vpFREE2 Forums

when is zero really zero?

#1

A famous scientist or mathmatician once said that if the probability of a certain event was less than 1 / 10e50 than such event can be counted on to never occur. Maybe someone is more familiar with that quote.
      At first I argued with my friend that showed me this quote. But now I understand just how unimaginable that a number as big as 10 to the 50th power is, like drawing a particlular grain of sand from a parsec full of sand.
      I started to ponder about the number of non winning hands one would have to play to reach that 10 e 50 mark. I tried to figure it out on my calculator but of course it said out of range. My old TI calculator went to 10 e 99 before going out of range.
      While this is of no practical value it is somewhat philsophical or inspirational to me and possible others. The answer may be surprising and show the magnitude of such a number. Can someone help?

Steve N

#2

<< A famous scientist or mathmatician once said that if the probability
of a certain event was less than 1 / 10e50 than such event can be counted on
to never occur. Maybe someone is more familiar with that quote.
      At first I argued with my friend that showed me this quote. But now I
understand just how unimaginable that a number as big as 10 to the 50th
power is, like drawing a particlular grain of sand from a parsec full of
sand.
      I started to ponder about the number of non winning hands one would
have to play to reach that 10 e 50 mark. I tried to figure it out on my
calculator but of course it said out of range. My old TI calculator went
to 10 e 99 before going out of range.
      While this is of no practical value it is somewhat philsophical or
inspirational to me and possible others. The answer may be surprising and
show the magnitude of such a number. Can someone help??>>

Surprisingly, only 190 consecutive losing hands of 9/6 Jacks or Better
played optimally will get you to 10^-50 probability.

Cogno

#3

Hi there

What would be the odds against drawing 8, 10, and 12 consecutive losing
hands? How about having 10 loser hands in a row 3 times in a 500 game
session?

Cheers...Jeep

···

Surprisingly, only 190 consecutive losing hands of 9/6 Jacks or Better
played optimally will get you to 10^-50 probability.

Cogno

#4

Surprisingly, only 190 consecutive losing hands of 9/6 Jacks or Better
played optimally will get you to 10^-50 probability.

Cogno

Yes, but that's if you define it as "what are my chances of losing my
next 190 hands of 9/6 JOB"? The chance of losing 190 consecutive hands
hands in a lifetime of JOB is considerably higher than 10^-50, though
still pretty small I'm sure!

--Dunbar

#5

<<Yes, but that's if you define it as "what are my chances of losing my next
190 hands of 9/6 JOB"? The chance of losing 190 consecutive hands hands in
a lifetime of JOB is considerably higher than 10^-50, though still pretty
small I'm sure!>>

Right you are sir. But the context of the question was "never have a winning
hand in your lifetime."

Cogno

#6

<<What would be the odds against drawing 8, 10, and 12 consecutive losing
hands?>>

The next n hands will all be losers once in

8 - 128
10 - 429
12 - 1443

trials. As Dunbar pointed out, the odds of it happening sometime during an
extended run of hands is much greater.

<< How about having 10 loser hands in a row 3 times in a 500 game
session?>>

Somebody can check me on this, but my first crack at this is
1-binomdist(2,470,1/429,true) or about 1 in every 10 tries.

Cogno

#7

    A famous scientist or mathmatician once said that if the probability of a certain event was less than 1 / 10e50 than such event can be counted on to never occur. Maybe someone is more familiar with that quote.
     At first I argued with my friend that showed me this quote. But now I understand just how unimaginable that a number as big as 10 to the 50th power is, like drawing a particlular grain of sand from a parsec full of sand.
     I started to ponder about the number of non winning hands one would have to play to reach that 10 e 50 mark. I tried to figure it out on my calculator but of course it said out of range. My old TI calculator went to 10 e 99 before going out of range.
     While this is of no practical value it is somewhat philsophical or inspirational to me and possible others. The answer may be surprising and show the magnitude of such a number. Can someone help?

Steve N

I would argue with it, too. I'm surprised that a mathematician would
make such a statement without qualifying "never" as meaning something
non-mathematical like "not in our lifetimes" or "not in 1000
lifetimes." Strictly (and philosophically) speaking, which is how I
believe mathematicians should speak, an event which has a 1 / 10e50
(or 1 / 10e50000 or whatever) chance of occurring on each trial has
the same 0 chance of never occurring that an event which has a 1/2
chance of occurring on each trial has. If you played video poker for
eternity, your chance of hitting a million (or a billion or whatever)
royal flushes in a row at some point is not only 1, but you couldn't
even do it a finite number of times. I understand that mathematics
and philosophy were thought of as much more linked in previous
centuries than they are now.

#8

Cogno Scienti wrote:

<< How about having 10 loser hands in a row 3 times in a 500 game
session?>>

Somebody can check me on this, but my first crack at this is
1-binomdist(2,470,1/429,true) or about 1 in every 10 tries.

I don't think so. 1/429 as the probability of 10 losses in 10 hands
of 9/6 Jacks is cool. However, you're saying you're looking at 470
independent trials. Those 470 "trials" overlap in this scenario and
if one fails then there's a strong likelihood that the next will fail.

I'm on weak ground here, but I haven't found a satisfactory way of
approaching a problem like this short of brute force.

- Harry

#9

Hi Harry

So are we saying we can expect one 10 game loss streak in each pack
of 429 games?

Jeep
.
.--- In vpFREE@yahoogroups.com, "Harry Porter" <harry.porter@...>
wrote:

Cogno Scienti wrote:
> << How about having 10 loser hands in a row 3 times in a 500 game
> session?>>
>
> Somebody can check me on this, but my first crack at this is
> 1-binomdist(2,470,1/429,true) or about 1 in every 10 tries.

I don't think so. 1/429 as the probability of 10 losses in 10 hands
of 9/6 Jacks is cool. However, you're saying you're looking at 470
independent trials. Those 470 "trials" overlap in this scenario and
if one fails then there's a strong likelihood that the next will

fail.

···

I'm on weak ground here, but I haven't found a satisfactory way of
approaching a problem like this short of brute force.

- Harry

#10

whitejeeps wrote:

So are we saying we can expect one 10 game loss streak in each pack
of 429 games?

No. The basic interpretation of the 1/429 probability that a 10-hand
sequence won't produce any winners is that if you played 429
independent sequences of 10 hands (i.e., each hand is part of only one
sequence), you'd expect one sequence to yield a no-winner streak.

- H.