variations short.

Then there's the jokers, which add

8 variations, so we're only 12 short of a perfect solution

At this point, we have two options. We can either give up or we can

leverage that probability that I mentioned earlier, and say that we will hope that we don't ever get assigned any Q's or Z's, the rarest Scrabble tiles.(That's 12 combinations thrown out.) Under this system, we have very good chances of getting a transmittable combination.

However, we still have the issue of

Creating an actual table of values to compare the card combinations with the dice-tile combinations. I think my previous text has outlined that this is possible, but would be very time-consuming. I suppose senior citizens would have the time to apply my solutions.

In conclusion, our chances

Are excellent, but not perfect.

Of course, we have the occasional blank tile, but that isn't the transmission method's fault. It seems that a perfect solution is

impossible without some gimmick.

We need to identify $26 times 6 = 156$ states.

Let's assume we have a standard deck of $52$ cards plus $2$ jokers, and the packet has a front and back and top and bottom (likely discernable by the writing).

We need to choose one of the $54$ cards and then decide how to place it in the box to communicate these states.

Usually many cards of a deck may be categorised as being upright or upside-down by virtue of more pips being upright...

A simple example is the five of spades:

The central pip is "upright".

A slightly less obvious example is the eight of spades:

Here five pips are upright and three are upside-down, making the card "upright".

Assuming a standard design like this:

Where diamond pips have no such orientation and face cards have no means to tell which orientation they are in, we have the following $24$ cards which have such a discernable orientation: $A$, $3$, $5$, $6$, $7$, $8$, $9$ of Clubs or Hearts or Spades, the two jokers (assuming, like the bicycle deck shown, have an obvious way up) ...and the $7$ of diamonds.

All cards may also be face-up or face-down. Thus these $24$ cards have $4$ ways they may be placed into the packet, while the other $30$ cards have $2$ ways. This gives a total of $24 times 4 + 30 times 2 = 156$ ...just enough to try to decide upon and remember a mapping.