- Yes the probability of a large "random" number being divisible by N would be expected to be about 1/N.

The last 114 could be left off since it's divisible by 19. That is

96 1 v2 v3...................v113 v

96 1 v2 v3...................v113 v 114

have the same remainder when divided by 19. I did do a computer check just to make sure using random permutations and there's no measurable difference of the probabiliy (of numbers of the above form being divisible by 19) from 1/19.

--- In mathforfun@yahoogroups.com, ynineteen <no_reply@...> wrote:

>

> 1/19?

> You mean every 19th trail hits the target?

> here we have fixed values start with 96 1 and variable "v" and at the end we have fixed 114.

> 96 1 v2 v3...................v113 v114

> Valuve for variable cards is ame 1 to 114 but 96 is talen out before shuffling.

> 19 cycle here is locked with fixed number 114

> --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:

> >

> > There are too many permutations to calculate the exact probability by enumeration on a computer but I'm 99% sure that the probability of that 468 digit number being divisible by 19 is very close to 1/19.

> >

> > In other words it's virtually a random number mod 19. This could be verified using a Monte Carlo program.

> >

> >

> > --- In mathforfun@yahoogroups.com, Edward Moore <emoore06905@> wrote:

> > >

> > > He's talking about string the digits of the cards together - count the number of digits between 1 and 114, and double it.Â I get 468, not 458.

> > >

> > > Ed

> > >

> > >

> > >

> > >

> > > ________________________________

> > > From: clooneman2000 <clooneman2000@>

> > > To: mathforfun@yahoogroups.com

> > > Sent: Wed, April 28, 2010 1:09:09 PM

> > > Subject: [MATH for FUN] Re: What would be % chances to win?

> > >

> > > Â

> > > So the second set of cards contains numbers from 1 to 114? Now forgive me if I'm misunderstanding this, but don't the numbers from 1 to 114 total to 6555, and not 458?

> > >

> > > --- In mathforfun@yahoogro ups.com, ynineteen <no_reply@ .> wrote:

> > > >

> > > >

> > > >

> > > >

> > > > Either set are printed 1 to 114.

> > > > one set in row from one to 114 and other set shuffled

> > > > I forgot to indicate that we pick number 96 before shuffelling.

> > > > 96 can be front or after fixed card number 1

> > > > 96 1 or

> > > > 1 96

> > > > Thank you for your concern

> > > >

> > > > --- In mathforfun@yahoogro ups.com, "clooneman2000" <clooneman2000@ > wrote:

> > > > >

> > > > > What are the values contained on the other set of 114 cards?

> > > > >

> > > > > --- In mathforfun@yahoogro ups.com, ynineteen <no_reply@> wrote:

> > > > > >

> > > > > >

> > > > > > There is 114 cards on the table in one row set from 1 to 114 in order,

> > > > > > All the card are displaced for inserting another card . Now we have

> > > > > > another set of 114 cards which we shuffle it and insert one at the times

> > > > > > starting before card or after . We insert all cards and now we have

> > > > > > large figure 458 digits . If this figure by multiple of 19 (114=19x6)

> > > > > > we are winner.

> > > > > >

> > > > > > 1-Would you show what would be % chances to win with one trial.

> > > > > >

> > > > > > 2- It it possible to use computer program to find the winner figure?

> > > > > >

> > > > > > Thank you

> > > > > >

> > > > > >

> > > > > >

> > > > > >

> > > > > >

> > > > > > [Non-text portions of this message have been removed]

> > > > > >

> > > > >

> > > >

> > >

> > >

> > >

> > >

> > >

> > >

> > >

> > > [Non-text portions of this message have been removed]

> > >

> >

> - --- In mathforfun@yahoogroups.com, "bogaduck" <pmaxotzen@...> wrote:
>

I tried it that way first before ynineteen specified the other way:(96 1 v2 v3...................v113 v 114) and it made no difference (using random permutations). The fraction divisible by 19 is always about 1/19

>

>

> --- In mathforfun@yahoogroups.com, "video_ranger" <video_ranger@> wrote:

> >

> > Yes the probability of a large "random" number being divisible by N would be expected to be about 1/N.

> >

> > The last 114 could be left off since it's divisible by 19. That is

> >

> > 96 1 v2 v3...................v113 v

> > 96 1 v2 v3...................v113 v 114

> >

> > have the same remainder when divided by 19.

> >

>

> It probably doesn't alter your general reasoning but those two above generally do not have the same remainder.

>

> eg if 96 1 v2 v3...................v113 v

> has a remainder of 8

> then 96 1 v2 v3...................v113 v 114

> has a remainder of 1.

>

> The fact they both have a remainder of some sort is the most important thing however.

>

> Not sure whether the option of placing the second pack of cards AFTER the first; to give

>

> 1 v1 2 v2 ...... 113 v 114 v alters the 1/19 option

>

> makes any difference.

>

> Peter

>