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what is the probability of... #4

 
 
Reply Fri 2 Sep, 2011 08:16 am
1. What is the probability of the player matching 3 numbers from a 10 number lottery (0...9) w/o repetition, given the conditions:
- player and lottery pick 6 numbers
- player and lottery always have a repeated number, for example:

123455
123455

2. What is the probability of the player matching 3 numbers from a 10 number lottery (0...9) w/o repetition, given the conditions:
- player and lottery pick 6 numbers
- player always has a repeated number,
- lottery has always two repeated numbers,
for example:

123455
124455
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w0lfshad3
 
  1  
Reply Fri 2 Sep, 2011 12:15 pm
@w0lfshad3,
Is this the answer to 1?
C(5,5)C(10-5,0)/C(10,5)=1/252 + (1/252)*(1/5)=6/1260=1/210
or just
(1/252)*(1/5)=1/1260?

And this to 2?
C(5,4)C(10-5,0)/C(10,4)=(5/210)*(2/5)=2/210?
w0lfshad3
 
  1  
Reply Sat 3 Sep, 2011 04:25 pm
@w0lfshad3,
1. is (1/252)*(1/5)=1/1260

2. is hard; take 2 simplified, it's still hard:
What is the probability of the player matching 3 numbers from a 10 number lottery (0...9) w/o repetition, given the conditions:
- player and lottery pick 6 numbers
- player always has a repeated number,
- lottery always has a repeated number,
for example:

123455
123455
w0lfshad3
 
  1  
Reply Sat 3 Sep, 2011 05:02 pm
@w0lfshad3,
Here there's many matches of 3. Because it's a match of 6/6 it means that there are C(6,3)=20 matches of 3... I think
w0lfshad3
 
  1  
Reply Wed 7 Sep, 2011 09:06 am
@w0lfshad3,
This is a conditional probability problem.

I experimented on matching all numbers between ABCD;EE and ABCD;EE
Hypothetic (calculated) probability: 1/1260

Real probability: ~1.75/1260
0.0013608333 experiment
0.00138888(8) guess = ~1.75/1260=1/720=1/6!=1/P(10,3)?=1/(1x2x3x4x5x6)

Anyone knows how to calculate this properly?
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