Question

1. In another problem, I had four movies D, C, S and R and a preference ordering for three people who were voting on which one to go see. Let’s use this one instead:

person 1: D R C S

person 2: S R D C

person 3: C S D R

(a) Suppose the voting method was to choose between the pairs D, C versus S, R, and then choose between whichever pair wins. Who would win?

(b) Suppose instead they started with a choice between D,S versus C,R. Who would win?

(c) Suppose they started the third way, a choice between D, R and C, S. Who would win?

(d) When we make these predictions about who would win, are we assuming that the players know each others’ preferences, or are we assuming only that they know their own?

(e) Which movie would they go to if they used the Borda count method? (There might be ties, of course. Then they’d flip a coin or roll dice among the highest scorers.)

Answer 1

a. If voting method is to choose between DC V/s SR as pair,

	DC	SR
Person 1	1
Person 2		1
Person 3	1

As stated above, DC will win the voting as either D or C are preferred choice for person 1 & 2. Following the majority wins principle, DC will be the choice.

Now, to choose between D & C, below is the table which explains D will be the primary choice. If we remove S & R movies from equation and just notice the preference of persons between D & C; it is evident that Person 1 & 2 will prefer watching D over C.

	D	C
Person 1	1
Person 2	1
Person 3		1

b. if voting is between DS & CR

	DS	CR
Person 1	1
Person 2		1
Person 3		1

On the other hand, if voting between DS v/s CR takes place, CR will win as R is the second most preferred choice for Person 1 & 2 whereas C is the topmost choice of person 3.

c. If voting method is to choose between DR V/s CS,

	DR	CS
Person 1	1
Person 2	1
Person 3		1

However, as noticed in the table above, even when choice is between DR and CS, 2 out of 3 persons will win choose DR over CS, as per their preferences.

d. When we make these predictions about who would win, we are assuming that the players only know their own choices, and not others. When we are collecting data of preferences and probabilities of which movie to be seen, the data is collected from Sample space. And on of the important rules of calculating probability is random sample which are mutually exclusive from each other. If person 1, 2 and 3 knows each other choices, then this exercise will be moot.

e. Borda count method:

Borda count method is a single method election voting system i.e. preferential in nature. The persons below will list their choices in order of preference. Votes can be counted by giving each candidate a number of points equal to the number of candidates (n) ranked lower than them, so that a candidate receives n − 1 point for a first preference, n − 2 for a second, and so on, with zero points for being ranked last (or left unranked) -- as marked in "Borda counts (X)" column below

Movies choice	1	2	3	Borda counts(X)	Count by 1	Count by 2	Count 3
1ST	D	S	C	3	D:3	S:3	C:3
2ND	R	R	S	2	R:2	R:2	S:2
3RD	C	D	D	1	C:1	D:1	D:1
4th	S	C	R	0	S:0	C:0	R:0

Hence total votes goes for 4 choices are:

D = D3+D1+D1+D0 = D5

S = S3+S2 +S0 = S5

C = C3+C1 = C4

R = R2+R2+R0= R4

As, we can voting has been won by D and S movies with 5 counts, whereas C & R had 4 counts.

Since, there are two highest scorers i.e. movies D & S, voters will flip a coin to decide which movies should they go for.

Let's assume voters will flip a coin once (n) and the outcomes(x) can 2 i.e. either D or S;

i.e. n=1, x=2 , probability is = n/x = 1/2 = 0.5 or 50%.