posted January 19, 2007 05:04 PM            

What are the probabilities of:

A: getting 3 ONEs in a row
B: getting a ONE 3 times only.
C. getting a ONE 3 times or more.
D. getting a single ONE, a single TWO and a single THREE in any order.
E. getting a single ONE, a single TWO and a single THREE in ascending order.

hint: Pascal's pyramid.

posted January 19, 2007 05:18 PM            

Added: it's the same thing as that old coin question: If you
flip a coin 10 times and, by some happenstance, it come up "heads"
all 10 times, what are the odds it will come up heads the 11th
time. Answer: 50/50.

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posted January 20, 2007 12:52 PM            

A binomial probability distribution for 5 trials:

A=1/6

B=1-A

(A+B)^5=1

Using Pascals Triangle (A+B)^5 expands to:

1*(A^5) + 5*(A^4*B) + 10*(A^3*B^2) + 10*(A^2+B^3) + 5*(A*B^4) + 1*(B^5)

questions B to E involve the third term: 10 * ( A^3 * B^2 )

posted January 20, 2007 04:23 PM            

well there are many who say this,

how-ever experimentation shows (proves?) different

posted January 20, 2007 05:00 PM            

quote:

---

how-ever experimentation shows (proves?) different

---

If the coin is unbiased then Mel is correct.

The probability of 11 consecutive heads is 0.5^11.

However, the probability of a head is 0.5 irrespective of its history. Coins don't have memories.

posted January 20, 2007 05:28 PM            

What is the probability of getting 4 heads in 11 flips of the coin ?

(H+T)^11 where H is 0.5 for heads and T=0.5 for tails.

in this case the 4th binomial term which we want is 165 * ( 0.5^4 + 0.5^7 )

Pascal's Triangle:
http://ptri1.tripod.com/

posted January 20, 2007 05:46 PM            

yes I understand that each toss has 50/50 probabillity,
how-ever, practice shows that the total counts will represent a bell curve,
so, given any one-sided streak the expected results must be weighted in order to reflect
a truer prediction,

many argue this, but experimentation shows different

B.

posted January 20, 2007 09:18 PM            

quote:

---

practice shows that the total counts will represent a bell curve

---

If we consider 100 tosses then it is a simple matter to calculate the probability of getting only one head, two heads, three heads up to 100 heads. If we then plot the probabilities we get a bell curve without putting our hands in our pockets for a coin.

quote:

---

so, given any one-sided streak the expected results must be weighted in order to reflect a truer prediction

---

quote:

---

but experimentation shows different

---

posted January 20, 2007 10:27 PM            

ask this, if we flip 100 times and we get 80% heads,
(which can be likely)

then we flip 1000 times, it's likelier that we get closer to 50%, no?

then if we flip 10,000 times, it's very likely we are near 50%, correct?

well if so, then the correct math MUST show (somehow?) , that a one-sided
deviance from norm will ultimately balance out from the opposing-sided deviance

no?

added;
I used to play a lot of keno for fun! and actually won quite a bit of money on
a regular basis, I wrote a program to analyse the results from years of posted
game results, and it showed that most often a number which had 1 in 4 odds to hit
per game, would often streak misses for up to 11 to 17 times in a row, (I beieve that the
longest lost streak that I recorded was around 90, and shockingly after a long (12 or more)
miss streak, the number would seem to win better than 2 to 1 for the next 10 or so games!!

I spent quite a while trying to convince myself that my results were coincedental, but honestly
if you think about it, it must diviate the other way at some point, and anyway I and a friend
who pretty much had a photographic memory would beat the games almost everytime we played

------------------
Washington DC Area
Borje's "Poff's" is likely the BEST tool for learning PB.. http://www.reonis.com/POFFS/index.htm
And a few PBTool's & Beginner Help:http://sweetheartgames.com/PBTools/JumpStart.html
& Another Good Resource: http://www.fredshack.com/docs/powerbasic.html

posted January 20, 2007 10:40 PM            

If you get 80 heads on the first 100 tosses then you have 60 more
heads than tails. If after 1000 tosses you still have 60 more
heads than tails, then you haven't "balanced" any "opposing-sided
deviance" yet the proportion of heads is now 530/1000 or 53%
which is certainly closer to 50%.

posted January 20, 2007 11:33 PM            

So, to your experience with keno. You were lucky, plain and simple. If I examined the stats I may well find that it was unusual but I very much doubt that it would be statistically significant.

A long lived trend does not tell me that a shift in the opposite direction is imminent, it tells me that something is wrong with the mechanics or the mechanics are not random.