Calculated member - Mode function

Question

Hello,
I need some help. I am quite new to MDX; I've created a cube and the measure needs to behave same as the MODE function in Excel (the following
link
describes what the function does). Apparently the only way of achieving this is by MDX (calling the MODE function directly from Excel is not an option for me). Practically the member should return a single value, that represents the price that appears
most times. There are 4 dimensions in my cube: dates, stores, products and client products. I need the following calculations:
1. MODE(price) for specific Date, Product and Client Product. There will be multiple lines returned for this set, as there are multiple stores.
2. MODE(price) for specific Store, Product and Client Product. There will be multiple lines returned for this set, as there are multiple dates.
I found the below code online, that I've adapted to work against my cube, currently the query returns correct results for the first part, Mode(price) for multiple stores. I can change it to do the same for date. 
What I need is to create 2 calculated measures using the last select and I am unsure how this can be achieved. Any help will be much appreciated or any other approach in achieving this.
WITH 
--Count how often each value appears
MEMBER [Measures].[ValueCount] AS 
SUM(  Union([Dim Stores].Store].CurrentMember.Level.Members,       {[Dim Stores].[Store].CurrentMember} AS CurrentStore),
IIF( ([Dim Stores].[Store].CurrentMember, [Measures].[Price]) = 
(CurrentStore.Item(0).Item(0), [Measures].[Price]) , 1, null))
--Only get the items that appear the most
SET [MaxModes] AS     
ORDER(FILTER(NONEMPTY([Dim Stores].[Store].Members, {[Measures].[Price]}),    
[Measures].[ValueCount] = MAX(NONEMPTY([Dim Stores].[Store].[Store].Members, [Measures].[Price]),    
[Measures].[ValueCount])), [Measures].[Price], ASC)
SELECT   {[Measures].[Price]} on 0,        
[MaxModes]       
--Filter out the duplicates        
HAVING [MaxModes].CurrentOrdinal = 0 OR [Measures].[Price] <>             ([Measures].[Price], [MaxModes].Item([MaxModes].CurrentOrdinal - 2)) ON 1
FROM [old_Prices_v2]
WHERE {[Dim Date].[Date].&[2013-06-23T00:00:00]}*{[Dim Client Products].[Client Product].&[13]}*{[Dim Products].[Product].&[551]}

Answer 1

I'm describing below what exactly is required:
The cube has 4 dimensions (Date, Store, Product, Client Product) and one measure (Price). I need a measure to behave same as the MODE function in
Excel (value that occurs most often). Practically the member should return a single value, that represents the price that appears most times. 
While browsing the cube I need, for any combination of dimensions/hierarchies, to get the MODE of the price at its lowest grain (that is Date, Store,
Product, Client Product). No need to aggregate the measure in any way. 
For example if someone looks at a specific combination of Product & Client Product, the calculated measure would show the price that appears most
times against that Product+Client Product, in any of the stores and in any of the dates.
Example:
Date
Store Product
Client Product
Price
05-Feb-2013
Store1 Prod1
ClProd1 50
05-Feb-2013
Store1 Prod1
ClProd1 60
06-Feb-2013
Store2 Prod1
ClProd1 60
06-Feb-2013
Store2 Prod1
ClProd1 70
05-Feb-2013
Store1 Prod1
ClProd2 80
05-Feb-2013
Store1 Prod1
ClProd2 50
06-Feb-2013
Store2 Prod1
ClProd2 50
06-Feb-2013
Store2 Prod1
ClProd2 70
Case1: If one looks at Prod1 & ClProd1 then the MODE(Price) is 60 as there are 4 rows to be aggregated and the value that repeats most is 60
Case2: If one looks at Prod1 & ClProd1 & Store1 then the MODE(Price) is either 50 or 60 as there are 2 rows to be aggregated and both 50 and
60 repeat 1 time. I this case any of them can be returned (act like TOP 1 in TSQL)
Case3: If one looks at Prod1 & Store1 then the MODE(Price) is 50 as there are 4 rows to be aggregated and the value that repeats most is 50
Case3: If one looks at Prod1 & 06-Feb-2013 then the MODE(Price) is 70 as there are 4 rows to be aggregated and the value that repeats most is 70.
Any guidance will be much appreciated!
Thank you in advance.
Mihai