Chance of 5/2 or 2/5 is 1 in 6. See, e.g., http://mathlaoshi.com/2011/10/13/dominion-starting-hand-probabilities/.
Yep, I mildly screwed up the math. Did exactly what was at the above link to get 8.3%, but didn't double it. 8.3% is the chance of 5 coppers on turn 1, and also the chance of 5 coppers on turn 2, so the chance of either/or is indeed 16.6% = 1 in 6.
Sure, it can definitely be right to choose a strategy that you think is worse 90% of the time, if the amount it's worse by is not that high. Say I'm deciding between Trader or Monument. Maybe Monument wins against opponent's Mountebank 5% of the time, and beats the opponent's 4/3 55% of the time; meanwhile, Trader wins against Mountebank 30% of the time, and wins against a 4/3 54% of the time. I'm happy to give up that 1% chance of victory in the most-likely scenario in exchange for drastically improving my chances in case my opponent does have Mountebank.
This may be tangential for the Salvager thread, but it's more fun than the argument a few pages back, so I'll run with it until someone asks us to stop
Let's crunch some win probabilities for the scenario you described: Suppose card A gives you probability A
2 to beat 2/5, and A
3 to beat 3/4. Card B likewise gives probabilities B
2 and B
3. Since your opponent gets 2/5 in 1 game out of 6, and we're assuming you have 3/4, the overall win rates with A and B would be:
P
win(A) = 1/6 * A
2 + 5/6 * A
3P
win(B) = 1/6 * B
2 + 5/6 * B
3So we can calculate what the differences have to be in order to justify taking a card that does worse against 3/4 but better against 2/5. If that card is A, we want to know when P
win(A) > P
win(B). Multiply through by 6 and we have:
A
2 + 5 * A
3 > B
2 + 5 * B
3Now we bring in the differences. Let:
D
2 = A
2 - B
2 (since A
2 > B
2)
D
3 = B
3 - A
3 (since B
3 > A
3)
Substitute into our inequality and do a little more algebra and we get D
2 > 5 * D
3. For every percentage point you give up against 3/4, you need to gain 5 against 2/5 to break even, more than 5 to come out ahead. How often does that happen? I'm not sure. I'm a little dubious that two cards which are usually near-even would change your P
win by 0.25 against a specific opposing opening (the number from your example), but 0.05 sounds more reasonable, so maybe it happens more than I think. The hypothesis might even be testable if we wanted to data-mine game logs, but I think I'll stop short of writing any code for this
As for playing 3 vs 4 Coppers: this is only true if your opponent presses the "easy" Play Treasures button - for the first shuffle, at least, I almost always play my treasures one at a time, because the possibility that knowledge of my hand will affect my opponent's play is highest then.
If my opponent does that, I can see it and suspect mind games are afoot, but I still very rarely let it affect my play. Pretty much all the highest rated players that I've played against use the easy button on their first turns, so I suspect that successful trickery doesn't grant a big advantage. Therefore I play the odds as well as I can figure them, regardless of my opponent's copper antics.