1357111317 said:
Lets say you were sidecounting aces and there were 3 decks played and there were 15 aces left. This means that 9.6% of the cards remaining to be played are aces. If you had no other knowledge of the cards that had been played how would you calculate your advantage or disadvantage at this point? For simplicities sake 6D H17 DAS RSA.
To get the following EVs I started with 6 decks with rules H17, DOA, split to 4 hands on both aces and non-aces, DAS for non-aces, NDAS for split aces, 1 card to split aces, no surrender, full peek
1. Removed 12 of each rank (1-9) and 48 tens (156 cards left = half shoe)
2. Added 3 aces
3. In order to keep shoe = half shoe and aces=15, 3 non-aces need to be removed
4. Broke non-aces into 3 groups (2-5), (6-9), (10,J,Q,K)
5. Removed 1 from each group
6. Since 10,J,Q,K are equivalent, there are 16 possibilites of removal of 1 card from each group
2-6-10, 3-6-10, 4-6-10, 5-6-10, 2-7-10, 3-7-10, 4-7-10, 5-7-10
2-8-10, 3-8-10, 4-8-10, 5-8-10, 2-9-10, 3-9-10, 4-9-10, 5-9-10
EVs for resulting 156 card slug with 15 aces with 1 ten and each differing sets of non-tens removed are (using basic strategy):
2-6-10 +.2227%
3-6-10 +.2439%
4-6-10 +.2965%
5-6-10 +.3307%
2-7-10 +.1444%
3-7-10 +.1668%
4-7-10 +.2173%
5-7-10 +.2528%
2-8-10 +.0543%
3-8-10 +.0744%
4-8-10 +.1255%
5-8-10 +.1613%
2-9-10 -.0074%
3-9-10 +.0126%
4-9-10 +.0648%
5-9-10 +.1003%
Full shoe EV (basic strategy): -.5514%
Average EV for above: +.1538%
Additional aces increases EV ~.7%
Actual compositions could vary wildly from above, some yielding +EV and some -EV but the more wild a composition is, the less likely it is to occur. The most likely composition is what is represented above.