# Do the vast majority of Rat-a-tat Cat rounds end very quickly?

My son was gifted this game today. Given that we play a variety of complex games in our family (including Bridge), we're very surprised how much we like it and the various skills it brings into play (probabilistic assessment, poker face, memory, etc.). We're also surprised at how rounds usually end so quickly, as the rules provide for getting to the end of the deck. We have yet to go half way through a deck.

Is this typical, or are we missing something to the strategy?

Our probabilistic reasoning:

A deck consists of 4 each of 0 through 8, 9 9's, and 3 each of the power cards. So an average dealt hand has 20 points, with 10 points visible (I'm counting the power cards as 5's given that you'll have to draw randomly from the deck at the end for each of these).

A typical starting hand might have a low number and a high number known upon deal. If your first draw is a low number, you replace your high numbered card with that one and then you call for the end of a round. Why? Because you have, say, 5 points on your ends and an assumed 10 points (it's random) in the middle, for perhaps 15 total.

You have no knowledge of other hands but on average each will be 20, and with one more draw will be reduced by 4 points to 16 (on average). Not a huge edge, but definitely an edge. And this only widens as the sum of your two end cards drops further below 5.

Am I missing something? Or is it correct to call for the end of the round if the sum of your two end cards is 5 or fewer points after you draw the first card of the round?

The reasoning gets a little more complicated for subsequent rounds as other players' hands improve, but similarly if you know that 2 of your cards are very low, or better yet 3 within the first few rounds, then you call for an end, right?

-

Yes, you should call for and end or knock-out as it is commonly called.

Your reasoning is sound. With 54 cards in the deck, 9 of which are special non-point cards, and four cards each of 0-8 plus nine 9s, the average hand is 20 points. There in no way of knowing what your opponents hands are worth, except in the rare case that you draw 1 of 3 peek cards and then you draw 1 of 3 swap cards. (In the children's variant, all players two outermost cards cards are revealed).

Since you cannot know what your opponents have, it is best to call for the end of the round when you know that the sum of 2 of your cards is less than 7, or 3 of your cards are less than 12. (Edit: My initial calculations didn't take into account that your opponents get to look at one more card, and reduce there score as well.)

The only thing that might change this strategy slightly (and I am not certain if it does change the optimal strategy), is as the number of players increases the chances that one or more of your opponents draws a Swap card during the knockout round also increases. If your opponents know that you are using this strategy of only knocking out when you have < 7, or < 12 when you replace a hidden card with a drawn card, they will be sure to Swap cards with you (most likely they will swap with the card that you drew to replace your other card). They will know that they have a better chance of getting a '4' or less from you, then from drawing from the deck.

As for conclusion that each player would take less than six turns on average, I am not certain. I haven't worked out the math for it, but the number of turns on average would differ depending upon the number of players. If there are more players playing with the < 10 points sum total of known cards, then the chances that at least one of those players has a low hand increases.

This analysis was only performed for the first turn. I suppose you could calculate the AVG number of points that you would reduce your score per turn, and use that to figure out when to go out on later rounds. For the first round, with only looking at 3 cards, you will reduce your points on average by 2.63 points. This result was arrived at by looking at a million random 3 cards from the deck (with wilds removed, so the odds might be slightly different), and choosing the lowest 2 cards to keep for a single player. I did not examine what happens as the number of players increase, but my guess would be that as the number of players increase, the chances that at least one other player is able to reduce their score by more than the average increases (i.e. with two players, it is probable that one player reduces their score by more than 2.63 and the other by less than 2.63. For three players or more, the odds that someone is able to drop a '9' or other high number for a '3' or other low number increases, and that lucky person will have on average less than your total.

-
Your answer cites <10 while I had calculated <6 (based on the idea that a typical last draw by opponents will drop their score of 20 by 4 (i.e. a 9 gets replaced by 5). Can you elaborate as to why 10 as opposed to 6? – Joe Golton Mar 13 '12 at 13:43
@JoeGolton, I did make a mistake in my calculation. If we only look at the deck without the wild cards, then I think on average your opponents will be able to reduce their score by 2.63 points. I might work on the more correct answer (using wilds), but not soon. The Peek card is worse than useless during a KO round (you don't even get to swap a card with the deck), the swap card is probably easy to handle programically always stealing a card you swapped with the deck, but the Draw 2 card is more complicated (you have to calculate the odds that giving up your first draw might improve with 2nd. – user1873 Mar 13 '12 at 16:59
Thanks for correcting. Interesting that my guess of "4" was so far off from your 2.63 derived from simulation. BTW, by my interpretation of the rules, you can swap the peek card into your hand if you want. So a "peek" card is therefore not "worse than useless." You can replace a 9 with it, which in turn leads to a random draw, whose average will be around 5. Not sure if that means the 2.63 increases slightly. – Joe Golton Mar 13 '12 at 17:08
@JoeGolton, I meant that during the KO round drawing a peek card is useless, the rules say "When you draw a Peek card, show it and then peek at any one of your cards." You clearly cannot choose to replace one of your outermost known '9' cards with a drawn Peek card. An alert opponent with good memory would remember where you placed that drawn card and accuse you of cheating. – user1873 Mar 13 '12 at 17:31
I just carefully reread the rules and you are correct - I was doing it wrong. For such a simple game, it is remarkable how poorly the rules are written. I've now read through them at least 4 times and every time I've discovered things I was doing incorrectly. – Joe Golton Mar 14 '12 at 1:49