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[methods of rationality]

Here is a quote from the scheme-article
"In Big Money type decks which only buy a few actions, Scheme can, essentially, act like a second copy of whatever flavor of terminal action you're using. By getting to top deck your terminal for use in consecutive hands, you reduce your collision chance to zero (or closer to 0 when you have blind draw). This, however, comes at a price. Whenever you draw your Scheme after your terminal, you only get to play the terminal once that reshuffle. Had the scheme been an actual second copy of the action, you'd have gotten two plays. Over the course of a game, the double terminal deck gets more plays of the terminal action than the scheme/terminal deck. So typically, you favor a second terminal over a scheme."
So basically, the effect of buying a scheme is that you can play your terminal more often. My question is, can we quantify this a bit more?
The simplest cases (which don't actually occur in real life) are when your total number of real cards in deck (i.e. excluding virtual cards such as schemes) is a multiple of 5 and your deck size doesn't change.
For example, If you have 15 cards in deck, 1 terminal and no schemes, you will play your terminal once before the reshuffle.
No, suppose you have 16 cards, 1 terminal and one scheme. You will have 3 hands till the reshuffle. If you draw the terminal in the first hand, you will play twice if and only if you also draw the scheme in the first hand, if you draw it in the second hand you will play it twice if and only if you draw the scheme in the first or second hands, if you draw it in the third hand you will play it only once. So you have a (aproximatly) (1/3)*(1/3)+(1/3)*(2/3) = 3/9 chance of playing the terminal twice. Thus the expected number of times you will play the terminal before the reshuffle is equal to 14/9.
The next case would be where you have 21 cards, 1 terminal and 1 scheme. If you...
1. Draw the terminal in the first hand, you will play it twice if and only if you also draw the scheme in the first hand (pr=1/4)
2. Draw it in the second hand, you will play it twice if and only if you draw the scheme in hands 1 or 2 (pr=1/2)
3. Draw it in the third hand, you will play it twice if and only if you draw the scheme in hands 1, 2 or 3 (pr= 3/4)
4. Draw it in the forth hand you will play it only once.
So the chances of playing it twice now equal (1/4)*(1/4)+(1/4)*(1/2)+(1/4)*(3/4)= 6/16. Thus the expected number of times you will play the terminal before the reshuffle is equal to 22/16.
Of course an actual game is much more complicated, and this leads to my question (which I suspect will only be answered with a simulator):
Let n and m be any 2 natural numbers, n>m-1. Consider a game of dominion where on turn 1 you buy a terminal action, and on turns 2 through m you buy scheme (if m=1 you don't buy any schemes). On all other turns you buy copper. What is the expected number of times will you have played the terminal action by turn n?

[methods of rationality]

Another interesting corollary of this is that you don't always want to return the terminal to your deck with the scheme. On the last turn before the reshuffle, not only will returning the terminal not help you play it twice before this reshuffle, but it will also greatly decrease your chances of playing it twice next reshuffle. This also means that the above question has 2 answers - 1 for when you always return the terminal, and a second for when you only return the terminal if it will increase your expected number of total plays.

[Cadence20]

Another interesting corollary of this is that you don't always want to return the terminal to your deck with the scheme. On the last turn before the reshuffle, not only will returning the terminal not help you play it twice before this reshuffle, but it will also greatly decrease your chances of playing it twice next reshuffle. This also means that the above question has 2 answers - 1 for when you always return the terminal, and a second for when you only return the terminal if it will increase your expected number of total plays.

This is sometimes but not always the case - if the deck is sufficiently small you would probably want to return the terminal nonetheless, as there is a significant chance of getting it on the reshuffle of the next one anyway (eg. at 14 cards)

Another instance is if you have a deck size divisible by 5 (eg. 15) - both the scheme and terminal fall to the last five cards. Probably want to play the terminal but not the scheme (not sure if the above mentioned rule would also affect this)

[jomini]

I might be misreading, but should you not also look at the chance that you say draw scheme on hand 1, top deck it, draw the terminal in hand 2 (play scheme -> terminal), top deck the terminal, and then play the terminal during hand 3? Likewise, if you play the terminal first THIS shuffle, then when you find scheme you can keep top decking until is ASSURED to hit during the next shuffle (allowing you to play the terminal twice THAT shuffle)?

[werothegreat]

I really only use Scheme for engine parts, anyway.  None of this hoity toity math stuff.

[methods of rationality]

I might be misreading, but should you not also look at the chance that you say draw scheme on hand 1, top deck it, draw the terminal in hand 2 (play scheme -> terminal), top deck the terminal, and then play the terminal during hand 3?

I did, that is called drawing the terminal in hand 2 - play it twice if and only if I drew scheme in hands 1 or 2

[jomini]

You need to look at more than one shuffle then. If you draw S (scheme) in H2 and T (terminal) in H1, yes you only play T once this shuffle, however you have 100% odds of drawing S in H1 the next shuffle.

S is equally likely in each hand at 5/16 and 1/16 for the split hand. T is 5/16 for each hand.
S1T1 - play T 2x - 9.77%
S1T2 - play T 2x - 9.77%
S1T3 - play T 1x, S1 is 100% odds next shuffle - 9.77%
S2T1 - play T 1x, S1 is 100% next shuffle - 9.77%
S2T2 - play T 2x - 9.77%
S2T3 - play T 1x - S1 is 100% odds next shuffle - 9.77%
S3T1 - play T 1x - S1 is 100% odds next shuffle - 9.77%
S3T2 - play T 1x - S1 is 100% odds next shuffle - 9.77%
S3T3 - play T 1x - S1 is 100% odds next shuffle - 9.77%
S4 - play T 1x - S1 is 100% odds next shuffle - 6.25%

Totals - play T 2x this shuffle - 29.31%; S1 assured next shuffle 70.69

If you missed this shuffle then next shuffle you get:
S1T1 - play T 2x - 33.33%
S1T2 - play T 2x - 33.33%
S1T3 - play T 1x, S1 is 100% odds next shuffle - 33.33%

So the odds that you will get T 3x or 4x over two shuffles is 76.63%. 8.59% of the time, you will play the terminal 4 times.

Now what if you had a second terminal instead?
Well the odds of drawing both T together are 28.57% a shuffle.

For the first shuffle you have a 71.43% chance of play T 2x.
For the second shuffle you have a 51.02% chance of playing T 4x, a 40.82% chance of playing T 3x, and 8.16 % chance of playing T 2x.

Note: I'm assuming for all calculations that playing the terminal multiple times is more valuable than playing it sooner in shuffle (e.g. monument) so scheme gets top decked if T is going to be shuffled in anyways. A better analysis would weight playing the terminal sooner against playing it more times (potentially).