CL Last 16 vs Barcelona (feb. 24 & mar. 18)

[squirtyflower]

Re: CL Last 16 vs Barcelona

No it isn't. That's the point. Last year was a dice throw and this year was another unrelated dice throw totally unrelated to last years result. There is no 1 in 30 chance.

indeed they are unrelated events as draws
but you still have to roll the die twice and get the same result each time, therefore 1/30

And there you have it. Every combination is exactly the same then. Barca/Barca. Barca/Real. Barca/Monaco. Real/Monaco. And so on. All exactly the same. No more or less chance of the combo being Barca/Barca than Barca/Monaco or Barca/Dortmund etc.......

i hope to god you don't teach maths

but if you do its probably in one of those inner city 'comprehensives'

 

[Tonster]

Re: CL Last 16 vs Barcelona

indeed they are unrelated events as draws
but you still have to roll the die twice and get the same result each time, therefore 1/30

And there you have it. Every combination is exactly the same then. Barca/Barca. Barca/Real. Barca/Monaco. Real/Monaco. And so on. All exactly the same. No more or less chance of the combo being Barca/Barca than Barca/Monaco or Barca/Dortmund etc.......

i hope to god you don't teach maths

but if you do its probably in one of those inner city 'comprehensives'

No but I understand it clearly, sorry you feel the need to make snide comments.
The chances of rolling a 6 is 1 in 6 then the chances of rolling another 6 is 1 in 6 (unrelated to the first roll) so a total chance then 1 in 36.
But the odds of rolling a 6 then a 2 or a 4 then a 3 are exactly the same. 1 in 36.
That's the point. That's probability if the second event isn't influenced by the first.

I didn't need grammar school to understand that.

 

[117 M34]

No it isn't. That's the point. Last year was a dice throw and this year was another unrelated dice throw totally unrelated to last years result. There is no 1 in 30 chance.

indeed they are unrelated events as draws
but you still have to roll the die twice and get the same result each time, therefore 1/30

And there you have it. Every combination is exactly the same then. Barca/Barca. Barca/Real. Barca/Monaco. Real/Monaco. And so on. All exactly the same. No more or less chance of the combo being Barca/Barca than Barca/Monaco or Barca/Dortmund etc.......

You are wrong! Or just very bad at explaining yourself.
I think you are wrong and don't understand the rules of probability.

 

[cleavers]

You are wrong! Or just very bad at explaining yourself.
I think you are wrong and don't understand the rules of probability.

I think he does, if you toss a coin 50 times, the chances of it being heads is 1 in 2, and its that every time you toss the coin, the previous result does not effect the next.

 

[Regular Joe]

Perhaps those arguing that to tonster is wrong could actually explain why they think that rather than making unsupported digs.

 

[Corky]

indeed they are unrelated events as draws
but you still have to roll the die twice and get the same result each time, therefore 1/30

And there you have it. Every combination is exactly the same then. Barca/Barca. Barca/Real. Barca/Monaco. Real/Monaco. And so on. All exactly the same. No more or less chance of the combo being Barca/Barca than Barca/Monaco or Barca/Dortmund etc.......

You are wrong! Or just very bad at explaining yourself.
I think you are wrong and don't understand the rules of probability.

He would be correct given his assumption of 1 in 5 and 1 in 6, however the assumption is wrong as other clubs were prevented from drawing more opponents than others on the other side of the draw,

It isn't very easy to work out.

 

[Tonster]

And there you have it. Every combination is exactly the same then. Barca/Barca. Barca/Real. Barca/Monaco. Real/Monaco. And so on. All exactly the same. No more or less chance of the combo being Barca/Barca than Barca/Monaco or Barca/Dortmund etc.......

You are wrong! Or just very bad at explaining yourself.
I think you are wrong and don't understand the rules of probability.

He would be correct given his assumption of 1 in 5 and 1 in 6, however the assumption is wrong as other clubs were prevented from drawing more opponents than others on the other side of the draw,

It isn't very easy to work out.

Actually as we were first out last year it was a straight 1 in 5 chance and second out this year (but the first game was Chelsea who we couldn't get) it meant for us again a straight 1 in 6 chance.
For other clubs the odds were slightly different agreed.

 

[Corky]

You are wrong! Or just very bad at explaining yourself.
I think you are wrong and don't understand the rules of probability.

He would be correct given his assumption of 1 in 5 and 1 in 6, however the assumption is wrong as other clubs were prevented from drawing more opponents than others on the other side of the draw,

It isn't very easy to work out.

Actually as we were first out last year it was a straight 1 in 5 chance and second out this year (but the first game was Chelsea who we couldn't get) it meant for us again a straight 1 in 6 chance.
For other clubs the odds were slightly different agreed.

Last year we had 5 clubs we could draw but Dortmund were the highest chance as they could only draw 5 clubs and the other 4 could draw 7.

You need to work out the probability as each ball is drawn like the lottery, as if Schalke and Bay Lev were drawn Dortmund stay and Barca could be drawn. I need a pencil sharpener.

 

[117 M34]

You are wrong! Or just very bad at explaining yourself.
I think you are wrong and don't understand the rules of probability.

He would be correct given his assumption of 1 in 5 and 1 in 6, however the assumption is wrong as other clubs were prevented from drawing more opponents than others on the other side of the draw,

It isn't very easy to work out.

Actually as we were first out last year it was a straight 1 in 5 chance and second out this year (but the first game was Chelsea who we couldn't get) it meant for us again a straight 1 in 6 chance.
For other clubs the odds were slightly different agreed.

was the question not about the chance of us drawing Barca in BOTH draws? Therefore you would need to multiply 5 by 6 which gives a 30/1 chance. This would be the same odds as saying what would be the chance of us getting Barca last year and real Madrid this year.
Yes, once last year's draw had been made and we got Barca, it then comes down to a 1 in 6 to see if we get them again but to answer the chance to get them in BOTH draws you would need to multiply the odds.

 

[Regular Joe]

The % chance of facing each team was well publicised before the draw and it was 15% for Barca, just under a 1 in 7 chance. That we drew Barca last year is statistically irrelevant.

A more relevant question is do we consistently draw harder teams than we could expect given a random draw, and without any research I suspect that is a big yes.