[Guest]

[Hi All. This is my first puzzle on this forum. Hope you all'll enjoy it.]

---START---

The following data is available for a general election held for the post for the Mayor in a small town in Yugoslavia.

• 
  
  
• Contestants: Roger and Andy
  
• Winner: Roger
  
• No. of votes polled: 1000
  
• No. of votes for Roger: 700
  
• No. of votes for Andy: 300
  

An interesting thing was observed during the counting of the votes. During the entire counting, Roger always remained "ahead" of Andy.

Calculate the probability of this happening, i.e. the counting happening in such a manner, in the following 2 cases.

1. Roger's votes were always “Greater than or equal to” Andy's votes.

2. Roger's votes were always “Greater than” Andy's votes.

---END---

[Edit: Corrected a minor typo.]

Edited July 17, 2009 by The Beta Guy

[Guest]

Good Puzzle!

Denote Roger's vote as R and Andy's as A

Then

The first vote must be R

The second must also be R

Now, after first vote for R is counted the next vote cant be A ---- this is done in 1 way

After 2 votes counted for R, A votes may be 0 or 1 --- in 2 ways

After 3 votes counted for R, A votes may be 0 or 1 or 2 --- in 3 ways

After 4 votes counted for R, A votes may be 0 or 1 or 2 or 3--- in 4 ways

and so on

After 300 votes counted for R, A votes may be 0 to 299 ---- 300 ways

After 301 votes counted for R, A votes may be 0 to 300 ---- 301 ways

After 302 votes counted for R, A votes may be 0 to 300 ---- 301 ways

After 699 votes counted for R, A votes may be 0 to 300 --- 301 ways

After 700 votes counted for R, A votes may be 300 --- 1 way

Total number of ways =300! * 301 * 399 = 301! * 399

Total number of ways in which the votes may be counted

Since A votes may be 0 to 300, no matter what the count for R, the total number of ways to count is

301! * 700

700 diff places where count of A may be 0 to 300 and last place where the count must be 300

Probability = 301!*399 / 301!*700 = 0,570

Now, after first vote for R is counted the next vote FOR A can be 0 or 1 ---- this is done in 2 ways

After 1 votes counted for R, A votes may be 0 or 1 --- in 2 ways

After 2 votes counted for R, A votes may be 0 or 1 or 2 --- in 3 ways

After 3 votes counted for R, A votes may be 0 or 1 or 2 or 3--- in 4 ways

and so on

After 300 votes counted for R, A votes may be 0 to 300 ---- 301 ways

After 301 votes counted for R, A votes may be 0 to 300 ---- 301 ways

After 302 votes counted for R, A votes may be 0 to 300 ---- 301 ways

After 699 votes counted for R, A votes may be 0 to 300 --- 301 ways

After 700 votes counted for R, A votes may be 300 --- 1 way

Total number of ways = 300! * 301 * 400 = 301! * 400

Probability = 301!*400 / 301!*700 = 0,5714

Edited July 17, 2009 by DeeGee

[Guest]

It's a nice puzzle

I wrote a program to solve this. After one million iterations the probability of the first case turned out to be 0.57163, and for the second case the probability is 0.39939

[Guest]

@ Rijk: Can u send me your program as a PM ?

[Guest]

@ Deegee & @ Rijk: For part '1', both of your answers are correct.

(PS: And i thought that this puzzle was tough!! )

[Guest]

@ Rijk: Can u send me your program as a PM ?

i did

[Guest]

@Rijk: The program was good. The approach is really innovative.

You have got both the answers correct.

@Degee: Your answer to the second part is incorrect!