PDA

View Full Version : Beyond Chance?



peteshimmon
2013-May-13, 07:20 PM
I like to do the British Thunderball draw as I
find I get 50% of my money back over time rather
than the measly 20% from the Lottery. But my
thunderball has not come up for three months
now and I am a bit narked. But looking into
recent results another number has been absent
for over 60 draws. With 14 numbers as the
thunderball you expect yours every 14 draws
on average. But obviously long runs without
showing are possible. So if 14 draws is one
sigma, how many sigma without showing is
reasonable and when does interferance look
likely?

I would like a set of statistical pictures
available on a website monitoring these draws.
Here a histogram of gaps between number shows
would help. These recent results are a long
way along the tail of any normal distribution.

Forgive me if I do not reply, the amount of
guff on reply pages is overwhelming my 12 year
old set top box. I will get updated IT one
day but the nasty code out there puts me off:)

cjl
2013-May-14, 04:43 PM
Assuming I remember my math correctly (and it's entirely possible I don't), there's a bit over a 1% chance that a specific number won't be pulled in 60 consecutive draws (assuming a 1/14 chance per draw). However (and this is the part that I'm a lot less sure about, given how long it's been since I've taken a probability class), the chance that some number exists that will not be drawn in 60 draws is over 16 percent. So, it doesn't sound that terribly unlikely at all.

orionjim
2013-May-15, 02:29 PM
I set up a simple Monte Carlo generator in Excel and ran 5500 draws and made this is a histogram of the results:
18447

As you can see itís close to cjlís 1% (I didnít see cjlís post till after I was done). As you can see there are a couple of really long runs.

Perikles
2013-May-15, 03:12 PM
Assuming I remember my math correctly (and it's entirely possible I don't), there's a bit over a 1% chance that a specific number won't be pulled in 60 consecutive draws (assuming a 1/14 chance per draw). I make that 1.172% (=13^60/14^60)

peteshimmon
2013-May-17, 04:32 PM
Good stuff thanks. I found a site listing
number frequencies etc and the biggest no
show was 87 draws for the number 7. A lucky
number some think. That is three large breaks
in number shows. Seems a lot for a game only
600-700 draws so far!

Ara Pacis
2013-May-17, 04:49 PM
Are you Rosencrantz or Guildenstern?

orionjim
2013-May-17, 05:16 PM
Good stuff thanks. I found a site listing
number frequencies etc and the biggest no
show was 87 draws for the number 7. A lucky
number some think. That is three large breaks
in number shows. Seems a lot for a game only
600-700 draws so far!

Not really when you consider that itís 700 * 14. I say this because I assume you are looking at all of the values 1 to 14.

peteshimmon
2013-May-18, 03:22 PM
Me and my memory! Let me correct myself after
checking that website again. The smallest largest
gap (heh heh) is 31 for the number 11 and the
largest is 89 for the number 7. That is for
472 draws in the modern Thunderball game so far.
Well 89 seems excessive but I accept it is chance.

Now I must try to understand shakesphearian comments.

Ivan Viehoff
2013-May-21, 03:07 PM
Now I must try to understand shakesphearian comments.
Shakespeare won't help you very much. It was a reference to Tom Stoppard's play Rosencrantz and Guildenstern are Dead. See the summary of Act 1 here http://en.wikipedia.org/wiki/Rosencrantz_and_Guildenstern_Are_Dead

Losing half my money wouldn't encourage me to buy lottery tickets. But you are right about the main lottery. Although the prizes are about 45% of the total stake, a lot of that prize fund goes into larger prizes, which you are unlikely to win at all during a lifetime's playing at moderate stakes. Thus the experienced return for the great majority of players is much less than 45%.

Hlafordlaes
2013-May-21, 03:32 PM
Been, what, 30 years since I though about this. I thought you could not chain independent events to get a cumulative probability. A spaceship, for example, might have n number of parts that can fail, and they are all in play simultaneously, so we can say the chance of nothing failing is p=(1-r)n, where r = reliability. Because this number is low, spaceships use redundant systems to gain reliability.

But can you chain up independent events as in the OP? I was under the impression you could not.

Perikles
2013-May-21, 05:07 PM
But can you chain up independent events as in the OP? I was under the impression you could not.Yes you can. If you throw one die n times, the probability of a particular outcome is the same whether you throw n dice once. The timing makes no difference to the probabilities.

HenrikOlsen
2013-May-21, 05:14 PM
You can treat the sequence as an outcome and calculate chances for various patterns regardless of whether there are dependencies or not.
It's actually a lot easier to do when there aren't.

Ivan Viehoff
2013-May-22, 10:57 AM
But can you chain up independent events as in the OP? I was under the impression you could not.
Quite the reverse: multiplying probabilities works only when the events are independent. If they aren't independent, then you have to employ Bayes' Theorem, and often the conditional probabilities required aren't available.

peteshimmon
2013-May-27, 06:14 PM
Just thought I would come back to this subject
after looking up results from the original
thunderball game that started fourteen years
ago. It ran for 964 draws and had a "no show"
gap of 133 draws for number 10. What are the
chances of that? If I could understand Perikles
formula I would know but the upward thingie
confuses me.

Looking at the playslip, number 10 as the
thunderball looks to be in a position that
would be chosen in a hurry. But I must not
have suspitious thoughts must I. And I note
that orionjims excellent histogram ran for
5000 draws to get the big gaps.

Is it a Poisson distribution or something?

Perikles
2013-May-27, 07:18 PM
It ran for 964 draws and had a "no show"
gap of 133 draws for number 10. What are the
chances of that? If I could understand Perikles
formula I would know but the upward thingie
confuses me. I was using a notation ^ meaning power, so 10^3 = 1000

The probability of the no-show for a 10 is very easy to calculate. The probability in the first week that it is not a ten is 13/14. So the probability of the same in 2 weeks is (13/14)^2 which is 13 squared divided by 14 squared. The probability of this happening for n weeks is (13 to the power n divided by 14 to the power n) but unless you have a computer device to calculate that for n > 100, it gets very tedious.

You have to look at these statistics in the right way. It's no good saying that the absence of a 10 in the draw for 100 weeks is so unlikely that it just can't happen, because the chance that something unusual happening is very much larger.

If you were, a priori, to bet with somebody that the 10 does appear at some point during that number of weeks, you would be incredibly unlucky to lose the bet to the point of disbelief. That is totally different to the probability of examining the outcomes and finding that something extreme has happened.