So you want an algorithm which returns a random number N between 1 and the total number of tickets (inclusive), and then tells you who holds the Nth ticket in the dataset?

You could generate a number for each name which is the sum total number of tickets for themselves plus everyone else before them in the list, and then find the name corresponding to the maximum individual-total number which was still smaller than N.

Example: You have four names: Alice, Bob, Charlie, and Dani - I'm going to assume you put these in Column A. They have 8,2,5, and 3 tickets respectively (for a sum total of 18 tickets). I'm going to assume these are in Column B, and their totals are in column C. Their matrix looks like this:

(You can fill column C with "=SUM($B$1:INDIRECT("rc[-1]",0))" if you like; that'll populate it.)

The random roll from 1-18 comes up with "11". Let's see how that plays out:

In D1, put:

=INDEX(A:A,MATCH(1,INDEX(C:C>=11,0),))

This will find the first value in the "C" column which is greater than or equal to 11, which is "15" in row 3. Then it will match that to the corresponding value in the "A" column and spit the correct name out: Charlie.

This requires a little bit of probabilistic reasoning.

As the comments have suggested, given a length n of a row, the probability that this row will win the lottery is comb(n, 15) / comb(25, 15). Therefore, we can construct an array of probabilities, p, for rows of length 15-20:

p = np.array([comb(n, 15) / comb(25,15) for n in range(15, 21)])

Assuming that the rows are indeed randomly generated and independent from one another, we can calculate the probability that the sheet has a row that wins the lottery:

Therefore, we need to count the number of rows of each length in the data sheet (how many rows with 15 entries, how many with 16, etc). Here is a function that can do that:

def CountRowLength(df):
    """Returns a dictionary with keys corresponding to row length and values corresponding to the number of times they occur"""
    answer = dict()
    for _, row in df.iterrows():
        rowLength = len(row.dropna().tolist())
        if rowLength in answer.keys():
            answer[rowLength] += 1
        else:
            answer[rowLength] = 1
    return answer

Therefore, all that remains is to convert this dictionary to an array/list and plug it into the formula from earlier:

# Convert the dictionary to a list
dictionary = CountRowLength(df)
lengths = [dictionary[i] if i in dictionary.keys() else 0 for i in range(15,21)]

# The formula from earlier
total_probability = 1 - np.product([(1-p[i])**lengths[i] for i in range(len(p))])
print(f"The probability that a row from this sheet has a lottery win is {total_probability}")

Then you can take the reciprocal to find the odds.

Try this:

=IF($J$1="All",COUNTIF($B$2:$H$5,J2),SUMPRODUCT(--(INDEX(($A$2:$A$5=$J$1)*$B$2:$H$5,)=J2)))

All does a simple COUNTIF.

While Wed/Sat iterates through the columns to find the correct rows and values to return a count.

EDIT: Found a shorter formula.