

A very interesting question with a game like this is what is the best score you can guarantee. Note that guarantee means what you can get no matter what (it does not mean what will you get 99% of the time or anything like that).
For 2 players I found out that you can not guarantee a score better than 14.
I think the simplest argument is this: Say that all the ones are the bottom 15 cards of the deck. It means that just before the first one is drawn, the two players have a combined 10 cards in their hands. One card must be discarded (or played) to draw the first one. At this point there are just 9 cards left in their combined hands that are not ones and the deck only contains ones. Even if they manage to play the 9 cards and the 5 different ones, they still only get a score of 14.
I believe the rules define that you need a score of 15 to win the game, so it means that no matter how good you get at the game, you can never have a strategy where you win every time (if you shuffle and the 15 ones end up at the bottom cards, you cannot get a score more than 14).
I think you can use a somewhat similar proof technique for higher player counts although I think the number you get might be slightly higher.
Here we have shown an upper bound on the best score you can guarantee. It would also be very interested to have a lower bound, that is a result of the form
With 2 players, you can always get a score of x.
Here a higher x means a stronger result. I think it will be hard to show a proof of this for x much higher than 5.

Robb Effinger
United States Bothell Washington

I think you might be able to guarantee a score of 13 in a 2er game, if that's your goal. At very least, 10 should be easy.
This requires an extremely artificial convention whose goal is to collect 1 two of each color, and one three of 4 colors. The basic form of play would look something along the lines of "If someone draws a 2 or 3 that isn't already held, clue it by giving a color clue. That card is now locked. The other player will then discard, keeping the clue count at 8" Number clues will occasionally have to be given opportunistically (At very least, when a player draws their 3rd two/three that needs to be held). At the start of the game players can spend clue tokens freely to lock down cards, since you're going to discard enough cards in the middle of the game to get back to 8.
Ideally, the 2s would be all collected into one hand, and all the threes in another. Since we aren't hitting our ideal situation, instead, the twos will be split 23 at the time that the last 15 cards are hit. (This can be guaranteed. If there are two twos as cards 16 and 17, for example, then the player who needs a 2 to fill their hand will be the only one drawing, the other player just cluing). (EDIT: This is ideal, because if all the 2s are in one hand then the other player can draw the 1s, which means after they play the 1 the two will play without needing to spend a clue to get back into the first player's hand. And if all the threes are also consolidated, then they will often play for free as well)
So going into the last 15 cards, you have 8 clues and 5 unique 2s, 4 unique 3s in the hands. I believe the following sequence of play is the worst case. On lines without a (discard), assume a clue had to be given so that the player who held the card could get the turn to play it. There are a couple places where both players have to be holding, at worst, the same one, and so they both have plays and a clue does not need to be spent to pass the turn.
15 cards, all 1s (Discard) 14 cards, 1 played 13 cards, 1 played, dup drawn 12 cards, 12 played, 2 dups drawn (Discard) 11 cards, 12 played, 2 dups 1 uniq drawn 10 cards, 12 1 played, 3 dups drawn 9 cards , 12 12 played, 4 dups drawn 8 cards , 12 123 played, 4 dups 1 unq drawn 7 cards , 12 123 1 played, 5 dups drawn 6 cards , 12 123 12 played, 6 dups drawn 5 cards , 12 123 123 played, 6 dups 1 unq drawn 4 cards , 12 123 123 played, 7 dups 1 unq drawn (Discard) 3 cards , 12 123 123 1 played, 8 dups drawn 2 cards , 12 123 123 12 played, 8 dups 1 unq drawn 1 card , 12 123 123 123 played, 9 dups 1 unq drawn 0 cards , 12 123 123 123 1 played play the 2, for 12 123 123 123 12 played
It's very possible I've messed something up, this was done kinda quickly and without much verification. But I stand by guaranteeing at least 10, since holding the 5 unique twos feels easy, and they'll all get played.



It was an interesting post, thanks a lot for participating.
However, note that to prove your statements "You can guarantee a score of 13 in a 2er game" you need a more strict argument. You can not assume that the 15 ones are on bottom of the deck. You need a strategy that works for any permutation of the cards.
I think you have some good ideas for the argument but it has to be more strict. I would already be very interested in a formal argument for
"You can guarantee a score of 13 in a 2er game" since it is not clear to me at all.

Robb Effinger
United States Bothell Washington

Muemmelmann wrote: It was an interesting post, thanks a lot for participating. However, note that to prove your statements "You can guarantee a score of 13 in a 2er game" you need a more strict argument. You can not assume that the 15 ones are on bottom of the deck. You need a strategy that works for any permutation of the cards.
I think you have some good ideas for the argument but it has to be more strict. I would already be very interested in a formal argument for
"You can guarantee a score of 13 in a 2er game" since it is not clear to me at all.
Err, sorry  I thought about this briefly, but didn't post anything about it. If your goal is to guarantee at least 13, you could start by playing with this convention, and if anyone draws a 1 before the last 15 cards, you could immediately break out of this convention by saying "That is a 1". At which point, it plays, and you can go right back into the convention, but now you have the ability to store an extra known card (and when/if the appropriate 2s are found, they will also play).
Since this yields a strictly better result then the scenario where all the 1s are on the bottom of the deck, I think that all the 1s on the bottom of the deck (grouped by color) is the worst deck configuration possible.
Surely you agree that a score of at least 10 is possible, no matter the deck configuration, using the strategy outlined above (and, if a 1 is drawn earlier, playing it). Even in the scenario where all 10 2s are on the bottom of the deck, you'll still achieve a score of at least 10 if you just only care about 1s and 2s. (and in reality, you'll be able to do much better, since you'll be able to play 1s and save 3s)?

Sean McCarthy
United States Seattle Washington

Making it work for 1s not being at the bottom of the deck, while admittedly adding complication, is certainly the easy version. Just adding the rule that if there are 1s to play you clue them as 1s should take you most of the way there.



I would still love to see it written in a more precise manner. Imagine you have to write it in a way where a machine would be able to perform it. You have not defined what to do if the opponent starts with 3 blue ones and 2 blue two's for example. You need to have a strategy where it is defined what to do no matter what the game state is.
I would love to see a precise argument (for 10 or even for a lower value). Before we have that, we cannot really know



As an example le'ts see how we could show that you can always get a score of a lest 1 as two players.
Strategy: If it is your first turn: If the opponent has any ones, give him a hint pointing to all his ones. Otherwise give him any color hint.
If it is not your first turn: If you received a hint about a one you have, play it. Otherwise, if the opponent has any ones, give him a hint pointing to all his ones. Otherwise discard any card.
In this strategy, if you start with a one in your hand, you will clearly get a score of at least one. If you do not start with a one in your hand, you will spend hints in your two first turns (bringing you down to six). After that you will either discard a card (getting you one hint) or give the opponent a hint (paying one hint) after which he will guarantee the score of 1. You will have at least five hints at all times.
You could probably use a similar strategy to guarantee that you play all the blue cards (or maybe all the ones).
Edit: I did not write it but you just discard every time if one one has been played, since you got the score you aimed for.

Andy Latto
United States Foxboro Massachusetts

Muemmelmann wrote: As an example le'ts see how we could show that you can always get a score of a lest 1 as two players.
Strategy: If it is your first turn: If the opponent has any ones, give him a hint pointing to all his ones. Otherwise give him any color hint.
If it is not your first turn: If you received a hint about a one you have, play it. Otherwise, if the opponent has any ones, give him a hint pointing to all his ones. Otherwise discard any card.
In this strategy, if you start with a one in your hand, you will clearly get a score of at least one. If you do not start with a one in your hand, you will spend hints in your two first turns (bringing you down to six). After that you will either discard a card (getting you one hint) or give the opponent a hint (paying one hint) after which he will guarantee the score of 1. You will have at least five hints at all times.
You could probably use a similar strategy to guarantee that you play all the blue cards (or maybe all the ones). I believe this strategy fails if both players start with two pairs of ones of the same color in their hands. You need some way to stop a player from playing a second one of the same color.
In practice, this would not be hard to do; if someone plays a blue one, and they have another blue 1, clue their blue cards next to stop them from playing the other blue 1. But if you want a completely algorithmic solution, you have to cover this case.
The case where ones are distributed between the players might be a bit trickier. Let's suppose each player starts with 5 ones of different colors.
Player 1: These are your 1's (points to all cards) Player 2: These are *your* 1's (points to all cards) Player 1: plays the blue 1. Player 2: <probably needs to temporize with useless clue here, to give player 1 the chance to indicate which of player 2's 1's is blue>
So I think that a full proof would be tedious, since there are so many different cases to consider.



Thanks, I added that part. As soon as one one has been played you simply discard every time.

Robb Effinger
United States Bothell Washington

I started working on a full algorithm, it was tedious. So here's a simpler one, which you can extend if you care .

Algorithm for a score of 5:
If you have a one of an unplayed color, play it. If you have a one of a duplicate color, discard it. If your partner has a one of an unplayed color, clue it Otherwise, discard

Let's define the middle of the game as the point in time when a player has no unique ones in their hand. Once this point is reached, one player will discard. State 1: Now, the other player will either see a unique one, in which case they will give a clue (using the token just made available by the discard), or they themselves will discard, or, if at max clue tokens, give a random clue by number. If a 1s clue was given, then the player who recieves the clue knows to play (because they know the algorithm the other player is following, they know it's a unique one). This means they play their card, drawing a new one. This puts us back into State 1, with one less clue token. If a clue was not given, then we are once again back in state one, but with the other player being the active player, and with one more clue token (or eight tokens, if we're at our max). If a random clue was given this puts us back into State 1, with 7 clue tokens. The player who was clued will discard, putting us back into state 1, with 8 clue tokens.
If we have at least as many clue tokens when we enter state 1 as the number of ones we have to play, then we can play all the 1s.
Now, how to get to the middle of the game.
At the start of the game, we have 8 clue tokens and both players hold 5 cards, with an unknown number of 1s, and zero 1s played
State A (Number of 1s played + clue tokens remaining= 8): The first player starts by cluing the second player their unplayed ones, if they have any, by number (State B). Otherwise, discard if possible, otherwise, give another random clue (State E).
State B (Number of 1s played + clue tokens remaining = 7): Play a 1. Goto State C.
State C (Number of 1s played + clue tokens remaining= 8): At this point, either the other player has duplicate ones, or they don't, out of their 4 remaining cards from their opening hand. If they do, clue the duplicates by color (Transition 5). Otherwise, go to state A.
State D (Transition 5)  (Number of 1s played + clue tokens remaining= 7): Discard a duplicate one. Go to state C.
State E (Number of 1s played + clue tokens remaining >=7): Rerun the above state machine, with the second player acting as the first player. Replace the transition to state E with a transition to the middle of the game. Note that when make that transition, the number of clue tokens + number of unplayed 1s is going to be at least 6. (I think it's going to be more like 8, or 7 if you play all the 1s out of the first players hand, but 6 is trivial to prove, since the first time through AD you only lost one clue token)

If you go an you replace all the references to "random clues" with references to "unclued twos by color", and add a caveat stating not to discard known/unique 2s, you'll should be able to get 10 no problem (since we have at least 6 clues when we hit the middle of the game, (which is more like 8 if we haven't played any ones), which we can use to store 2s, and still have a clue kicking around to force a one play if needed. I beleive the number of stored 2s + unplayed ones + unused tokens should be basically constant (over 2 turns), unless you're able to play those stored 2s. err, oops, this is just wrong  storing a 2 doesn't cost you a clue over two turns, because you can just discard after the 2 is stored (unless your hand is full). So storing twos doesn't actually cost anything in the middle of the game, except once. So even easier to prove .
After that, you can extend it further to care about 3s. At which point things become trickier, because you'll need to handle the case of who can see what and what they know to determine if you should clue by color or number. But it shouldn't be too bad, because essentially, a) only one player is going to be drawing cards at a time, b) You can constantly clue them by color as cards are drawn, and c) if their hand becomes full of interesting cards, you can give them a number clue to indicate that, which will signify that their hand is full, and remove all ambiguity as to which cards are 2s and 3s, giving them the info they need to become the clue giver. (You may still need to fully clue that last card they drew, though? Although you may be able to handle that by discarding a duplicate if they ever clue it in your hand, or just letting them handle it by knowing what it is when the deck runs out)

Jay Ackerman
United States Palm Beach FLORIDA

Hmmm, you know that rules prohibit discarding if you have 8 clue available, right?
Robb wrote: I think you might be able to guarantee a score of 13 in a 2er game, if that's your goal. At very least, 10 should be easy.
This requires an extremely artificial convention whose goal is to collect 1 two of each color, and one three of 4 colors. The basic form of play would look something along the lines of "If someone draws a 2 or 3 that isn't already held, clue it by giving a color clue. That card is now locked.
I think these starting hands cause a problem. Player 1: 2R 2B 2G 2W 2Y Player 2: 2R 2B 2G 2W 2Y but I think it can easily be fixed by allow the 2/3's to be clued by color or number. I think you have enough clues to fully clue every card.

Robb Effinger
United States Bothell Washington

SliceOfBread wrote: Hmmm, you know that rules prohibit discarding if you have 8 clue available, right? Robb wrote: I think you might be able to guarantee a score of 13 in a 2er game, if that's your goal. At very least, 10 should be easy.
This requires an extremely artificial convention whose goal is to collect 1 two of each color, and one three of 4 colors. The basic form of play would look something along the lines of "If someone draws a 2 or 3 that isn't already held, clue it by giving a color clue. That card is now locked. I think these starting hands cause a problem. Player 1: 2R 2B 2G 2W 2Y Player 2: 2R 2B 2G 2W 2Y but I think it can easily be fixed by allow the 2/3's to be clued by color or number. I think you have enough clues to fully clue every card.
That hand would be actually pretty awesome, as player 1 could give the clue of "these are all your twos", and player 2 could discard, signalling that he knows that he holds no unique 2s . You probably need player 2 to hold on to some of those twos, so both players have the ability to discard, but they should more than enough clues to work that out between them.
And ya, I'm aware that you can't discard when at max clue tokens. That shouldn't pose an issue  players can give clues about 4s or 5s, and at worst, give a clue to save a card, and then later, fully clue it as discardable. In the worst casedeck (all the 1s at the bottom) you're going to have more clues then you need.

Robb Effinger
United States Bothell Washington

And ya, probably players will occasionally clue by number. But in general, by color is going to be more efficient, since you're only caring about the 2s and 3s, it will only take one clue to differentiate between them, so you want to make that clue as late as possible.

Eric Fletcher
United States California

All 1s on the bottom > max score is either:
5 (for the 1s) + Players x Hand Size {10, 15, 16, 20}  1 (the discard) = {14, 19, 20, 24} based on cards available OR 14 + players (max number of available turns after a 1 has been drawn) = {16, 17, 18, 19} So the max score for {2, 3, 4, 5} players in the basic game is: {14, 17, 18, 19}
All 2s on the bottom > max score is either: 5 (for 1s) + 5 (for 2s) + hand size > not a constraint OR 9 + players (max turns after drawing a 2) + 5 (you played 1s) Which is exactly the same results as all 1s.
All of one color at the bottom: equivalent to "all the 1s" since that case already accounted for having 3 1s of the same color on the bottom.
Anything that is not a 1 near the bottom just gives 1 extra turn, and 1 extra card, which would increase the possible score.



It is not clear that that score is achievable though, but you are right, it is an upper bound.


