Quote:
Originally Posted by pkdk
I think you read my post wrongly. If you pick 1 of 3 from x there is a 33.3333... recurring% chance of the left array alignment of the value 3.
This is a horrendous sentence.
But I think you mean that - given 3 sets of 3 numbers, each set containing 1, 2 and 3, randomly arranged, the odds of drawing a 3 from any GIVEN set is 1/3. This is true.
Quote:
But if I pick one from y there is 70.3703~% chance of getting a 3.
You have not defined x or y yet so this is very confusing. If Y is ONE set, then you're dead wrong. If Y is ALL THREE sets - then you're right.
Lets nail it down:
3 sets of 3
Set X: {1,2,3}
Set Y: {1,2,3}
Set Z: {1,2,3}
Each set is randomly shuffled.
The odds of drawing a 3 from X are 1 in 3. (33~%)
The odds of drawing a 3 from Y are 1 in 3. (33~%)
The odds of drawing a 3 from Z are 1 in 3. (33~%)
However the odds of drawing AT LEAST 1 3 from X + Y are not 33% + 33%.
They are, instead, more easily calculated by working it backwards based on odds of not getting a 3.
Odds of not getting a 3 in X = 2/3
Odds of not getting a 3 in Y = 2/3
Odds of not getting a 3 in either X OR Y : 2/3 * 2/3 = 44.4~%
Odds of getting at least 1 3 when drawing from X and Y = 1-44.4% = 56.6~%
Odds of not getting a 3 in either X or Y or Z = 2/3 * 2/3 * 2/3 = 29.629%
Odds of getting at least 1 3 when drawing from X, Y & and Z = 1 - 29.629% = 70.370~%