Removing duplicates from 'over partition by' - sql

Question

Im using over partition by clause to calculate peoples monthly figures.
A short example of my results:
Date Person Team Daily Figure Month To Date
24/09/17 James 2 50 200
24/09/17 James 2 50 200
25/09/17 James 2 50 200
25/09/17 James 2 50 200
I am calculating the monthly figure by partitioning the daily figure over the person and the month e.g.
CASE
WHEN
MONTH([DATE]) = MONTH(getdate())
THEN SUM(Figure)
OVER (PARTITION BY [Name],
MONTH(DATE]))
ELSE 0
END AS 'Month To Date'
The main issue im having is i only want to display todays daily figure but with the whole month value. I group the figure for each person and limit to today but to be able to group by person i need to SUM the month to date figure which obviously leaves me with
Date Person Team Daily Figure Month To Date
25/09/17 James 2 100 800
The daily figure is correct but its obviously summing the duplicated rows which gives me an incorrect figure.
The ideal result for today would be:
Date Person Team Daily Figure Month To Date
25/09/17 James 2 100 200
with no duplicated rows.
Has anyone got any advice on this? To basically remove duplicated partitioned rows so i just get the correct month to date figure grouped
UPDATE:
Each row is just an individual figure for each person. Nothing is grouped so each person could have atleast 20 separate rows for figures on each day

Answer 1

Something like this?
declare #t table (Date date, Person varchar(100), Team int, [Daily Figure] int);
insert into #t values
('20170924', 'James', 2, 50),
('20170924', 'James', 2, 50),
('20170925', 'James', 2, 50),
('20170925', 'James', 2, 50),
('20170801', 'James', 2, 80),
('20170802', 'James', 2, 80);
select Date, Person, Team, sum([Daily Figure]) as [Daily Figure],
sum(sum([Daily Figure])) over(partition by Person, Team, month(date)) as [month to date figure]
from #t
group by Date, Person, Team;