ajm

Its been a while since I have done any statistics, but surely the lottery cannot be a binomial distribution because each ball drawn affects the probability of the next, as there are less ***** left to choose from.

Hence probability of a particular ball being drawn is not 6/49 it is 1/49 or 1/48 or 1/47 depending on how many have already been drawn!

Claudius

Yes, but at each draw, the ***** all have the same probability of being selected, so it makes no difference to the end result (not saying you implied that, just pointing it out).

jasey

I think this applies here...


The s-wave Faddeev differential equations for systems Lambda np (for two spin states S=1/2 and S=3/2) and Lambda Lambda alpha are solved numerically on the base of cluster reduction method. The MTI-III potential model is used as the NN interaction. Model Lambda N and Lambda Lambda interaction potentials are set as VLambda N =VNN/2 and VLambda Lambda = 2VNN/3. The Lambda alpha interaction is described by phenomenologycal potentials. It is shown, that such simple models reproduce reasonably the binding energy of hypertriton 3LambdaH (EB=-2.37 MeV) and hypernucleus LambdaLambda6He (EB=-10.88 MeV). For doublet and quadruplet Lambda-d and Lambda-{Lambda5}He scattering the effective range expansion parameters are calculated. The similarity between Lambda -{Lambda5}He and n-d systems is discussed.

ajm

Yes, I was just asking a techie question of Sprint_Chief

Obviously none of this stuff changes the fact that you cannot predict which ***** are going to come up!

jasey

I predict It wont be mine or yours this weekend.

Hope I'm wrong .

ajm

Well in my case there is 100% chance you are right, because I never enter the godforsaken thing!

(although theoretically I suppose there is an infinitessimally small chance I may enter by mistake! )

Sprint Chief

Hi ajm

No, this isn't the case. You are considering a single number coming out of the machine as an event - I am considering the event to be whether a number is one of the six drawn as a single event. If you see it like this, each number is equally likely to be one of the six and unaffected by the draw order etc.

ajm

Ah, I see!

SJ_Skyline

I've said it once, I'll say it again:

It's all a load of *****!!

jasey

Go on - get yourself a lucky dip - I feel Lucky for ya

ps all you do is walk into the shop and say "Lucky dip for tonight please" - Six lovely random numbers and no Lambda(np-3)'s anywhere to be seen.

Sprint Chief

Blimey a lot of other posts got squeezed in there! I got a bit distracted so there was a delay between typing the post and actually submitting it! Claudius gave the right answer though, glad someone is keeping up

Jasey m8 I think you'll find the reverse tangential coefficient of your multi dimensional hypercube has not been recalibrated correctly against the third hypothesis as laid down by Einstein in his paper Maths for dummies. You want to check that

jasey

Got a Link

Sprint Chief

I don't think a link would help understand it, I find ten or eleven pints usually makes things *much* clearer.

paulr

so the larger numbered ***** arent heavier cos they have more ink on them and gravitate to the bottom?

jasey

That's this bit "LambdaLambda6He (EB=-10.88 MeV). " of my formula - Well spotted

Jerome

To throw a different slant on this, on the very first lottery, I did 6 lines and didn't pick the same number twice. Not a single number came up in the draw. What is the probability of this happening?

jasey

Just think - If you'd done two more you'd probably be a millionaire .

Sprint Chief

Jerome

Probability of picking 36 numbers and not one of them coming up:

Probability of first number out of the machine not coming up is 13/49
second number 12/48
third number 11/47
fourth number 10/46
fifth number 9/45
sixth number 8/44

The product of these gives you the odds

Comes to be approx 0.012% or approx thousand to one

If you did a couple more you'd almost certainly have twenty quid to your name anyway