Maths question ?

[bookiebasher]

Yeah , never thought of that , we're getting there slowly.

Thanks Trev , far more helpful than the other Trev , he's a right stick in the mud.

[RED-DOG]

Just play those numbers yourself until you get your money back.

No need to thank me.

[bookiebasher]

Just play those numbers yourself until you get your money back.

No need to thank me.

Are you related to Typhoon Trev by any chance ?

[bookiebasher]

You need to provide a complete breakdown of the number of spins and the times each number came up. 

Ok , I am on the case and will report for duty in the morning hopefully 

Right I have some data from the sessions they gave me. There are a few more losing days still to be logged to

see if it was the same chap but the data for the numbers I gave earlier are...

538 spins with  5... 24 times ; 17....22 times ; 35...19 times.

So , those 3 numbers came out a total of 65 times out of 538 spins.

If my maths is correct you would expect any 3 numbers to come out 43.63 times and any one number 14.54 times ,

obviously over a much larger sample.

So can any maths expert work out the probability using the above figures ?

[DMorgan]

Stolen from here: http://stats.stackexchange.com/questions/102769/how-do-you-calculate-the-standard-deviation-for-roulette

That is for an american roulette wheel but we'll go with it for now

standard deviation = root 33.207 = 5.7625

Your man landed 65/538 with expectation of 44/38, so the difference was 21 hits which is 21/5.7625 =  3.644 standard deviations away from the mean over his sample



Which puts you pretty far down the chart, so if they took 1,000 machines and they all had 538 spins and were ordered by amount paid out, yours was the 5th highest paying machine this time if it had two zeroes

With only one zero your edge is smaller so his results won't be quite as far away from the mean. My intuition is that it won't change it by very much since the second zero only represents one of 37 possible outcomes but I'm not sure on that.

Its pretty gross but it is (just about) within normal bounds assuming that the calcs are correct.

Having looked at the numbers the 50p was a bit harsh so apologies for that, I'm a little sensitive about people running bad this month

Click to see full-size image.

[bagel]

add all the numbers up on a roulette wheel and it comes to 666

stay the **** away, the devils game

[TightEnd]

add all the numbers up on a roulette wheel and it comes to 666

stay the ****  away, the devils game

Bookiebasher is a provider of the service, not the player....

[Woodsey]

add all the numbers up on a roulette wheel and it comes to 666

stay the ****  away, the devils game

Bookiebasher is a provider of the service, not the player....

The devil operates in mysterious ways...... Makes sense in this case as he is masquerading as a good lad 

[bookiebasher]

add all the numbers up on a roulette wheel and it comes to 666

stay the ****  away, the devils game

Bookiebasher is a provider of the service, not the player....

The devil operates in mysterious ways...... Makes sense in this case as he is masquerading as a good lad 

Thought you were banned

[bookiebasher]

Stolen from here: http://stats.stackexchange.com/questions/102769/how-do-you-calculate-the-standard-deviation-for-roulette

That is for an american roulette wheel but we'll go with it for now

standard deviation = root 33.207 = 5.7625

Your man landed 65/538 with expectation of 44/38, so the difference was 21 hits which is 21/5.7625 =  3.644 standard deviations away from the mean over his sample



Which puts you pretty far down the chart, so if they took 1,000 machines and they all had 538 spins and were ordered by amount paid out, yours was the 5th highest paying machine this time if it had two zeroes

With only one zero your edge is smaller so his results won't be quite as far away from the mean. My intuition is that it won't change it by very much since the second zero only represents one of 37 possible outcomes but I'm not sure on that.

Its pretty gross but it is (just about) within normal bounds assuming that the calcs are correct.

Having looked at the numbers the 50p was a bit harsh so apologies for that, I'm a little sensitive about people running bad this month

Click to see full-size image.

Hey no worries Darren , thanks for your input , much appreciated . Will make it a £1 next time I post 

[bookiebasher]

All I basically wanted to know was how bad I ran .

The company supplying the machines said it was normal "variance"

I disagreed but could not argue my case with correct statistical analysis .

[bagel]

add all the numbers up on a roulette wheel and it comes to 666

stay the ****  away, the devils game

Bookiebasher is a provider of the service, not the player....

well aware of that, was a tongue in cheek observation.

calm down 

[TightEnd]

You ran worse than bad but not impossibly bad. Google three sigma events....

[bookiebasher]

You ran worse than bad but not impossibly bad. Google three sigma events....

Oh no , that's going to get me really involved and I am slightly drunk at present so will investigate

 " three sigma events "  in the morning.

Come on then , probability , anyone ? 1000-1 , 5000-1 ?

Where's Woodsey when you need him ?

[Woodsey]

You ran worse than bad but not impossibly bad. Google three sigma events....

Oh no , that's going to get me really involved and I am slightly drunk at present so will investigate

 " three sigma events "  in the morning.

Come on then , probability , anyone ? 1000-1 , 5000-1 ?

Where's Woodsey when you need him ?

Funnily enough I'm out of the office this week and at least a couple of sheets to the wind already 😄