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Darkbeer
09-17-2005, 01:15 PM
http://disneyland.disney.go.com/disneyland/en_US/calendar/specialEvents/detail?name=CoolestRideSpecialEventDetailPage

Here are the basics...

Each day.. (and a day is from 3:01 PM the prior day to 3 PM that day) 7 "winning" game pieces will be placed in the stack of non-winning pieces (overall, 1.8 million, or about 58,000 per day average, of course slow weekdays will have less attendance).

All folks 3 and older have a chance to play, so a family of four would get 4 game pieces.

Each of the seven daily winners get a pair of Deluxe AP's, and then get to participate at about 4:30 PM at the Main Entry Plaza for a chance to win a car.... Each of the 7 winners will draw a key, one of which will open the car. If less than 7 folks turn in their winning piece, then they will remove one "non-winning" key for each no-show.

There are NO other prizes except for the up to 7 daily winners of the pair of AP's, and the one car a day.

The turnstile CM's know, as stated on their computer screen, if it is your first entry of the day, and then those folks will get a game piece. Any additional entry (ParkHopper or return) does not get a game piece. The non-purchase clause allows someone to go to Guest Relations next to the DCA turnstiles and pick up a game piece, limit of one per day.

The game piece has a value of about 33 cents, based on the prizes compared to the odds. That value will go up on slow weekdays (better odds), but more than likely not more than a $1.

The best hope is for some REALLY bad weather, then show up at the park on that day. Otherwise, it won't be worth it to make a special trip to the park to just pick up a game piece. (The cost to get there in the vast majority of cases is more than $1.)

So let's look at the "odds" of winning. First off, there is no way to exactly determine the odds, since it is based on how many tickets are handed out in a "Prize Day", that is from 3:01 PM the day before to 3 PM the day of the giveaway.

The odds mention 1,800,000 tickets. We have no idea if they will use all 1,800,000, which works out to 58,065 people a day on average. We know that is too high for a normal off-season weekday. Last year (per the AB/ERA estimates) Disneyland Park had 13.3 million, and DCA had 5.6 million for a total of 18.9 million a year, or a average of 1,575,000 a month. Now we know the really peak times are Summer, and the last couple of weeks in December, where they have high numbers 7 days a week. In October, most weekdays are a lot less busy. So while October has some special events and promotions (such as Haunted Mansion Holiday), it seems that the total months attendance would be LESS than the monthly average. Then we need to factor in the boost Disneyland has received due to the 50th, but some of that can be offset by the drop in attendance that has been seen at DCA. So, I think the 1.8 million tickets is Disney's HIGHEST expectation of tickets needed, if everything, including this promotion creates a large spike in attendance.

To further this discussion, we need to make some presumptions....

Let's drop the 1.8 million to a more reasonable number, and say 1.5 million in attendance (Still close to last year's monthly average), and I think I am using a number that might be even a bit high.

Now we need to look at Special events....

October 1st and 2nd - Unofficial Gay Days

October 2nd - CHOC walk (These folks will not get a scratch-off ticket for the walk, but many of the folks will go to the park after the DtD ending event)

October 9th - Sunday of a Three Day Weekend

October 10th - Federal Holiday

Mickey's Halloween Event at DCA shouldn't have much of an effect, those tickets do not get a game piece, and most of these folks will be dressed up, and won't be able to go to Disneyland first.

Other than that, we should be able to break down attendance into a few catagories

Monday thru Thursday - Slow off-season days, with the exception of the Holiday - 16 "regular"

Friday - 3 "regular"

Saturday - 3 "regular"

Sunday - 3 "regular"

One more factor, the first day of the contest, no tickets will be handed out from 3:01 PM to closing the day before, only from Park Opening (aka 8:30 AM) to 3 PM, but since it is Saturday and one of the Gay Days, that won't make much of the difference. And anyways, the vast majority of folks have made their first entry into the park by 3 PM.

So, we need to break down the 1.5 million into these catagories, so here is an educated guess ...

Saturday, October 1st - Disneyland 67,000, DCA 13,000 = 80,000

Sunday, October 2nd - Disneyland 60,000, DCA 18,000 = 78,000

Friday, October 7th - Disneyland 60,000, DCA 12,000 = 72,000

Saturday, October 8th - Disneyland 65,000, DCA 15,000 = 80,000

Sunday, October 9th - Disneyland 62,000, DCA 15,000 = 77,000

Monday, October 10th - Disneyland 50,000, DCA 10,000 = 60,000

Regular Friday - Disneyland 47,000, DCA 10,000; 57,000 x 3 = 171,000

Regular Saturday - Disneyland 55,000, DCA 12,000; 67,000 x 3 = 201,000

Regular Sunday - Disneyland 50,000, DCA 12,000; 62,000 x 3 = 186,000

Regular Weekday (Mon-Thur) Disneyland 25,000, DCA 6,500; 31,500 x 16 = 504,000

Ok, that works out to 1.5 million guests for the month of October, and of course these numbers can change due to weather or other fators, it should be a fairly representive number...

So lets look at the "revised odds"....

We still have 271 "winning" cards, 7 per day

We still have a total prize value of $629,486.

But using the more realistic 1.5 million number, your odds of winning a prize has gone down to 1 in 5,535 to win a pair of AP's, and 1 in 48,387 to win a car on average, and the average game piece has a value of 42 cents.

So let's break it down per "catagory"... Of course, the odds of winning the car is 1 in "the daily attendance" (the first number).



Saturday, October 1st - 80,000, Winning Game Piece - 1 in 11,429, Game Piece value - 25 cents


Sunday, October 2nd - 78,000, Winning Game Piece - 1 in 11,143, Game Piece value - 26 cents


Friday, October 7th - 72,000, Winning Game Piece - 1 in 10,286, Game Piece value - 28 cents


Saturday, October 8th - 80,000, Winning Game Piece - 1 in 11,429, Game Piece value - 25 cents


Sunday, October 9th - 77,000, Winning Game Piece - 1 in 11,000, Game Piece value - 26 cents


Monday, October 10th - 60,000, Winning Game Piece - 1 in 8,571, Game Piece value - 34 cents


Regular Friday - 57,000, Winning Game Piece - 1 in 8,143, Game Piece value - 36 cents


Regular Saturday - 67,000, Winning Game Piece - 1 in 9,571, Game Piece value - 30 cents


Regular Sunday - 62,000, Winning Game Piece - 1 in 8,857, Game Piece value - 33 cents


Regular Weekday (Mon-Thur) - 31,500, Winning Game Piece - 1 in 4,500, Game Piece value - 64 cents



So, as you can see, going a slow day does help your odds, but still don't go to the park just to win the car.

One more number, less say we have a day of really bad weather, and say only 14,000 folks show up...

Odds of getting a winning game piece, 1 in 2,000, Game Piece Value would be $1.45.

So go have fun in the park, and if you get lucky, you get lucky...

Maus
09-17-2005, 01:47 PM
I think my head is going to explode. :rolleyes:

I'm planning on going Oct 16. Everyone else stay away. :p

Wendi
09-17-2005, 01:54 PM
I hope I win... but I certainly won't get my hopes up or anything!!!

So, are you saying that if you have a winning gamepiece at 3:30pm on Friday, you will need to stay until Saturday afternoon if you want to vie for the car??? So, if your last day of vacation is Friday... they'll keep you an extra day? (I know I wouldn't leave if I had a 1:7 (or better) chance of winning a car).

Oh... and if you've already got an AP, can you give it to someone else or use it towards your next AP after the current one expires???

Wendi
09-17-2005, 02:01 PM
Just another tidbit:

A Second Chance to Win!
Beginning October 1, 2005, register with disneyland.com for the "second chance" drawing and your opportunity to win an all-new 2006 Honda Civic from Honda!

Darkbeer
09-17-2005, 02:02 PM
I hope I win... but I certainly won't get my hopes up or anything!!!

So, are you saying that if you have a winning gamepiece at 3:30pm on Friday, you will need to stay until Saturday afternoon if you want to vie for the car??? So, if your last day of vacation is Friday... they'll keep you an extra day? (I know I wouldn't leave if I had a 1:7 (or better) chance of winning a car).

Oh... and if you've already got an AP, can you give it to someone else or use it towards your next AP after the current one expires???

The rules state if you can't be there for the "ceremony/key contest", Disney will provide a "stand-in", so you would get your 1 in 7 chance too win the car(or maybe better if their is a no-show or two, and some folks might just throw it away before scratching it, or don't speak english and won't understand the game piece).

The AP's will more than likely be certificates, which can be transferred to others, or hanged onto to be used by you at a later date.

Darkbeer
09-17-2005, 02:04 PM
Just another tidbit:

A Second Chance to Win!
Beginning October 1, 2005, register with disneyland.com for the "second chance" drawing and your opportunity to win an all-new 2006 Honda Civic from Honda!

This I don't understand, the ONLY way a car would be available in the Second Chance Drawing (and the AP's are NOT part of the Second Chance) is if ALL 7 winning pieces are not turned in one the same day, which is very unlikely. If that happens, then the second chance will award that car... NOT very likely.

But it won't hurt to register (Only once per person) AFTER October 1st....

olddumbguy
09-17-2005, 02:41 PM
Are game pieces given for paid child admissions, or just adult admissions? If my 4 year old daughter won, would I be able to win the car myself?

olddumbguy
09-17-2005, 02:42 PM
My last post really makes me look like a greedy bastard, doesnt it?

vmjess
09-17-2005, 02:55 PM
I don't think you looked bad in your post- you'll be your daughters taxi for at least the next 12 years so if she wins the car it is only right that you get to drive it! :~D

hlbtimes2
09-17-2005, 03:39 PM
Are game pieces given for paid child admissions, or just adult admissions? If my 4 year old daughter won, would I be able to win the car myself?


Not only do you get it, but you get to pay the taxes on it!




(yes, I know that is not really a bonus.)

hlbtimes2
09-17-2005, 03:40 PM
I knew there was a reason I planned a trip for mid Oct! Counting down the days until I win free AP's and a new car....................

Hank919
09-17-2005, 06:09 PM
Well I want my share of the chances

SpacedOutCM
09-18-2005, 12:41 AM
On the billboard, looks like Disney still uses the word Esplanade to refer to the area between Disneyland and Disney's California Adventure.

cstephens
09-18-2005, 09:44 AM
For Disneyland's a car (Honda Civic) a day giveaway in October, there's a display in the esplanade between DL and DCA that includes a smaller (compared to the real one) model of Space Mountain which has a Honda Civic on either side. Next to the model is a video screen that shows video of various present and past attractions (I didn't sit through the whole cycle, so I don't know what the whole thing says), which means even though the Block Party Bash cone is gone, the esplanade is still a very loud place.

Here's a picture of the sign:
http://www.colddeadfish.net/images/DLR/critg2.jpg.

Here's a picture of the display:
http://www.colddeadfish.net/images/DLR/critg1.jpg.

Here's a closeup of the "license plate" on the cars:
http://www.colddeadfish.net/images/DLR/critg3.jpg.

hlbtimes2
09-18-2005, 10:09 AM
If I win a car I'm going to try for the "spacmtn" vanity plates. I wonder if anyone here in WA has that already? (thinking out loud).

I asked my 3 year old if he won if I can have his car. He said "NO, I know how to drive!" LOL

Alex S.
09-18-2005, 04:39 PM
Unless I am misreading Darkbeer's analysis, it appears he is assuming that they will distribute the winning 7 tickets among the actual number of cards issues in a given day.

I don't think that is likely. I would assume the 1,800,000 are evenly spread across the 31 days and then the 7 are randomly distributed among each days alotment. Meaning that there isn't any guarantee that all 7 will be distributed in any given day (or, even, that any of the 7 tickets will be distributed on a given day if random chance puts all 7 at the end of the alotment).

If this is the case then, the level of attendance doesn't affect your chance of getting one of the 7 tickets but low attendance would increase the chance of winning the car if you first win a ticket (since odds are bettre that all 7 will not be given out).

Also affecting the odds considerably are what percentage of the park guests are from countries other than the U.S., Canada, and Mexico. Other countries are not eligible and someone from that country getting a winning ticket would not be able to cash in on it (so unless they passed it on to someone eligible that ticket would be tossed).

MrTomMorrow
09-18-2005, 04:44 PM
Wow, great analysis. The odds of winning are still small, even on a slow day, but hope somebody from MousePlanet wins!

Darkbeer
09-18-2005, 05:06 PM
Unless I am misreading Darkbeer's analysis, it appears he is assuming that they will distribute the winning 7 tickets among the actual number of cards issues in a given day.

I don't think that is likely. I would assume the 1,800,000 are evenly spread across the 31 days and then the 7 are randomly distributed among each days alotment. Meaning that there isn't any guarantee that all 7 will be distributed in any given day (or, even, that any of the 7 tickets will be distributed on a given day if random chance puts all 7 at the end of the alotment).


Also affecting the odds considerably are what percentage of the park guests are from countries other than the U.S., Canada, and Mexico. Other countries are not eligible and someone from that country getting a winning ticket would not be able to cash in on it (so unless they passed it on to someone eligible that ticket would be tossed).

It is seven tickets per DAY, as it is KEY for the daily 4:30 PM key ceremony at the Main Entry Plaza...

From the Official Rules....




Each day during the Game Period, seven (7) winning Game Pieces will be randomly distributed amongst all non-winning Game Pieces.






If you have a Game Piece indicating that you are a winner on that same date, you MUST report to either City Hall in DisneylandŽ park, or Guest Relations in Disney's California AdventureŽ park, prior to 4 pm. If you have a winning Game Piece indicating that you are a winner for the following day, you must report to Guest Services at least 30 minutes prior to the park closure. Failure to report to the specified location by the specified times may result in forfeiture of participation in the Final Round, which will be held each day during the Game Period at 4:30 pm.






Each day October 1-31st, all eligible First Prize Winners will be invited to participate in the Final Round scheduled to take place at 4:30 pm. All guests with Winning Game Pieces must check in at either City Hall in DisneylandŽ park, or Guest Relations in Disney's California AdventureŽ park no later than 4:00 pm on the scheduled event day. During the Final Round, each participant will be randomly distributed one (1) of up to seven (7) keys and the chance to play for the Daily Grand Prize Package [One (1) Honda Civic, as detailed under the "Prize" heading.] The automobile will be awarded to the Final Round participant in possession of the winning key.






Grand Prize Packages (31): Up to one (1) 2006 Honda Civic with standard factory equipment will be awarded per day via the Final Round during the Game Period (31 days). Each Grand Prize has an ARV of $17,100. Odds of winning in the Final Round will depend on the number of people participating, but are no greater than 1:7.



And I have a lot of photos of the Main Entry Plaza display, plus a lot more at my photo website (link below)....

Alex S.
09-18-2005, 05:51 PM
But it doesn't say that all game pieces for that day will be distributed.

The only way they could guarantee that all 7 winners would be distributed is if they had someone manually inserting the winning tickets into the mix through the day, which either sounds like a very expensive proposition (paying for independent people to do/observe this) or somewhat prone to fraud.

And not to mention it would be random.

Darkbeer
09-18-2005, 06:26 PM
But it doesn't say that all game pieces for that day will be distributed.

The only way they could guarantee that all 7 winners would be distributed is if they had someone manually inserting the winning tickets into the mix through the day, which either sounds like a very expensive proposition (paying for independent people to do/observe this) or somewhat prone to fraud.

And not to mention it would be random.

I agree that there are some issues, but it is clear that a MAXIMUM of 7 Winning Game pieces will be in the stack of distributed tickets for a "game day", that is from 3:01 PM from the prior day, to 3 PM the current day.

The ONLY way this will work is to have ticket stacks for EACH day. There are two ways to do this. One is to use the Disney estimates made a few weeks in advance, and then at the game piece "factory", they are inserted into the stack of game pieces, and then shrink wrapped and marked for that day (a slow weekday having a lot less pieces than a busy weekend day), and if needed, just add more "non-winning" tickets to the stack of tickets if the crowd is larger than normal. If not all tickets are distributed, then the rules are designed to take into consideration the fact that not all 7 tickets are turned in. And the second chance if ALL 7 are not turned in, but Disney and Honda does NOT want that, as the 4:30 ceremony would be cancelled.

The more likely situation is that a independent third party controls the game pieces, and around Noon, gets a estimate from Disney for the amount of guests expected (along with the estimated "non-purchase" tickets) for the following day. This third party prepares the allotment by inserting the winning tickets into the lot, then delivers the entire amount to the Front Gate Managers, who then divide the tickets and assign them to the gate CM's. The multiple levels should prevent any sort of tampering or fraud.

After the problems McDonalds had with its Monopoly game, I bet there are a lot of folks for all sides (Disney, Honda and the game piece manufacture Visionworks Promotion, though being based in Celebration, Florida, I am not sure how independent they are...) watching to make sure there can't be any real hanky-panky.

Alex S.
09-18-2005, 06:45 PM
Personally, I think the former of those two are more likely, but each of these things shifts the odds. And the rules do not give a hedge on the odds as they normally would if there (and as they are hedged in the 1:7 case). It is stated simply that the odds are 1:1.8 million.

So, what says to me is that 1.8 million game pieces have been printed and that 217 winners have been distributed among the daily alotments.

Now, while they may adjust the daily alotments for exptected daily attendance, it would mean that as Disney falls short of 1.8 million for the month it means more of the 217 will not be distributed. But this does not change overall odds.

But it is certainly possible I'm wrong. I tried to find a phone number for Visionworks Promotions but haven't been successful yet (and a reverse address lookup doesn't produce any likely candidate). As for their independence, if search their names you'll find that they do run contests for non-Disney entities but I don't know know if Disney is their primary customer.

Darkbeer
09-18-2005, 07:07 PM
The biggest issue is the fact that the rules state (Paragraph 5) "are no greater than 1:7".

If they just printed 217 winning tickets, and inserted them into 1.8 million tickets, then more than likely some days more than 7 tickets would be distributed within a 24 hour period (a busy day). This would not work.

But for example, I know that Lottery Scratcher tickets are packaged as "bricks", usually around 200 tickets equals a "brick". For Disney, it would make more sense to do bricks of 500 or 1000. The independent person that works for Visionworks could be given a box of 217 "bricks", in which one winning game piece has been placed somewhere within that set of tickets. Then when they get the estimated attendance for the the next day, he/she takes the amount of needed tickets (say 50,000 expected, with a slight margin of overestimation to prevent the tickets running out). If there were bricks of 500, he/she would take 93 bricks that had no winning ticket, plus add the 7 "winning" bricks in a random placement within the order. Since all the bricks look the same, the DLR managers would just hand out the bricks to the turnstile CM's, and that should prevent fraud for multiple levels.

But I don't think we will get to know more until the actual game starts on October 1st....

TowerofTerror
09-18-2005, 07:37 PM
great odds thanks for the breakdown

Alex S.
09-18-2005, 08:00 PM
I'm not saying that they randomly inserted the 217 throughout the entire 1.8 million.

I was saying that they divided the 1.8 million into 31 lots and then inserted 7 into each lot and that they're doing this at the point of manufacture rather than having a person do it each day at the park.

It is possible that they've adjusted each days alotment of game pieces based on expected attendance, but if so that it is on an assumption that almost certainly will overestimate each day's attendance (as falling short would either mean you stop giving out any game cards until 3 p.m. or you take cards from another days alotment, or there is a pool of extra non-winners; and besides, it would mean some awkward shuffling of game cards from location to location as you start to run out).

If they've just created 31 equal alotments then individual daily attendance has no impact on daily odds. If they created 31 unequal alotments then the actual odds vary from day to day (as you presented) and I believe that they would be required to disclose such (since they are essentially running the same contest 31 times, not one cumulative contest).

But either way, there are fair odds that one any given day at least one of the winning tickets will not be distributed and that at least one will be distributed to a non-eligible player (someone from outside of U.S., Canada, and Mexico).

Unless they are going to the trouble of having someone on site each day intermingle the winning cards and ensuring that they get distributed. Which is certainly possible but strikes me as unlikely. But I've been wrong before.

I also wonder how they'll manage the 3:00 p.m. shiftover from one batch of game cards to the next. Will they actually hold up all the open gates at the same moment to switch over? (Not that many gates are open at a normal 3 p.m.)

Darkbeer
09-19-2005, 07:42 AM
If they've just created 31 equal alotments then individual daily attendance has no impact on daily odds.

This is not true.... While the odds of winning a pair of AP's would not be influenced by attendance, the chance to win the car is better on slow weekdays than a busy day.

Let's say that they take the full 1,8 million tickets and divide it up into 31 equal stacks of 58,065 tickets. That means that a winning ticket would be handed out once for each 8,295 tickets handed out. But let's use my "approximate" guest counts.

On a slow weekday (31,500) only about 4 winning tickets would be handed out, and that would make the odds of winning the car 1 in 4 if you have a winning ticket, and still 1 in 31,500 for that day overall, since one person at the park should win the car (unless nobody turns in a winning ticket). Remember, the rules state if less than 7 people turn in a winning ticket, they will remove a "non-winning" key for each no-show. On a busy day, all 7 winning tickets would be distributed, and then the odds would be 1 in 58,065 to win the car.