I have been doing some calculations for outer chain strategy.
Suppose you have set up an outer chain of 3x6s in the first 4 racks, which is a pretty standard play, if you follow Dynasty strat.
In Rack 5 you get one 6, which is easy to place.
Now in Rack 6 you get another 6, which you will probably place like this (POSITION 1):
Today I want to suggest another strategy, which is to place like this (POSITION 2):
The first placement is the natural placement and allows you to create a full chain if 2x6s come in the racks 7-8.
The second placement is a bit risky and could possibly have you end up with a broken 6-chain as a sacrifice to maximize possibility to get a full chain of 6s.
If only one more 6 arrives in rack 7-8, you place like this:
And you hope to get a 6 in rack 9 and complete a 5-zone chain, while "normal" strategy would have created a 6-chain covering only 4 zones.
Let´s compare POSITION 1 and 2.
If rack 7 and 8 produce 2x6s or more (51.55%), both strategies will win and 5-zone the outer chain.
If rack 7 and 8 produce 1x6 (32.30%), POSITION 1 will yield a 4-chain of 6s. POSITION 2 will depend on rack 9 to produce at least one 6, make a full chain and beat POSITION 1.
Probability for at least 1x6 in rack 9 = 59.81%
So:
POSITION 1 will win 0.3230*0.4019= 12.98%
POSITION 2 will win 0.3230*0.5981=19.32%
If rack 7 and 8 produces 0x6 (16.15%), POSITION 1 will yield a 4-chain of 6s, while POSITION 2 will depend on rack 9 to achieve the same result.
POSITION 1 will win 0.1615*0.4019= 6.49%
POSITION 2 will tie 0.1615*0.5981=9.66%
So:
POSITION 1 will win 12.98+6.49=19.47%
POSITION 2 will win 19.32%
And there is a tie 51.55+9.66=61.21%
The strategies are practically equivalent. If the second 6 appears in rack 6, you can follow either strategy.
Actually, in the daily challenge, you should choose POSITION 2, because you will have a better chance to outplay the field who will probably choose for POSITION 1.
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Now suppose a slightly different scenario: the second 6 comes in rack 7, not in rack 6. Still we compare both strategies, but now with only two racks to come.
If rack 8 produces 2 or more 6s, there will be a tie = 19.62%
If rack 8 produces 1x6 (40.19%):
POSITION 1 will win if 0x6 appear in rack 9 = 0.4019*0.4019=16.15%
POSITION 2 will win if any 6 appears in rack 9 = 0.4019*0.5981=24.04%
If rack produces 0x6 (40.19%)
POSITION 1 will win if 0x6 appear in rack 9 = 0.4019*0.4019=16.15%
POSITION 2 will tie if any 6 appears in rack 9 = 0.4019*0.5981=24.04%
So in this scenario:
POSITION 1 will win = 32.30%
POSITION 2 will win = 24.04%
There will be a tie = 19.62+24.04=43.66%
So in this scenario, the risky strategy is inferior (but still very playable if you aim for a DC win). If the second 6 appears in rack 7, you are better of following standard strategy.
Of course all this is looking at the position in a vacuum: other things might influence the probabilities (bonus, other chains etc).
For example if your board after 5 racks looks like this:
And you get a 5 and a 6 in rack 7, you might even consider placing like this:
And 5-zone the 5s while having decent possibilities to 5-zone the 6s as well.