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There are approx 3000 members.
What are the chances of being drawn 2 weeks in a row?
Quote: RSDoes each member have a 1/3000 chance of winning? Or is it a "the more you play the more entries you earn" kind of drawing? Is there only one winner or are there multiple "winners"? How many weeks?
Each member has the same chance of winning
Only 1 drawn per week
Whats the chance of winning 2 consequtive weeks?
100% when it's rigged somehow.Quote: DanthemanThe local club i frequent has a members badge draw every thursday.
There are approx 3000 members.
What are the chances of being drawn 2 weeks in a row?
Quote: DanthemanEach member has the same chance of winning
Only 1 drawn per week
Whats the chance of winning 2 consequtive weeks?
How many weeks?
If it's an infinite amount of weeks, which it appears to be since you haven't mentioned anything....then the chance 1 person wins 2 in a row is 100%.
I mean i won 2 weeks in a row, what are the chances of this?
Is it 1/3000x1/3000 ?
For the chance of me winning 2 weeks in a row?
if it's a specific/distinct 2 weeks, then it's 1 / (3000^2)
Quote: DanthemanYes infinate amount of weeks.
I mean i won 2 weeks in a row, what are the chances of this?
Is it 1/3000x1/3000 ?
For the chance of me winning 2 weeks in a row?
What does the winner win?
Quote: RSif it's an infinite amount of weeks, then it's a 100% probability.
if it's a specific/distinct 2 weeks, then it's 1 / (3000^2)
If I figured this out correctly, the probability of it happening at least once in N weeks is:
1 - (qN + combin(N,1) p qN-1 + combin(N-1,2) p2 qN-2 + combin(N-2,3) p3 qN-3 + combin(N-3,4) p4 qN-4 + ... + combin(N/2,N/2) pN/2 qN/2)
(or ... + combin((N+1)/2,(N-1)/2) p(N-1)/2 q(N+1)/2 if N is odd)
where p = 1/3000 and q = 2999/3000
There "should be" a way to simplify the sum, but I don't see it at the moment.
William Feller did not have one, inclusion-exclusion was used back in the 1600s but requires lots of math that no one enjoysQuote: ThatDonGuyThere "should be" a way to simplify the sum, but I don't see it at the moment.
now we have Excel, and others like, can easily do this in one column (i have shown B4)
here i did it in more than one column
results
i used 100 years max (infinite is too large)
weeks | years | p | 1 in |
---|---|---|---|
2 | 0.04 | 1.11111E-07 | 9,000,000 |
3 | 0.06 | 2.22185E-07 | 4,500,750 |
4 | 0.08 | 3.33259E-07 | 3,000,667 |
5 | 0.10 | 4.44333E-07 | 2,250,563 |
6 | 0.12 | 5.55407E-07 | 1,800,480 |
7 | 0.13 | 6.66481E-07 | 1,500,417 |
8 | 0.15 | 7.77555E-07 | 1,286,082 |
9 | 0.17 | 8.88629E-07 | 1,125,328 |
10 | 0.19 | 9.99703E-07 | 1,000,297 |
11 | 0.21 | 1.11078E-06 | 900,270 |
16 | 0.31 | 1.66615E-06 | 600,187 |
19 | 0.37 | 1.99937E-06 | 500,158 |
31 | 0.60 | 3.33225E-06 | 300,097 |
46 | 0.88 | 4.99836E-06 | 200,066 |
91 | 1.75 | 9.99666E-06 | 100,033 |
181 | 3.48 | 1.99932E-05 | 50,017 |
361 | 6.94 | 3.99859E-05 | 25,009 |
901 | 17.33 | 9.99617E-05 | 10,004 |
4504 | 86.62 | 0.000500042 | 2,000 |
the trend is...
more weeks...
higher chance of success at least 1 time
of course, if you win one week,
the chance that the next week gives you 2 in a row
=
1 in 3000
the exact same if you have won 3 weeks in row and going for 4 weeks in a row
chance = 1 in 3000
also,
could happen more than one time if the draw is fair
Sally now on vacation (always on vacation)
Yahoo!
Quote: RSif it's an infinite amount of weeks, then it's a 100% probability.
if it's a specific/distinct 2 weeks, then it's 1 / (3000^2)
So that about 1 in 9million that I would win the next 2 weeks in a row?
Quote: DanthemanSo that about 1 in 9million that I would win the next 2 weeks in a row?
Yes. And about 1 in 3000 for anyone to win back to back weeks.