Newsgroup sci.stat.math 12437

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On 13 Jan 1997 02:25:48 GMT, Craig Franck  wrote:
>>>consecutive drawings if you buy 1 ticket per draw.  And even after 168
>>>draws, there still is 0.042398 probability of having lost all draws.
>>
>>That's saying 95.76 % chance of winning (>= 3) if playing the same 1 ticket
>>in 168 consecutive draws. IMHO an important conclusion from this would be:
>> Playing the same 1 ticket in x consecutive draws is better than playing 
>> x different tickets (or a wheel) in 1 draw. 
>>Isn't it?
>
>No. If the odds of winning a game are 1 in 54 million, then if you 
>buy 54 tickets, your odds of winning are 1 in 1 million. 
Sorry, I don't know what you exactly mean. What has this 54 million todo?
(I think you mean the "6/54" game, don't you?)
Then, why "1 in 1 million", shouldn't this be 54 in 54 million?
>If you buy 1 ticket for 54 drawings, your odds of winning never get above 
>1 in 54 million. 
You mean the single probability, but what about the probability over 
this period (ie. after 54 drawings). There is a well known formula for 
calculating of this, which I (and also Normand) have used to get the
above probability values.
I got also an interessting email from a statistician:
:Date: Mon, 13 Jan 1997 15:36:43 -0500 (EST)
:Message-ID: <970113150300_38246188@emout13.mail.aol.com>
:To: bm373592@muenchen.org
:Subject: Re: keno
:Content-Type: text
:
:In a message dated 97-01-11 10:08:54 EST, you write:
:
:<< In Lotto 6/49 the probability for at least 3 matching numbers is
:1/53.6551. My conclusion was: if 54 randomly chosen different tickets are
:played in 1 drawing   then one should expect at least 3 matching numbers.
:Does this hold in probability terms? And, is this the same as playing 1 fixed
:ticket (ie. always the same numbers) in 54 consecutive (random) drawings? >>
:
:For the first question it depends on what you mean by "expect." If you buy 54
:tickets your number of tickets with 3 matches has an approximate Poisson
:distribution with a mean of 1. That means the probability of no winning
:tickets is 36.79%, some remaining probabilities are shown below:
:
:0	36.79%
:1	36.79%
:2	18.39%
:3	6.13%
:4	1.53%
:5	0.31%
:6	0.05%
:
:On average you will have one winning ticket but you can see you will often
:have 0, 2 or 3 (4 or more is unlikely).
:
:Yes, it is the same as playing a fixed ticket in many drawings.
:
Any comments?
We have to seperate the things:
 - probability of 54 fixed  tickets in  1 drawing
 - probability of  1 fixed  ticket  in 54 drawings
 - probability of 54 fixed  tickets in 54 drawings
 - probability of 54 random tickets in  1 drawing
 - probability of  1 random ticket  in 54 drawings
 - probability of 54 random tickets in 54 drawings
(there are no duplicate tickets in each case)
Can someone calculate some or all of them?

On 14 Jan 1997 23:24:01 -0500, rhoads@sceloporus.rutgers.edu (Glenn Rhoads) wrote:
>bm373592@muenchen.org (Uenal Mutlu) writes:
>
>>LOTSIM - Simulation-Program for all pick-X type Lottery Games
>
>>[text deleted]
>
>> Draw numbers are generated by the standard RNG, ie. the rand()
>> function. Seed (srand(time)) is done once at pgmstart.
>
>Is this in C?
It's C++
>You should note that the pseudo-random number generator in most
>implementations of C is flawed.  C returns an unsigned integer but
>the rightmost bits of the number returned are NOT RANDOM!  Suppose
>C returns an integer in the range from 1 through 4,000,000,000 and
>you want to convert this to a number in the range from 1 through 49.
>Typically, programmers use the formula x = (n mod 49) + 1 to convert
>n, the number returned by pseudo-random number generator, to the
>number x of the desired range.  This is a bad thing to do in C.
>This formula emphasizes the bits on the right end, namely those bits
>that are not random.  To use the C's random number generator properly,
>you have to first get rid of the rightmost bits.  (e.g. x >>= 8 will
>get rid of the 8 rightmost bits)  If you want to do some serious
>simulations with lots of samples, you really shouldn't use any
>language's built-in generator and instead use a stronger generator.
>
>-- Glenn Rhoads
Thanks for the info, but I doubt this being still true with the 
newer C++ compilers. Let me know if someone knows if it is also the 
case with the Borland C++ compiler v4.52 in 32-bit target mode. 
Nevertheless I'll check the workaround you gave, though I don't know 
why even 8 (!) bits should be affected by this, whereas only 1 bit is 
required for the sign... 

jedhudson@cix.compulink.co.uk wrote on 15th jan
> Given the infinite XY plane 
> and a uniform random distribution of points on it with a density of N 
> per unit area,  what is the probability distribution of the distance 
> between each point and its nearest neighbor.
Problem now solved.  It has a very simple form:-
p(R) = k.R.exp(-R^2/density^2)
ie. the derivative of a Gaussian which takes about 5 lines to derive.
It is also easy to find the pdf of distance to N-th nearest neighbour.
JEH