![]() Also the "2x" changes the average number of DS's a bit - so lets say, you have about 20 tries for 2 shuffles and 1.2 DS - discounting the 2 free tries from the shuffle that would be 180 stars for 1.2 DS or 150 stars/DS. Ignoring the "show two", you will find the DS only in about half the cases - that means that in 19 tries there are about 2 shuffles and one DS - the "show two" sligthly increases the number of DS's as it shows you which presents not to take as they do not have the DS or even as they include the shuffle or it might show you exactly where to find the DS. I generally like that approach, but it would have to be more expensive.Ĭurrently the average tries to find the shuffle would be about 9.5. Even the big money spenders aren't very likely to spend 2995 diamonds for a set of rewards they have no interest in.Įdit - my 6th and 7th reindeer rewards are both items I would be prepared to spend diamonds on, upgrades for the train and tactition tower, but not 2995 of them. Looking at the selection I have so far available from the reindeers I don't want any of the five items so I can quite easily see a situation where it's full without having a single prize I want. Same ten stars to delete one of the prizes. There shouldn't be a limit as players who spend diamonds for extra turns will doubtless be after particular daily prizes and so will need to be able to use the shuffle many many times on the day the prize they want is available. However, I'd be very happy to pay 10 stars to use it when I wanted without winning those stars back. Click to expand.With the way the shuffle function works now we get the 10 stars that we pay to unwrap it back as a prize, so it is effectively free now.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |