Personally I find it a very risky strategy, you have no idea what your opponent has or whether they can't get out. Until you meld you can't pick up from the discard pile.
In a multi-point game it is even more risky

I thought that you had to meld with your own tiles first?

So do I until I realise that my opponent is never going to play anything...at which point I've almost always already lost because they'll just sit back until they can meld everything using what I've already put down

surely it's a personal choice how you play, why wait until all the tiles are dealt if you don't like that style?
I always meld as soon as I can, irrelevant of how my opponent choices to play

it seems pointless to me to play in that style where you do nothing until all the tiles are dealt, and I'd rather not play in those games...but each to their own

Seems a pointless variation to me

Please can we find a way to either create a Full Deal variation of Rummikub, or have some kind of option on the game that both players are just going to wait until all tiles are dealt before playing anything. It’s a huge waste of time if both players are just going to spend 2 weeks picking tiles

Personally I’d rather it was a separate variation and something was done in the main Rummikub game to stop players doing it. I just don’t see the skill in waiting for all the tiles to be dealt then dropping 40 tiles at once and finishing the game in one hit…you might as well just flip a coin!

I think that would require a complicated simulation for a multiplayer game. It would be very interesting though. Modeling of discards vs. cards kept would be a problem I think best solved through machine learning.

I meant a game where no player is able to meld

I don't know how you define an unwinnable game. An intial card distribution that can't be made to win with cards from the stock?

I wonder if it's possible for a game to have all players with an unwinnable hand and unable to meld? Probably more likely when more than 2 players in a game

Getting in here late, but the easiest (?!) way to find the odds in such complex situations, which would defy easy expression in enumerative combinatorics, would be to run a Monte Carlo simulation. I have done this a lot of times for various games and if you run maybe ten million hands you'll get something pretty close to the correct odds. "All" it requires is some computer programming knowledge and experience. (No, I'm not volunteering to run this one.)

That would apply to the odds of getting a hand with no intial meld. Getting a totally unwinnable hand requires playing out entire games, and depends on the opponent's play too, so the problem is too broad to solve except with sophisticated machine learning techniques.

Thanks but those are the rules for a game of straight rummikub rather than American Rummikub, the latter requiring a lowewr number to meld and a higher chance of playable combinations. That said it would still be a lot of work to consider all possible combinations--plus combinations of combinations--to meld and then come up with a formula to calcu late the odds. Very interesting though--thanks for that info!

no idea, but eager to find out. I suspect it's quite a complex formula to work it out given that;

• Rummikub is a Rummy-style game played with tiles. It's basically 2 decks of cards (colors instead of suits) plus 2 jokers, for a total of 106 tiles.
• The game starts with each player drawing 14 tiles.
• total possible hands is 10^17. Choose 14) = 106! / ((106-14)! * 14!) =~ 10^17.

some of the odds for various things are posted here, as I said rather complex
https://math.stackexchange.com/questions/3710304/odds-of-an-unwinnable-game-of-rummikub

Thanks but there's so many possible playable combos that lead to 21 or more that it would be rather difficult to do--all the more so with the general uncertainty over how often one's opponent will go out before one is able to meld (it would have to be some sort of average). I was just hoping that such info might come from the game creator, whoever that might be.

You will probably need to work out the odds of being dealt 14 playable tiles out of the initial pool of tiles.

OK--but again, if there is anyone who can explain this matter to me, please respond--thanks!

Ok, thx 🙂

No offense taken. That said, Please understand that my question was not directed at you or anyone else personally. Again I merely seek someone who can help me understand this paerticular aspect of the odds of this game, and I would deeply appreciate anyone who would asasist me in doing so--thanks!

I am not laughing at your question just the responses because I too have played games where I couldn’t make a move. I stumbled upon the fact that you could remove any tile you would like by looking at prior moves because there were too many rules to read.

Of course we've all experienced this to a certain extent but this is still a serious question--if anyone can provide the answer I believe it could help all of us understand and/or help us calculate the odds, thanks!

Thanks for the laugh. I needed that!

I don't know the odds but, I've been there, Done that.

Does anyone know the odds of (1) being dealt an opening hand where one cannot meld, and
(2) being dealt a hand where you lose by never being able to meld?