Roulette probability

There was a time back in school when I could figure that out with little trouble.

Now...the best I can do is say: Very, very, very high to 1 against.

No, but the expected number of spins to get all 37 number is slightly less than 155.5.

Can you put that in odds to 1 for me please

Might sound harsh but if you don't know the answer, don't tell me what you know. Cheers

Harsh, and stupid, because when you post like that, other people besides your intended recipient will read what you wrote and think "**** him! - I know the answer but I won't bother telling him!"

If someone reads my question and can answer it , I would be great full , if they don't know the answer there is no point in commenting . You might as well tell me your favourite colour or where you go on holiday

It's "grateful". I don't have a "favourite colour", and I generally go to Spain for my holidays - Catalunya this year but sometimes Andalucia. I may go to Galicia next year.

Cheers for that, do you have any other useless info to give me? Completely unrelated to the question I first asked.

My shoe size is 9 1/2 and my favourite tipple is single malt Scotch. I prefer brunettes, and I hate Apple products. I believe the recent US government shutdown has made that country a laughing stock. The majority of Belgium's railways are electrified at 3000 volts DC.

The transactional interpretation of quantum mechanics (TIQM) describes quantum interactions in terms of a standing wave formed by retarded (forward-in-time) and advanced (backward-in-time) waves. It was first proposed in 1986 by John G. Cramer, who argues that it helps in developing intuition for quantum processes, avoids the philosophical problems with the Copenhagen interpretation and the role of the observer, and resolves various quantum paradoxes. More recently he has also argued TIQM to be consistent with the Afshar experiment, while claiming that the Copenhagen interpretation and the many-worlds interpretation are not.

The existence of both advanced and retarded waves as admissible solutions to Maxwell's equations was explored in the Wheeler–Feynman absorber theory. Cramer revived their idea of two waves for his transactional interpretation of quantum theory. While the ordinary Schrödinger equation does not admit advanced solutions, its relativistic version does, and these advanced solutions are the ones used by TIQM.

In TIQM, the source emits a usual (retarded) wave forward in time, but it also emits an advanced wave backward in time; furthermore, the receiver also emits an advanced wave backward in time and a retarded wave forward in time. The phases of these waves are such that the retarded wave emitted by the receiver cancels the retarded wave emitted by the sender, with the result that there is no net wave after the absorption point. The advanced wave emitted by the receiver also cancels the advanced wave emitted by the sender, so that there is no net wave before the emitting point either. In this interpretation, the collapse of the wavefunction does not happen at any specific point in time, but is "atemporal" and occurs along the whole transaction, and the emission/absorption process is time-symmetric. The waves are seen as physically real, rather than a mere mathematical device to record the observer's knowledge as in some other interpretations of quantum mechanics.

Perfect, thank you. This was the first time I had used this forum and it has been so helpful

I get:
423030204689513726210072936292160920801561080787791024317213402193644748599415737827021280728152092307740231928582175719278979195970121497021775872000000000 / 6609557828843866774348296857793615320986068325257944996730965130260195627493490637048004105256563742994070037769599882399012397170569200279466412758131334001

This is greater than 1/16, but less than 1/15.

Thank you

Ok, Mark, how did you solve this? I ran a Monte Carlo and got the same percentage so I know you are right, but when I did the first principles solution I got something completely different.

Using the classic stars and bars technique, the number of ways to put 100 balls into 37 bins should be (136!)/100!/36!. If you exclude zeros it should be (99!)/63!/36!. This leads to a ridiculously low result. What am I doing wrong here?

1) You're including distributions with empty bins, which means you're counting too many since all 37 numbers have to be used. Fixing this will make your answer even smaller.

2) You're not taking into account order of arrival. For instance (assuming 3 balls in bin 2), the 3 balls in bin 2 could have arrived there as the first 3 balls, the last 3 balls, or some other combination. There are C(100,3) ways those balls could have arrived in that bin. Fixing this will drastically increase your answer.

I wrote a Python script to count the number of 100-character strings that use all 37 characters of a 37-character alphabet. Here's an example with 5 spins of the wheel and 3 possible outcomes:

First, identify all partitions of 5 into 3 subsets:
a) 3 1 1
b) 2 2 1

Then determine all of the character permutations for each distribution:
a) 3!/(1!*2!) = 3 (because there is one 3 and two 1s in the distribution)
b) 3!/(2!*1!) = 3 (because there are two 2s and one 1 in the distribution)

Then, calculate the number of 5-character strings based on each distribution:
a) 5!/(3!*1!*1!) = 20
b) 5!/(2!*2!*1!) = 30

Finally, sum the a-products and the b-products to get 3*20 + 3*30 = 150.
Divide that by 3^5=243, which is the total number of possible outcomes to get 150/243.

FYI, there are 1,496,203 partitions of 100 into 37 subsets.

Turns out the numerator is computed via theorem 5-2 in this paper:
http://www.elcamino.edu/faculty/gfry/210/DistributeBallsBoxes.pdf

That would have taken less time to compute.

So why isn't theorem 4 the correct one to use for k "indistinguishable" balls in the denominator and theorem 5-4 correct for the numerator? That's what I was using and it is clearly wrong, but the balls are indistinguishable, right?

The balls are distinguishable because there is an order to the outcomes at the roulette wheel. The denominator must be 37^100 because that's how many possible outcomes there are.

Just to clarify, if I did it 17 times it is likely to come in 16 times?

On average, you'll get all 37 number in 100 spins once every 15 or 16 times.