EFD contest (500 EFD): Math problem - part 1

[mrproper]

Starting math contest with this simple problem.
- we have 100 people in a room
- we give each one a chance to roll a zocchihedron
- we offer the top 3 rolls a list of prizes described by values: A, B and C.
- if two people draw the same number, a re-roll is required and it's value is added as 1/100th of the initial roll.

The questions are:
1. What are your chances to win a prize, if you are one of those 100 people?
2. What prize value will you statistically win on average if you take values A, B and C, for the first 3 ranks, and 0 for the rest?

 

[Moonfish]

23..............................

 

[khaos]

Let's try

my chance is uhm.. very small ^^
average price money is A+B+C/100

 

[mrproper]

... updated first post to show there are 2 questions ...

First complete correct post gets the EFD (no edits please). Everyone else goes to round 2.

 

[GoNi]

1. 3%
2. (a+b+c)/3

 

[mrproper]

1. 3%
2. (a+b+c)/3

Sorry, just realized this is wrong

 

[Colbey Kal]

1) 3%
2) ((a+b+c)/3)*0.03

 

[mrproper]

1) 3%
2) ((a+b+c)/3)*0.03

First correct solution

You get a prize, with a chance of 3%, but the average prize is an average of A, B and C, thus, ((A+B+C)/3)*(3/100).

 

[mrproper]

Continued to https://www.planetcalypsoforum.com/...00-efd-math-problem-part-2-a.html#post2267406

 

[Deity A]

sure 3% is right for the first one?
1/100 +
1/99 +
1/98
=0,0303050917336632

since you cannot win more then one prize...

else you get
1/100
1/100
1/100
=
3/100

Ok, I can be (and think I am completely wrong here; since only 3 persons will win, but for some reason I get this result)

 

[Satans]

It's over 9000 !!

 

[mrproper]

sure 3% is right for the first one?
1/100 +
1/99 +
1/98
=0,0303050917336632

since you cannot win more then one prize...

That would be correct if those that did not win would get a reroll for each subsequent prize. In some math problems, that is the case. For this one, a single roll from everyone, and all prizes given at once.

 

[Deity A]

That would be correct if those that did not win would get a reroll for each subsequent prize. In some math problems, that is the case. For this one, a single roll from everyone, and all prizes given at once.

thx for enlighting me; I know my logic was flawed; but could trace it (too tired or still drunk I guess)