Quote: atrainI am asking what the probility of 2 distinct numbers is in four spins
4 spins
unique/repeats
4/0 (37/37)*(36/37)*(35/37)*(34/37) = 0.845754447
3/1 [(37/37)*(36/37)*(35/37)*(3/37)]+[(37/37)*(36/37)*(2/37)*(35/37)]+[(37/37)*(1/37)*(36/37)*(35/37)] = 0.149250785
2/2 [1- (0.845754447 + 0.149250785 + 0.000019742)] = 0.004975026 or 1 in 201
... or [(37/37)*(36/37)*(2/37)*(2/37)] + [(37/37)*(1/37)*(36/37)*(2/37)] + [(37/37)*(1/37)*(1/37)*(36/37)]
1/3 (37/37)*(1/37)*(1/37)*(1/37) = 0.000019742
The math can get very messy very quickly with more spins and more repeat numbers.
Quote: guido1114 spins
unique/repeats
4/0 (37/37)*(36/37)*(35/37)*(34/37) = 0.845754447
3/1 [(37/37)*(36/37)*(35/37)*(3/37)]+[(37/37)*(36/37)*(2/37)*(35/37)]+[(37/37)*(1/37)*(36/37)*(35/37)] = 0.149250785
2/2 [1- (0.845754447 + 0.149250785 + 0.000019742)] = 0.004975026 or 1 in 201
... or [(37/37)*(36/37)*(2/37)*(2/37)] + [(37/37)*(1/37)*(36/37)*(2/37)] + [(37/37)*(1/37)*(1/37)*(36/37)]
1/3 (37/37)*(1/37)*(1/37)*(1/37) = 0.000019742
The math can get very messy very quickly with more spins and more repeat numbers.
so what u r saying if u chose the numbers 1 and 2 u have a 1 in 201 chance of that happening in 3 spins
Quote: atrainQuote: guido1114 spins
unique/repeats
4/0 (37/37)*(36/37)*(35/37)*(34/37) = 0.845754447
3/1 [(37/37)*(36/37)*(35/37)*(3/37)]+[(37/37)*(36/37)*(2/37)*(35/37)]+[(37/37)*(1/37)*(36/37)*(35/37)] = 0.149250785
2/2 [1- (0.845754447 + 0.149250785 + 0.000019742)] = 0.004975026 or 1 in 201
... or [(37/37)*(36/37)*(2/37)*(2/37)] + [(37/37)*(1/37)*(36/37)*(2/37)] + [(37/37)*(1/37)*(1/37)*(36/37)]
1/3 (37/37)*(1/37)*(1/37)*(1/37) = 0.000019742
The math can get very messy very quickly with more spins and more repeat numbers.
so what u r saying if u chose the numbers 1 and 2 u have a 1 in 201 chance of that happening in 3 spins in a row is that correct
Quote: atrainQuote: guido1114 spins
unique/repeats
4/0 (37/37)*(36/37)*(35/37)*(34/37) = 0.845754447
3/1 [(37/37)*(36/37)*(35/37)*(3/37)]+[(37/37)*(36/37)*(2/37)*(35/37)]+[(37/37)*(1/37)*(36/37)*(35/37)] = 0.149250785
2/2 [1- (0.845754447 + 0.149250785 + 0.000019742)] = 0.004975026 or 1 in 201
... or [(37/37)*(36/37)*(2/37)*(2/37)] + [(37/37)*(1/37)*(36/37)*(2/37)] + [(37/37)*(1/37)*(1/37)*(36/37)]
1/3 (37/37)*(1/37)*(1/37)*(1/37) = 0.000019742
The math can get very messy very quickly with more spins and more repeat numbers.
so what u r saying if u chose the numbers 1 and 2 u have a 1 in 201 chance of that happening in 3 spins
No.
This just shows the probability of each possible outcome in 4 spins. Where any number can be unique or a repeat.
For 2 numbers like 1,2 ( that you choose before the actual spins)
(2/37)^3 for 3 spins ... 0.000157937/ or 1 in 6,331
(2/37)^4 for 4 spins ...0.0000085372/ or 1 in 117,135
and so on.
I see you have that question also answered in another thread.
Good luck!
... the first two results can be <any> and <any>, so probability = 1
... the next two have to be 2/37 and 2/37. So ...
... wouldn't the probablilty be 1 * 1 * 2/37 * 2/37 = ~0.3%?
Quote: ItsCalledSoccerI'm no statistician, but ...
... the first two results can be <any> and <any>, so probability = 1
... the next two have to be 2/37 and 2/37. So ...
... wouldn't the probablilty be 1 * 1 * 2/37 * 2/37 = ~0.3%?
No,
The first spin probability would be (37/37) or 1
Now since we have 1 unique number the second spin would either be a (1/37)...a repeat or a (36/37)... a unique number
So, without choosing any numbers, the results of 4 consecutive spins would be:
4 spins: all possible outcomes probability
unique/repeats
4u/0r (37/37)*(36/37)*(35/37)*(34/37) = 0.845754447 (all 4 unique numbers ie: 1,2,3,4)
3u/1r: The 1 repeat number could happen on the 4th, 3rd or 2nd spin only.
[(37/37)*(36/37)*(35/37)*(3/37)]+[(37/37)*(36/37)*(2/37)*(35/37)]+[(37/37)*(1/37)*(36/37)*(35/37)] = 0.149250785
2u/2r [1- (0.845754447 + 0.149250785 + 0.000019742)] = 0.004975026 or 1 in 201
... or [(37/37)*(36/37)*(2/37)*(2/37)] + [(37/37)*(1/37)*(36/37)*(2/37)] + [(37/37)*(1/37)*(1/37)*(36/37)]
1u/3r (37/37)*(1/37)*(1/37)*(1/37) = 0.000019742 (all same number ie: 1,1,1,1)