Point Salad: Valuing Cards I

 

Point Salad.  Three cards, cabbages only.

Card A:  two points per cabbage.

Card B:  five points per two cabbages.  No rounding or partial credit.

Card C:  eight points per three cabbages.  No rounding or partial credit.

If you had your choice of one of these, which would you take?  Maybe not a super-difficult or interesting question, but we can have a little bit of fun with it.  Answer:  

If you expect to have three or fewer cabbages by the end, take card A.

If you expect to get more than 17 cabbages, take C.

Otherwise, take B.

And all right, there are only 18 cabbage cards in the game, which doesn't say much for card C.  And don't get me wrong -- there's nothing wrong with C, and it's a good play if you are long on cabbages.  But if you are forced to take exactly one of these three, go with B.

Reasoning

The math is a little annoying, but here's a rough way of looking at it.  And this isn't the most clear explanation, so if it doesn't make sense to you, my apologies, and you can just skip over it.

If you allow fractions, the cards score like this:

A:  2 / cab

B:  2 and a half / cab

C:  2 and two-thirds / cab

So far so good.  Comparing A and B, because you need whole multiples, the scoring for A can move ahead because for odd numbers of cabbages, A can score an extra card while B is waiting to reach that next multiple of 2.  So we need to figure out when we have enough cabbages so that the extra 1/2 point per cabbage makes up for a missing card.  Algebraically, we're trying to solve

1/2 * n = 2

which works out to n = 4.

To find the crossover point between B and C, it's a similar calculation.  Except now, C is only buying you an extra 1/6 points per card, and you have to get enough extra points to cover potentially two cabbage cards, or five points, while you are waiting to hit that multiple of three.  The formula becomes:

1/6 * n = 5

which works out to n = 30.

These numbers show the crossover points where you are guaranteed that the larger card wins out.  In reality, the advantage goes back and forth with the exact number of cabbages.  If you have three, then C beats B which beats A.  For four, B beats A beats C.  For five, B ties A beats C.  And so it goes.  But on average, B is going to be a better bet until you reach 17 cards, when C starts to take over.

Data

Here's a spreadsheet showing how many points you get for having however many cabbages, under the different scoring cards:  https://docs.google.com/spreadsheets/d/1Hb8oNnZ29ZwKpESkDVMxeGEW0DozEKl83MHCXbWyf-s

Not gonna lie, it's pretty boring.

Further Questions

I only have more questions about everything, but in particular:

1.  How can you estimate how many cabbages (or other specific vegetable) are you going to have the opportunity to draw, given where you are in the game?

Certainly, it depends on whether there are other cabbageheads in the game trying to swipe your harvest.  The math isn't hard, but this deserves its own post.  To be continued.

Strategy Takeaways

The five points-per-two card is generally stronger than the two-per-one and the eight-per-three.

Further Information

I started playing Point Salad after hearing about it on one of my favorite podcasts, Decision Space, and listening to their community talk about it on their Discord channel.  You can listen to the episode here.

Puns

Moving the wordplay to an appendix so as not to groan out people who don't care for this sort of thing.  Anyways, I hope this information will help you get a head in the game, and I also hope you slaw what I just did there.  Feel free to comment, as two heads are better than one.