How much your turn hands the other player
Your data
The one rule behind all of this is that you keep playing while the counter is still on your side, and the turn passes the instant it crosses over. Whatever it crossed by is what the other player starts their turn holding.
Nothing here is tied to one game: any game with a shared counter that hands the turn over the moment it tips works out the same way.
Results
| Plays that fit, in your order | — |
| What the other player starts with, in your order | — |
| Plays that fit, cheapest first | — |
| What the other player starts with, cheapest first | — |
| The least you could ever hand over | — |
| Plays that fit while handing over that least | — |
Your turn, play by play
| Play number | What it costs | Where the counter lands | Whose turn it is after |
|---|
Read down the last column and the rule stops being frightening: nothing is being taken from you turn by turn, you are simply choosing how far past zero to go, and the amount you go past is the amount you are handing over. It is the last play of the turn that sets it, not the total you spent.
That is why the order matters so much for something so simple. The same three or four plays, rearranged, can hand over a single point or half a dozen, and the plays you never got to make are still sitting in your hand either way.
The uncomfortable part is in the last four rows, because the two things you want are usually different orders. Cheapest first always squeezes the most plays out of a turn, and it very often hands over more than you had to. Getting one more play down has a price, this page quotes it, and whether that price is worth paying depends on what the play does, which is the half of the decision the arithmetic will never see.
What decides how much the other player starts with?
Only the last play of your turn. You keep playing while the counter is on your side, and the moment it crosses, the turn passes and whatever it crossed by is what they get.
The total you spent does not appear anywhere in that. Four cheap plays that land you one past zero hand over one; a single expensive play that lands you six past zero hands over six.
Does the order of my plays really change anything?
It changes both of the things you care about, and usually not in the same direction. Cheapest first always fits the most plays into a turn, and it very often gives away more than you needed to.
With seven on the counter and plays costing one, two, three and four, cheapest first gets all four down and hands over three. Giving up one play brings the handover down to one.
| Counter | Costs | Plays that fit | Handed over |
|---|---|---|---|
| 7 | 1, 2, 3, 4 | 4 | 3 |
| 7 | 1, 2, 3, 4 | 3 | 1 |
| 4 | 3, 3, 1 | 3 | 3 |
| 4 | 3, 3, 1 | 2 | 2 |
Is cheapest first the right habit then?
It is the right habit for squeezing plays out of a turn, and that was checked rather than assumed: across twenty thousand random hands it never lost to any other order on the number of plays.
It is not the right habit for keeping the handover small, and it lost on that in about four hands out of ten. Which of the two you want is the part of the decision the arithmetic cannot make for you.
What does the page assume?
That every play on your list is legal when its turn comes, that costs are fixed, and that nothing gives the counter back to you part way through the turn.
Anything that moves the counter the other way, or that changes what a play costs once the turn has started, sits outside it.
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