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atrain
atrain
atrain
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November 17th, 2010 at 4:34:38 PM permalink
of two numbers coming up on a european roulette wheel 4 times in a row
atrain
atrain
atrain
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November 17th, 2010 at 4:40:14 PM permalink
like as in 1 in lets say a thousand or whatever the actual chance is
PapaChubby
PapaChubby
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November 17th, 2010 at 5:14:30 PM permalink
Are you asking: What is the probability that on four consecutive spins of the wheel, that only two distinct numbers would result during those four spins? It can be any two numbers, and not two that are selected in advance of the spins?
atrain
atrain
atrain
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November 17th, 2010 at 5:24:20 PM permalink
I am asking what the probility of 2 distinct numbers is in four spins
rdw4potus
rdw4potus
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November 17th, 2010 at 5:39:58 PM permalink
That is the odds of 1 number hitting * the odds of another number hitting * the odds of any other number hitting * the odds of any other number hitting. So, 1/37*1/37*36/37*36/37. So, if I were to play my birth date (10 and 17), I should expect those two numbers to both hit once each within 4 spins about 1 in 1400 times.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
guido111
guido111
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November 17th, 2010 at 5:47:06 PM permalink
Quote: atrain

I 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.
atrain
atrain
atrain
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November 17th, 2010 at 7:33:50 PM permalink
Quote: guido111

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.



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
atrain
atrain
atrain
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November 17th, 2010 at 7:34:40 PM permalink
Quote: atrain

Quote: guido111

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.



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

guido111
guido111
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November 19th, 2010 at 11:51:38 AM permalink
Quote: atrain

Quote: guido111

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.



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!
atrain
atrain
atrain
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November 19th, 2010 at 1:33:44 PM permalink
yeah i was wondering big time thank u
ItsCalledSoccer
ItsCalledSoccer
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November 19th, 2010 at 1:44:18 PM permalink
I'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%?
atrain
atrain
atrain
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November 19th, 2010 at 3:04:02 PM permalink
that may be correct i am no statistician
guido111
guido111
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November 19th, 2010 at 3:20:23 PM permalink
Quote: ItsCalledSoccer

I'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)
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