Hi Hardrockers,
        With the lottery looming large on the Hardrock calendar, I would like
to remind everyone that each year we select a piece of original art or a
photograph that we think captures the essence of the Hardrock Hundred for our
finishers print and promotional posters. If you have a piece of art or a
picture you would like us to consider please send it to us NO LATER THAN Fri.
Jan 18th.
   Please send your submissions to me at this e-mail address or if you want us
to have a hard copy send it to Hardrock Hundred , 195 Ball Lane, Durango, CO
81301.
Happy Holidays Everybody!!
Dale

Dale Garland
RD, Hardrock Hundred Endurance Run

Merry Christmas Dale!

--- On Wed, 12/5/12, Dale Garland <hardrock100@...> wrote:

From: Dale Garland <hardrock100@...>
Subject: [hr100] Hardrock finishers print for 2013
To: hr100@yahoogroups.com
Date: Wednesday, December 5, 2012, 9:09 PM
















 









       Hi Hardrockers,

        With the lottery looming large on the Hardrock calendar, I would like

to remind everyone that each year we select a piece of original art or a

photograph that we think captures the essence of the Hardrock Hundred for our

finishers print and promotional posters. If you have a piece of art or a

picture you would like us to consider please send it to us NO LATER THAN Fri.

Jan 18th.

   Please send your submissions to me at this e-mail address or if you want us

to have a hard copy send it to Hardrock Hundred , 195 Ball Lane, Durango, CO

81301.

Happy Holidays Everybody!!

Dale



Dale Garland

RD, Hardrock Hundred Endurance Run



























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

Maybe I missed it but what time does the lottery start on the 16th.
Thanks
Warren

Don't know I did not put in

Clark Fox

----- Reply message -----
From: "warren" <jacksorbetter4@...>
To: <hr100@yahoogroups.com>
Subject: [hr100] Time
Date: Mon, Dec 10, 2012 12:35 pm
Maybe I missed it but what time does the lottery start on the 16th.

Thanks

Warren











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

9 am MST, but it will probably be 10 am before we start pulling tickets, as it
takes a while to cut up a gazillion little pieces of paper.



- Blake



----- Original Message -----


From: "warren" <jacksorbetter4@...>
To: hr100@yahoogroups.com
Sent: Monday, December 10, 2012 12:35:56 PM
Subject: [hr100] Time

 




Maybe I missed it but what time does the lottery start on the 16th.
Thanks
Warren




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

Thank you,
Warren



________________________________
  From: "bpwrlc@..." <bpwrlc@...>
To: hr100@yahoogroups.com
Sent: Monday, December 10, 2012 8:06 PM
Subject: Re: [hr100] Time


 


9 am MST, but it will probably be 10 am before we start pulling tickets, as it
takes a while to cut up a gazillion little pieces of paper.

- Blake

----- Original Message -----

From: "warren" <jacksorbetter4@...>
To: hr100@yahoogroups.com
Sent: Monday, December 10, 2012 12:35:56 PM
Subject: [hr100] Time

 

Maybe I missed it but what time does the lottery start on the 16th.
Thanks
Warren

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




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

Blake can we volunteer to help and maybe earn an extra ticket or two for the
lottery? Any chance can help, really want to run HR 2013.

John Constan

Desire nothing and achieve everything in life!

On Dec 10, 2012, at 9:06 PM, bpwrlc@... wrote:

>
>
> 9 am MST, but it will probably be 10 am before we start pulling tickets, as it
takes a while to cut up a gazillion little pieces of paper.
>
> - Blake
>
> ----- Original Message -----
>
> From: "warren" <jacksorbetter4@...>
> To: hr100@yahoogroups.com
> Sent: Monday, December 10, 2012 12:35:56 PM
> Subject: [hr100] Time
>
>
>
> Maybe I missed it but what time does the lottery start on the 16th.
> Thanks
> Warren
>
> [Non-text portions of this message have been removed]
>
>


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

A gazillion?  I was figuring you were only going to pull a kajillion.  Be
grateful it could have been a katriliion.

If I had a bit more time and a little more of that youth that is quickly waning
it would have been a gazillion plus one......

Best of luck to all the owners of those little pieces of paper in the hat!
[cid:image002.jpg@...]
Jerry Bloom
Manager, Metrology Department
P.O. Box 13222, MS 5594
Sacramento, CA 95813
T: 916.355.2082
M: 916.508.2760
F: 916.355.6163
jerry.bloom@...



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

Okay, perhaps I exaggerated.  It's really only 2775 tickets in the three
lotteries put together.



- Blake



----- Original Message -----


From: "Jerry S Bloom" <jerry.bloom@...>
To: hr100@yahoogroups.com
Sent: Wednesday, December 12, 2012 8:01:05 AM
Subject: [hr100] Re: Time

 




A gazillion? I was figuring you were only going to pull a kajillion. Be grateful
it could have been a katriliion.

If I had a bit more time and a little more of that youth that is quickly waning
it would have been a gazillion plus one......

Best of luck to all the owners of those little pieces of paper in the hat!
[cid: image002.jpg@... ]
Jerry Bloom
Manager, Metrology Department
P.O. Box 13222, MS 5594
Sacramento, CA 95813
T: 916.355.2082
M: 916.508.2760
F: 916.355.6163
jerry.bloom@...

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




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

Blake,

when will you guys post on the website the ticket distribution amongst
wanna-be's? As in, how many each of us has, and how many people in each
pool?

OLga

On Wed, Dec 12, 2012 at 9:13 AM, <bpwrlc@...> wrote:

> **
>
>
>
>
> Okay, perhaps I exaggerated.  It's really only 2775 tickets in the three
> lotteries put together.
>
> - Blake
>
>


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

That should be up later today.



- Blake



----- Original Message -----


From: "Olga King" <olgav100@...>
To: hr100@yahoogroups.com
Sent: Wednesday, December 12, 2012 8:16:02 AM
Subject: Re: [hr100] Re: Time

 




Blake,

when will you guys post on the website the ticket distribution amongst
wanna-be's? As in, how many each of us has, and how many people in each
pool?

OLga

On Wed, Dec 12, 2012 at 9:13 AM, < bpwrlc@... > wrote:

> **
>
>
>
>
> Okay, perhaps I exaggerated. It's really only 2775 tickets in the three
> lotteries put together.
>
> - Blake
>
>

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




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

 
  Blake,
 
  Will be wild looking at the odds.  Looks like approx 1784 on the never
started group.  I think that is.05% per drawing or 1.75% chance per ticket in
simple math without taking out for the odds changing after someone is drawn. 
 
 80% of Vets and 37% of else group will get in.  Didn't count how many never
started group put in but their overall odds will be low.  Think I wish we had
given them 40 spots.   Sorry about the deluge of questions and emails you will
be getting.  Maybe you can put a statement out when you post numbers not to
email about the lottery unless it is a challenge of the number of tickets people
have.   Speaking of that you still have to add the volunteer tickets. 
 
Ken


________________________________
From: "bpwrlc@..." <bpwrlc@...>
To: hr100@yahoogroups.com
Sent: Wednesday, December 12, 2012 8:30 AM
Subject: Re: [hr100] Re: Time

 


That should be up later today.

- Blake

----- Original Message -----

From: "Olga King" <mailto:olgav100%40gmail.com>
To: mailto:hr100%40yahoogroups.com
Sent: Wednesday, December 12, 2012 8:16:02 AM
Subject: Re: [hr100] Re: Time

 

Blake,

when will you guys post on the website the ticket distribution amongst
wanna-be's? As in, how many each of us has, and how many people in each
pool?

OLga

On Wed, Dec 12, 2012 at 9:13 AM, < mailto:bpwrlc%40comcast.net > wrote:

> **
>
>
>
>
> Okay, perhaps I exaggerated. It's really only 2775 tickets in the three
> lotteries put together.
>
> - Blake
>
>

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

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




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

Hi Blake (and everyone),

Before you get showered with questions by those of us who are trying
to make sure our ticket allocation is accurate, can you please clarify
something:

First-timers appear to get 2^n but NOT get a ticket for being in the lottery
this year (in 2012).  Is that correct? (And then get additional service tickets
if appropriate, etc.)

Thanks,

Kari

Yes, you are correct.   The first-timers do not get an additional ticket for
applying this year.  n is the number of times you have previously applied and
met the qualification standard.



- Blake


----- Original Message -----


From: "Kari Fraser" <karifraser@...>
To: "\"“hr100-yahoogroups.com“\"" <hr100@yahoogroups.com>
Sent: Thursday, December 13, 2012 12:09:55 AM
Subject: [hr100] Lottery Clarification for First-Timers

 




Hi Blake (and everyone),

Before you get showered with questions by those of us who are trying
to make sure our ticket allocation is accurate, can you please clarify
something:

First-timers appear to get 2^n but NOT get a ticket for being in the lottery
this year (in 2012). Is that correct? (And then get additional service tickets
if appropriate, etc.)

Thanks,

Kari


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

Didn't see any discussion of our odds yet.  638 tickets in the "Everyone
Else" group. 70 slots... assuming drawn tickets are set aside...

1 ticket = 11.61%
2 = 23.22%
3 = 34.84%
4 = 46.45%
5 = 58.06%
6 = 69.67%

... is that right?


joe lea
chapel hill, nc


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

Joe,

    Realizing that my probability skills are extremely rusty and could stand
correction by the more math capable on the list, I'm getting about a 37.25%
chance of you getting in with four tickets in the "Everyone Else" category.  In
calculating 638 C 70, my thought process was as follows:

(638 - 4) / 638 = 99.373% chance of not getting in on the first draw
(637 - 4) / 637 = 99.372% chance of not getting in on the second draw
(636 - 4) / 636 = 99.371% chance of not getting in on the third draw
...
(569 - 4) / 569 = 99.297% chance of not getting in on the 70th draw

    The product of these percentage chances of not getting in is 62.749%, which
would leave a 37.251% chance of getting in (smoking my 7.556%).  Regardless of
the accuracy of my math, good luck with the draw!

Chris "Four Tickets, But The Wrong Category" Martin


________________________________
  From: Joe Lea <joelea@...>
To: bpwrlc@...
Cc: hr100@yahoogroups.com
Sent: Thursday, December 13, 2012 8:42 AM
Subject: [hr100] Odds?


 
Didn't see any discussion of our odds yet.  638 tickets in the "Everyone
Else" group. 70 slots... assuming drawn tickets are set aside...

1 ticket = 11.61%
2 = 23.22%
3 = 34.84%
4 = 46.45%
5 = 58.06%
6 = 69.67%

... is that right?

joe lea
chapel hill, nc

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

Joe's odds are a little higher than I get at the higher ticket numbers.
Below is what I get for all three lottery pools.
I followed more or less the same methodology as Chris Martin describes,
except after each winner's draw I removed the average number of tickets/person,
not just one ticket (because once a person wins, the rest of their tickets are
essentially removed too).
No guarantee these are right -- I'm definitely no expert in probability.
 
Vets:
#tickets chances(%)
 5            62.1
 6           68.9
 7           74.5
 8           79.2
 9           83.0
 10         86.1
 11         88.7
 12         90.8
 13         92.5
 14         93.9
 16         96.0
 17         96.8
 18         97.4
 
Never:
 #tickets chances(%)
 1            2.0
 2            3.9
 3            5.8
 4            7.7
 5            9.5
 8          14.8
 9          16.5
 10        18.2
 16        27.5
 32        47.6
 33        48.6
 64        72.8
 
Else:
#tickets chances(%)
 1          12.7
 2          23.9
 3          33.6
 4          42.1
 5          49.6
 6          56.1



>________________________________
> From: Joe Lea <joelea@...>
>To: bpwrlc@...
>Cc: hr100@yahoogroups.com
>Sent: Thursday, December 13, 2012 8:42 AM
>Subject: [hr100] Odds?
>
>
> 
>
>Didn't see any discussion of our odds yet.  638 tickets in the "Everyone
>Else" group. 70 slots... assuming drawn tickets are set aside...
>
>1 ticket = 11.61%
>2 = 23.22%
>3 = 34.84%
>4 = 46.45%
>5 = 58.06%
>6 = 69.67%
>
>... is that right?
>
>joe lea
>chapel hill, nc
>
>[Non-text portions of this message have been removed]
>
>
>

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

Hardrock runners -



A list of the number of tickets each applicant has in Sunday's Hardrock Hundred
lottery has been put online at http://hardrock100.com/HR-tickets-121212.pdf



Our intention in doing this is to give you the chance to help spot errors. 
We've double- and triple-checked our lists, but there are still occasional
problems when runners have the same names, compound names, or have changed their
name.



If you think we have made a mistake, please email me at
bwood@...<mailto:bwood@...> and
bpwrlc@...<mailto:bpwrlc@...>, and let me know how many tickets
YOU think you should have and why (e.g. "I was a DNS in 2010 and 2011 and DNFed
in 2012").  Please do not just say "I think you've made a mistake" or "I don't
understand your calculation".



Note that there are actually three lotteries:



- "Veterans" for anyone with >= 5 finishes.  You get one ticket for each finish.
There are 35 slots in this lottery.



- "Never-started" for anyone who has never started a Hardrock.  You get 2^N
tickets, where N is the number of times you have applied and met the
qualifications before this year.  There are 35 slots in this lottery.  This
provides a very strong bias for those who have been on the wait list multiple
years without getting in.



- "Everyone-else" for everyone not in the other two lotteries.  Tickets are
allocated the same as we did in the past few years, and the algorithm can be
found via the wayback machine at
http://web.archive.org/web/20101218033901/http://hardrock100.com/hardrock-lotter\
y.php  There are 70 slots in this lottery.



In addition, in all three lotteries, the top-5 men and women finishers from 2012
get one add'l ticket each.  There are also "service" tickets added in for those
who put in two days labor at our trail work weekend last year, captained an aid
station last year, or had a major role in course clearing last year.





- Blake



Blake Wood

President and lottery wonk, Hardrock Board of Directors


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

Ah, I took a glass half full view and summed the probability that I would
get in on each pull, which was 46.45% for my 4 tickets in the everyone else
category. I'm sure that's off though because of the variable number of
tickets that are invalidated after each new lottery winner (and maybe other
reasons).

Alas, Dale informs me that it is a moot discussion for my odds since I am
on the wrong (albeit, *winning*) side of the DUKE v UNC basketball rivalry.

joe lea
chapel hill, nc


On Thu, Dec 13, 2012 at 9:27 AM, Christopher Martin <
christopherhallmartin@...> wrote:

> **
>
>
> Joe,
>
>     Realizing that my probability skills are extremely rusty and could
> stand correction by the more math capable on the list, I'm getting about a
> 37.25% chance of you getting in with four tickets in the "Everyone Else"
> category.  In calculating 638 C 70, my thought process was as follows:
>
> (638 - 4) / 638 = 99.373% chance of not getting in on the first draw
> (637 - 4) / 637 = 99.372% chance of not getting in on the second draw
> (636 - 4) / 636 = 99.371% chance of not getting in on the third draw
> ...
> (569 - 4) / 569 = 99.297% chance of not getting in on the 70th draw
>
>     The product of these percentage chances of not getting in is 62.749%,
> which would leave a 37.251% chance of getting in (smoking my 7.556%).
>  Regardless of the accuracy of my math, good luck with the draw!
>
> Chris "Four Tickets, But The Wrong Category" Martin
>
> ________________________________
> From: Joe Lea <joelea@...>
> To: bpwrlc@...
> Cc: hr100@yahoogroups.com
> Sent: Thursday, December 13, 2012 8:42 AM
> Subject: [hr100] Odds?
>
>
>
>
> Didn't see any discussion of our odds yet. 638 tickets in the "Everyone
> Else" group. 70 slots... assuming drawn tickets are set aside...
>
> 1 ticket = 11.61%
> 2 = 23.22%
> 3 = 34.84%
> 4 = 46.45%
> 5 = 58.06%
> 6 = 69.67%
>
> ... is that right?
>
> joe lea
> chapel hill, nc
>
> [Non-text portions of this message have been removed]
>
>
>


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

The problem is that when you have a draw where people can have more than one
ticket the math is not so simple.

Suppose there are 1000 tickets in the pot and that people have either
1,2,4,8,16,32,or 64 tickets. I am considering the never been in pool where the
number of tickets is 2^n, and  therefore you get 1,2,4,8,16,32,... tickets for
n=0,1,2,3,4,5,...

If you have 4 tickets your chance of being selected in the first round is
4/1000.
If you are not selected your chance of being selected in the second round is not
necessarily 4/999.
It is 4/(1000-m), where m is the number of tickets of the person selected in the
first round, but the problem is that you don't know in advance what m is, it can
be 1,2,4,8,16,32,or 64. To get the chances of getting in you have to consider
all the possibilities. The same applies for each round. It becomes a mess
quickly.

I believe Blake has approximated this in the past with monte carlo simulations.
You could also just work through all the options, but regardless the simple
methods presented here will provide an underestimate of the odds of getting in
since they assume only one ticket is taken out per round, but it will in general
be more.

Best,

Brian

-----Original Message-----
From: hr100@yahoogroups.com [mailto:hr100@yahoogroups.com] On Behalf Of Jeff
List
Sent: Thursday, December 13, 2012 7:45 AM
To: hr100@yahoogroups.com
Subject: Re: [hr100] Odds?




Joe's odds are a little higher than I get at the higher ticket numbers.
Below is what I get for all three lottery pools.
I followed more or less the same methodology as Chris Martin describes, except
after each winner's draw I removed the average number of tickets/person, not
just one ticket (because once a person wins, the rest of their tickets are
essentially removed too).
No guarantee these are right -- I'm definitely no expert in probability.

Vets:
#tickets chances(%)
  5            62.1
  6           68.9
  7           74.5
  8           79.2
  9           83.0
  10         86.1
  11         88.7
  12         90.8
  13         92.5
  14         93.9
  16         96.0
  17         96.8
  18         97.4

Never:
  #tickets chances(%)
  1            2.0
  2            3.9
  3            5.8
  4            7.7
  5            9.5
  8          14.8
  9          16.5
  10        18.2
  16        27.5
  32        47.6
  33        48.6
  64        72.8

Else:
#tickets chances(%)
  1          12.7
  2          23.9
  3          33.6
  4          42.1
  5          49.6
  6          56.1

>________________________________
> From: Joe Lea <joelea@... <mailto:joelea%40gmail.com> >
>To: bpwrlc@... <mailto:bpwrlc%40comcast.net>
>Cc: hr100@yahoogroups.com <mailto:hr100%40yahoogroups.com>
>Sent: Thursday, December 13, 2012 8:42 AM
>Subject: [hr100] Odds?
>
>
>
>
>Didn't see any discussion of our odds yet. 638 tickets in the "Everyone
>Else" group. 70 slots... assuming drawn tickets are set aside...
>
>1 ticket = 11.61%
>2 = 23.22%
>3 = 34.84%
>4 = 46.45%
>5 = 58.06%
>6 = 69.67%
>
>... is that right?
>
>joe lea
>chapel hill, nc
>
>[Non-text portions of this message have been removed]
>
>
>

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

Isn't this a hypergeometric distribution?

So in Excel:

=HYPGEOMDIST(1,70,4,658)
Tom

--
Tom Stockton
(505) 660-8591


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

Easier to think the chance is 50%.

You either get in - or you won't.

Deep breaths until then.
Olga

On Thu, Dec 13, 2012 at 9:04 AM, Crone, Brian Keith <bcrone@...> wrote:

> **
>
>
> The problem is that when you have a draw where people can have more than
> one ticket the math is not so simple.
>
> Suppose there are 1000 tickets in the pot and that people have either
> 1,2,4,8,16,32,or 64 tickets. I am considering the never been in pool where
> the number of tickets is 2^n, and therefore you get 1,2,4,8,16,32,...
> tickets for n=0,1,2,3,4,5,...
>
> If you have 4 tickets your chance of being selected in the first round is
> 4/1000.
> If you are not selected your chance of being selected in the second round
> is not necessarily 4/999.
> It is 4/(1000-m), where m is the number of tickets of the person selected
> in the first round, but the problem is that you don't know in advance what
> m is, it can be 1,2,4,8,16,32,or 64. To get the chances of getting in you
> have to consider all the possibilities. The same applies for each round. It
> becomes a mess quickly.
>
> I believe Blake has approximated this in the past with monte carlo
> simulations. You could also just work through all the options, but
> regardless the simple methods presented here will provide an underestimate
> of the odds of getting in since they assume only one ticket is taken out
> per round, but it will in general be more.
>
> Best,
>
> Brian
>
> -----Original Message-----
> From: hr100@yahoogroups.com [mailto:hr100@yahoogroups.com] On Behalf Of
> Jeff List
> Sent: Thursday, December 13, 2012 7:45 AM
> To: hr100@yahoogroups.com
> Subject: Re: [hr100] Odds?
>
>
>
>
> Joe's odds are a little higher than I get at the higher ticket numbers.
> Below is what I get for all three lottery pools.
> I followed more or less the same methodology as Chris Martin describes,
> except after each winner's draw I removed the average number of
> tickets/person, not just one ticket (because once a person wins, the rest
> of their tickets are essentially removed too).
> No guarantee these are right -- I'm definitely no expert in probability.
>
> Vets:
> #tickets chances(%)
> 5 62.1
> 6 68.9
> 7 74.5
> 8 79.2
> 9 83.0
> 10 86.1
> 11 88.7
> 12 90.8
> 13 92.5
> 14 93.9
> 16 96.0
> 17 96.8
> 18 97.4
>
> Never:
> #tickets chances(%)
> 1 2.0
> 2 3.9
> 3 5.8
> 4 7.7
> 5 9.5
> 8 14.8
> 9 16.5
> 10 18.2
> 16 27.5
> 32 47.6
> 33 48.6
> 64 72.8
>
> Else:
> #tickets chances(%)
> 1 12.7
> 2 23.9
> 3 33.6
> 4 42.1
> 5 49.6
> 6 56.1
>
> >________________________________
> > From: Joe Lea <joelea@... <mailto:joelea%40gmail.com> >
> >To: bpwrlc@... <mailto:bpwrlc%40comcast.net>
> >Cc: hr100@yahoogroups.com <mailto:hr100%40yahoogroups.com>
> >Sent: Thursday, December 13, 2012 8:42 AM
> >Subject: [hr100] Odds?
> >
> >
> >
> >
> >Didn't see any discussion of our odds yet. 638 tickets in the "Everyone
> >Else" group. 70 slots... assuming drawn tickets are set aside...
> >
> >1 ticket = 11.61%
> >2 = 23.22%
> >3 = 34.84%
> >4 = 46.45%
> >5 = 58.06%
> >6 = 69.67%
> >
> >... is that right?
> >
> >joe lea
> >chapel hill, nc
> >
> >[Non-text portions of this message have been removed]
> >
> >
> >
>
> [Non-text portions of this message have been removed]
>
>
>
>
>
>


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

Anyone know if all ultrarunners are statistics junkies or just Hardrockers? I am
amazed at the math, guys! I take more of a sports fan point of view to the
lottery: I sit on my couch peeking at the lottery through my fingers afraid to
get my hopes up and my heart rate faster than if I were approaching the finish
line  with less than a minute to go.

Sent from my iPhone

On Dec 13, 2012, at 10:04 AM, "Crone, Brian Keith" <bcrone@...> wrote:

> The problem is that when you have a draw where people can have more than one
ticket the math is not so simple.
>
> Suppose there are 1000 tickets in the pot and that people have either
1,2,4,8,16,32,or 64 tickets. I am considering the never been in pool where the
number of tickets is 2^n, and therefore you get 1,2,4,8,16,32,... tickets for
n=0,1,2,3,4,5,...
>
> If you have 4 tickets your chance of being selected in the first round is
4/1000.
> If you are not selected your chance of being selected in the second round is
not necessarily 4/999.
> It is 4/(1000-m), where m is the number of tickets of the person selected in
the first round, but the problem is that you don't know in advance what m is, it
can be 1,2,4,8,16,32,or 64. To get the chances of getting in you have to
consider all the possibilities. The same applies for each round. It becomes a
mess quickly.
>
> I believe Blake has approximated this in the past with monte carlo
simulations. You could also just work through all the options, but regardless
the simple methods presented here will provide an underestimate of the odds of
getting in since they assume only one ticket is taken out per round, but it will
in general be more.
>
> Best,
>
> Brian
>
> -----Original Message-----
> From: hr100@yahoogroups.com [mailto:hr100@yahoogroups.com] On Behalf Of Jeff
List
> Sent: Thursday, December 13, 2012 7:45 AM
> To: hr100@yahoogroups.com
> Subject: Re: [hr100] Odds?
>
>
>
>
> Joe's odds are a little higher than I get at the higher ticket numbers.
> Below is what I get for all three lottery pools.
> I followed more or less the same methodology as Chris Martin describes, except
after each winner's draw I removed the average number of tickets/person, not
just one ticket (because once a person wins, the rest of their tickets are
essentially removed too).
> No guarantee these are right -- I'm definitely no expert in probability.
>
> Vets:
> #tickets chances(%)
> 5 62.1
> 6 68.9
> 7 74.5
> 8 79.2
> 9 83.0
> 10 86.1
> 11 88.7
> 12 90.8
> 13 92.5
> 14 93.9
> 16 96.0
> 17 96.8
> 18 97.4
>
> Never:
> #tickets chances(%)
> 1 2.0
> 2 3.9
> 3 5.8
> 4 7.7
> 5 9.5
> 8 14.8
> 9 16.5
> 10 18.2
> 16 27.5
> 32 47.6
> 33 48.6
> 64 72.8
>
> Else:
> #tickets chances(%)
> 1 12.7
> 2 23.9
> 3 33.6
> 4 42.1
> 5 49.6
> 6 56.1
>
> >________________________________
> > From: Joe Lea <joelea@... <mailto:joelea%40gmail.com> >
> >To: bpwrlc@... <mailto:bpwrlc%40comcast.net>
> >Cc: hr100@yahoogroups.com <mailto:hr100%40yahoogroups.com>
> >Sent: Thursday, December 13, 2012 8:42 AM
> >Subject: [hr100] Odds?
> >
> >
> >
> >
> >Didn't see any discussion of our odds yet. 638 tickets in the "Everyone
> >Else" group. 70 slots... assuming drawn tickets are set aside...
> >
> >1 ticket = 11.61%
> >2 = 23.22%
> >3 = 34.84%
> >4 = 46.45%
> >5 = 58.06%
> >6 = 69.67%
> >
> >... is that right?
> >
> >joe lea
> >chapel hill, nc
> >
> >[Non-text portions of this message have been removed]
> >
> >
> >
>
> [Non-text portions of this message have been removed]
>
>
>
>
>


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

So in Excel:

=HYPGEOMDIST(1,70,4,658)  = 30%

where

70 is the number of spots

4 is the number of tickets

658 is the number of tickets in the lottery

1 ticket: 10%

2 ticket: 19%

3 ticket: 26%

4 ticket: 30%

5 ticket: 34%
....




On Thu, Dec 13, 2012 at 8:07 AM, Tom Stockton <tbstockton@...> wrote:

> Isn't this a hypergeometric distribution?
>
> So in Excel:
>
> =HYPGEOMDIST(1,70,4,658)
> Tom
>
> --
> Tom Stockton
> (505) 660-8591
>



--
Tom Stockton
(505) 660-8591


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

every year, in Los  Alamos, (home of the Hardrock Lottery brain trust),
it's been sort of traditional to get together
around the lottery time for a party of sorts...  It has also been known
to happen that certain alcoholic beverages might
be specially labeled for the event... ie, "Hardluck Ale" beer, and
"Solace in a Shot" Tequila bottles.  Notice the leaning towards
"losing" in the lottery.

I thought you all might enjoy the special beer label we created for last
year's event.
https://picasaweb.google.com/110082310728448110031/HardrockLottery?authuser=0&au\
thkey=Gv1sRgCOKci72r4dClGA&feat=directlink

On 12/13/2012 8:35 AM, Tom Stockton wrote:
>
> So in Excel:
>
> =HYPGEOMDIST(1,70,4,658) = 30%
>
> where
>
> 70 is the number of spots
>
> 4 is the number of tickets
>
> 658 is the number of tickets in the lottery
>
> 1 ticket: 10%
>
> 2 ticket: 19%
>
> 3 ticket: 26%
>
> 4 ticket: 30%
>
> 5 ticket: 34%
> ....
>
> On Thu, Dec 13, 2012 at 8:07 AM, Tom Stockton <tbstockton@...
> <mailto:tbstockton%40gmail.com>> wrote:
>
> > Isn't this a hypergeometric distribution?
> >
> > So in Excel:
> >
> > =HYPGEOMDIST(1,70,4,658)
> > Tom
> >
> > --
> > Tom Stockton
> > (505) 660-8591
> >
>
> --
> Tom Stockton
> (505) 660-8591
>
> [Non-text portions of this message have been removed]
>
>



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

This is obviously just nervous thumb twiddling before the lottery, as random
chance can be very cruel or very kind.
 
But one possible way to evaluate the predictions is to see if the
calculated number of entrants equals the actual number of entrants accepted in
a lottery.
This can be found by multiplying the odds (divided by 100) by the number of
entrants for each ticket group, and then finding the sum of these numbers. Of
course a correct answer doesn't prove that the distribution of odds among
ticket classes is correct, but it's a good sign.
 
So for the everyone else lottery (for which the correct answer is 70)
the odds found so far give:
 
Source        entries predicted
Jeff                   68.4
Joe                    74.1
Tom                  50.4  (including the 6 ticket group not
reported by Tom = 37%)
 
So none of them appear to be correct.
 

>________________________________
> From: Tom Stockton <tbstockton@...>
>To: hr100@yahoogroups.com
>Sent: Thursday, December 13, 2012 10:35 AM
>Subject: Re: [hr100] Odds?
>
>
> 
>
>So in Excel:
>
>=HYPGEOMDIST(1,70,4,658)  = 30%
>
>where
>
>70 is the number of spots
>
>4 is the number of tickets
>
>658 is the number of tickets in the lottery
>
>1 ticket: 10%
>
>2 ticket: 19%
>
>3 ticket: 26%
>
>4 ticket: 30%
>
>5 ticket: 34%
>....
>
>On Thu, Dec 13, 2012 at 8:07 AM, Tom Stockton <mailto:tbstockton%40gmail.com>
wrote:
>
>> Isn't this a hypergeometric distribution?
>>
>> So in Excel:
>>
>> =HYPGEOMDIST(1,70,4,658)
>> Tom
>>
>> --
>> Tom Stockton
>> (505) 660-8591
>>
>
>--
>Tom Stockton
>(505) 660-8591
>
>[Non-text portions of this message have been removed]
>
>
>

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

So, we can fix my numbers by simply adding 4 more slots to the Everyone
Else group. All in favor...?

joe lea
chapel hill, nc


On Thu, Dec 13, 2012 at 12:52 PM, Jeff List <jlist1@...> wrote:

> **
>
>
> This is obviously just nervous thumb twiddling before the lottery, as
> random chance can be very cruel or very kind.
>
> But one possible way to evaluate the predictions is to see if the
> calculated number of entrants equals the actual number of entrants accepted
> in a lottery.
> This can be found by multiplying the odds (divided by 100) by the number
> of entrants for each ticket group, and then finding the sum of these
> numbers. Of course a correct answer doesn't prove that the distribution
> of odds among ticket classes is correct, but it's a good sign.
>
> So for the everyone else lottery (for which the correct answer is 70)
> the odds found so far give:
>
> Source        entries predicted
> Jeff                   68.4
> Joe                    74.1
> Tom                  50.4  (including the 6 ticket group not reported by
> Tom = 37%)
>
> So none of them appear to be correct.
>
>
> >________________________________
> > From: Tom Stockton <tbstockton@...>
> >To: hr100@yahoogroups.com
> >Sent: Thursday, December 13, 2012 10:35 AM
> >Subject: Re: [hr100] Odds?
>
> >
> >
> >
> >
> >So in Excel:
> >
> >=HYPGEOMDIST(1,70,4,658) = 30%
> >
> >where
> >
> >70 is the number of spots
> >
> >4 is the number of tickets
> >
> >658 is the number of tickets in the lottery
> >
> >1 ticket: 10%
> >
> >2 ticket: 19%
> >
> >3 ticket: 26%
> >
> >4 ticket: 30%
> >
> >5 ticket: 34%
> >....
> >
> >On Thu, Dec 13, 2012 at 8:07 AM, Tom Stockton <mailto:
> tbstockton%40gmail.com> wrote:
> >
> >> Isn't this a hypergeometric distribution?
> >>
> >> So in Excel:
> >>
> >> =HYPGEOMDIST(1,70,4,658)
> >> Tom
> >>
> >> --
> >> Tom Stockton
> >> (505) 660-8591
> >>
> >
> >--
> >Tom Stockton
> >(505) 660-8591
> >
> >[Non-text portions of this message have been removed]
> >
> >
> >
>
> [Non-text portions of this message have been removed]
>
>
>


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

Hilarious, Deb!  You are too creative.

Did Drew enter HRH this year?  I don't know his last name and I'm too lazy to
page through for Drew's from TX.

I also have a creative project for you.  I bought some blue t-shirts and want to
make an Avon Hardrock shirt to give away to people staying at the Avon.  I was
thinking a "Team Avon" theme maybe mentioning Tom (as coach?) and some kind of
catchy phrase.  I'm not so creative so I wanted to run it by you.

Hope things are going well for you guys.  I miss you two!

Scott

On Dec 13, 2012, at 11:06 AM, Deb's Gmail wrote:

> every year, in Los Alamos, (home of the Hardrock Lottery brain trust),
> it's been sort of traditional to get together
> around the lottery time for a party of sorts... It has also been known
> to happen that certain alcoholic beverages might
> be specially labeled for the event... ie, "Hardluck Ale" beer, and
> "Solace in a Shot" Tequila bottles. Notice the leaning towards
> "losing" in the lottery.
>
> I thought you all might enjoy the special beer label we created for last
> year's event.
>
https://picasaweb.google.com/110082310728448110031/HardrockLottery?authuser=0&au\
thkey=Gv1sRgCOKci72r4dClGA&feat=directlink
>
> On 12/13/2012 8:35 AM, Tom Stockton wrote:
> >
> > So in Excel:
> >
> > =HYPGEOMDIST(1,70,4,658) = 30%
> >
> > where
> >
> > 70 is the number of spots
> >
> > 4 is the number of tickets
> >
> > 658 is the number of tickets in the lottery
> >
> > 1 ticket: 10%
> >
> > 2 ticket: 19%
> >
> > 3 ticket: 26%
> >
> > 4 ticket: 30%
> >
> > 5 ticket: 34%
> > ....
> >
> > On Thu, Dec 13, 2012 at 8:07 AM, Tom Stockton <tbstockton@...
> > <mailto:tbstockton%40gmail.com>> wrote:
> >
> > > Isn't this a hypergeometric distribution?
> > >
> > > So in Excel:
> > >
> > > =HYPGEOMDIST(1,70,4,658)
> > > Tom
> > >
> > > --
> > > Tom Stockton
> > > (505) 660-8591
> > >
> >
> > --
> > Tom Stockton
> > (505) 660-8591
> >
> > [Non-text portions of this message have been removed]
> >
> >
>
> [Non-text portions of this message have been removed]
>
>



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

Interesting tid bit for those of us in the "First Timers" lottery.
The top 26 ticket holders in that lottery have 641 tickets or 35.6% of that
entire lottery. A little top heavy this year.

Eric Lee, one of the lucky (or unlucky) 26


--- In hr100@yahoogroups.com, Tom Stockton <tbstockton@...> wrote:
>
> So in Excel:
>
> =HYPGEOMDIST(1,70,4,658)  = 30%
>
> where
>
> 70 is the number of spots
>
> 4 is the number of tickets
>
> 658 is the number of tickets in the lottery
>
> 1 ticket: 10%
>
> 2 ticket: 19%
>
> 3 ticket: 26%
>
> 4 ticket: 30%
>
> 5 ticket: 34%
> ....
>
>
>
>
> On Thu, Dec 13, 2012 at 8:07 AM, Tom Stockton <tbstockton@...> wrote:
>
> > Isn't this a hypergeometric distribution?
> >
> > So in Excel:
> >
> > =HYPGEOMDIST(1,70,4,658)
> > Tom
> >
> > --
> > Tom Stockton
> > (505) 660-8591
> >
>
>
>
> --
> Tom Stockton
> (505) 660-8591
>
>
> [Non-text portions of this message have been removed]
>

In the ticket allocation list, it mentions service tickets. I see no mention of
service tickets and how they are earned or allocated in the Lottery Process page
on the website.

What are service tickets? Why they they mentioned now after the application
period has ended?

Thanks,
Mario