Pointless Prize 2023 – Bicycle Bingo

[bigredboiler]

GAME #2 will start immediately following confirmation of a Game#1 winner….join anytime / read details below & post to let me know any questions

Final Updates & Reminders before the Game Starts (Monday, Jan 16):

• 4 people created invalid cards (see your name below)….so I created NEW bingo cards for you….see the Google Sheet to get your URL
• I’ll post each days “Object” (like a Bingo #) in-line on this forum + you can check it (and whole history of what’s been called) on the Google Sheet
• When I post the “Object” you need to find, I’ll attempt to clarify what you need to find/photograph to avoid any confusion….if you still have questions, post to the thread or private msg me
• You can find the object on your ride starting the day the object is posted/available for finding AND any day after that
• You can only find/claim ONE OBJECT PER DAY
• You must post image of your object found here in the forum for all to see
• You call “BINGO” by posting here in this forum thread.
• You will need to make your Strava available to me to validate that you won (you can let me know your Strava name here in the forum; anytime or wait until you win

===== EARLIER POST & INFO BELOW =======

Happy 2023 everyone, and this year I have a new game to try. It won’t start until January 16 to give people time to get a BINGO card if they want to play.

Basic premise: ** Rule Updated Jan 12: You can claim a Bingo spot on or after the day that “thing” is called **

• Daily (starting Jan 16) I will announce the proverbial “bingo number” to the forum (which will actually be an object/thing)
• If a player finds and photographs the “thing” on their ride that day (or a future day), they mark that spot on their Bingo card
  
• You can only claim one spot per day
• First player to find 5 things making “BINGO” wins

How to you get your BINGO card?
Option #1: Generate your own & post to this thread to share –> Click this link and then the “Generate Card” button that you will see on that page…the result will be the URL to your BINGO card.
Option #2: I will generate a card/URL for you and share to this thread & the Google Sheet for tracking the game. Just post asking me for a URL and then check the forum/Google Sheet for your URL.

*** NOTE *** If the card you generate contains a “FREE” Space, please generate a new card and post that to the forum and use for the game.
My apologies. I thought I chose setting without a Free space, but apparently, the random card generator still includes Free in the generator given that this has happened to a few people already.

Players who generated an invalid card: I created a new link for you…check the Google Sheet for the link you need to play the game:

• amanka
• Kbikeva
• HokieBeth
• LGRides

If you’re on this list and haven’t created a valid card by Jan 14/15, I will create a card/URL for you and post here + Google Sheet.

You can also check the Google Sheet I’ll use for tracking to confirm I have your valid URL reported for your card.

Key additional info (and I’ll probably add more later to clarify based on questions I get):

• You must post the URL of your unique Bingo card to this thread (I will create a Google sheet to share everyone’s card)
• You must post the photo of the object taken during your ride to this thread
• You must include the photo in your Strava ride for the day and #Bingo in your ride title
• First person to announce BINGO to this thread (and I verify their entries) wins
• If the game ends quickly, I’ll do another round (it’s all new so we’ll see how it goes)

Other thoughts:

• I plan to post the day’s Bingo# late the prior evening
• Most objects are likely self-explanatory, and I will provide a little more detail with each day’s announcement. Also feel free to message me if you have questions.
• There are 75 objects/items in total and 25 are randomly selected for each player’s Bingo card
• Here is the link to the Google sheet I’ll use to track what’s been called, and everyone’s card
• If you win, I’ll verify you submitted valid photos on the right day when item was called
• If you want to cheat and ruin the fun for everyone, I’m sure you can.
• Prize will likely be a gift card of some kind (last year’s winners were to breweries and bike shops)

Have fun and let me know any questions! -Kris

Topic

[camiller]

Christmas Decoration

[ATTACH=CONFIG]29979[/ATTACH]

[AlanA]

@LhasaCM 223478 wrote:

Since I’m too lazy to do this the rigorous (or right) way but still wanted the distraction – some hand wavy calculations in the other direction:
1. Each card has 25 words out of a bank of 75, so you have 1/3 of the available words on your card.
2. That means 2/3 of the words are missing from your sheet.
3. Let’s just assume that each day is like a weighted coin flip – you’ve picked heads, but you only have a 1/3 chance of landing a heads, and 2/3 chance of landing tails. (This is the worst assumption of all because it treats each day’s odds as an independent event, but it’s early enough that this isn’t completely out of this world.) What are the odds you only got 2 heads in 14 attempts with this “unfair” coin?

Then abusing this as a binomial distribution, the “expected” number of words chosen to date you would have on your card would be between 4 and 5. Having 2 (or fewer) would be somewhat unlikely, but still a bit more than 1:10 (just over 10.5% for having 0, 1, or 2 matches, with exactly 2 being 7.8%)

Where the more rigorous calculation gets annoying is that, over 14 days, there are a total of 91 combinations of ways that you could have 2 matches just in terms of “matched on days x and y” and 16,384 ways you can arrange the simple “did I match” tied to each day. There are plenty of websites that go into detail on regular Bingo calculations (EV for the number of turns before a BINGO is called is apparently 41), but the math here isn’t quite the same since I think it’s one bank of 75 words rather than 5 banks of 15 words each for each column.

Whoa! That’s deep.

[CBGanimal]

@Steve O 223475 wrote:

Hey statistics gurus – what are the odds that after 14 plays, only two appear on a random card?
’cause that’s me.

:confused:
.

Well I have you beat! I only have one and only got that yesterday! But I’m on my way to getting the next 4 in a row…just watch!

[drevil]

Flag (military)

[drevil]

@drevil 223505 wrote:

Flag (military)

[AlanA]

@CBGanimal 223502 wrote:

Well I have you beat! I only have one and only got that yesterday! But I’m on my way to getting the next 4 in a row…just watch!

Haha! I’m only playing this game because I saw Post Office on your card. After finding out I had a ‘free’ space, I had to change cards. I kept changing until I saw another with a Post Office (that’s the only reason I’m playing!).

I’m quite sure I won’t win (I rarely win), but I have 7 on my card!!

[bikingjenn]

@LhasaCM 223478 wrote:

Since I’m too lazy to do this the rigorous (or right) way but still wanted the distraction – some hand wavy calculations in the other direction:
1. Each card has 25 words out of a bank of 75, so you have 1/3 of the available words on your card.
2. That means 2/3 of the words are missing from your sheet.
3. Let’s just assume that each day is like a weighted coin flip – you’ve picked heads, but you only have a 1/3 chance of landing a heads, and 2/3 chance of landing tails. (This is the worst assumption of all because it treats each day’s odds as an independent event, but it’s early enough that this isn’t completely out of this world.) What are the odds you only got 2 heads in 14 attempts with this “unfair” coin?

Then abusing this as a binomial distribution, the “expected” number of words chosen to date you would have on your card would be between 4 and 5. Having 2 (or fewer) would be somewhat unlikely, but still a bit more than 1:10 (just over 10.5% for having 0, 1, or 2 matches, with exactly 2 being 7.8%)

Where the more rigorous calculation gets annoying is that, over 14 days, there are a total of 91 combinations of ways that you could have 2 matches just in terms of “matched on days x and y” and 16,384 ways you can arrange the simple “did I match” tied to each day. There are plenty of websites that go into detail on regular Bingo calculations (EV for the number of turns before a BINGO is called is apparently 41), but the math here isn’t quite the same since I think it’s one bank of 75 words rather than 5 banks of 15 words each for each column.

I just love the math/numbers/stats geeking out conversation. Thinking someone on this string has access to some stats software and knows how to model the problem? I am stinking at this game. Only five of the called items have appeared on my card.

Sent from my iPhone using Tapatalk

[Indiana]

Named pond/lake
[ATTACH=CONFIG]29986[/ATTACH]
Constitution Gardens Pond
(with apologies to the duck persistently voicing disapproval of a nighttime visitor)

[bigredboiler]

Jan 31 (Day #16): Blue USPS Drop Box

Additional clarifying info:
Hopefully you know what this is (and can find one

Reminders:

• you can also find each day’s “Bingo#” on the Google Tracking Sheet
• you can find all the rules on the first post in this thread

[bigredboiler]

@AlanA 223495 wrote:

Pond/Lake (named).

I had to recheck the requirements for this one. Nothing said it had to have water in it.

This is Burn Lake in Las Cruces, NM. Just to be safe, I’m including the sign.

[ATTACH=CONFIG]29977[/ATTACH]

[ATTACH=CONFIG]29978[/ATTACH]

Haha….no surprise out West (but sad — Absolutely meets requirements. Just checked it out on Google Maps….beautiful interstate location

[Steve O]

@LhasaCM 223478 wrote:

Since I’m too lazy to do this the rigorous (or right) way but still wanted the distraction – some hand wavy calculations in the other direction:
1. Each card has 25 words out of a bank of 75, so you have 1/3 of the available words on your card.
2. That means 2/3 of the words are missing from your sheet.
3. Let’s just assume that each day is like a weighted coin flip – you’ve picked heads, but you only have a 1/3 chance of landing a heads, and 2/3 chance of landing tails. (This is the worst assumption of all because it treats each day’s odds as an independent event, but it’s early enough that this isn’t completely out of this world.) What are the odds you only got 2 heads in 14 attempts with this “unfair” coin?

Then abusing this as a binomial distribution, the “expected” number of words chosen to date you would have on your card would be between 4 and 5. Having 2 (or fewer) would be somewhat unlikely, but still a bit more than 1:10 (just over 10.5% for having 0, 1, or 2 matches, with exactly 2 being 7.8%)

Where the more rigorous calculation gets annoying is that, over 14 days, there are a total of 91 combinations of ways that you could have 2 matches just in terms of “matched on days x and y” and 16,384 ways you can arrange the simple “did I match” tied to each day. There are plenty of websites that go into detail on regular Bingo calculations (EV for the number of turns before a BINGO is called is apparently 41), but the math here isn’t quite the same since I think it’s one bank of 75 words rather than 5 banks of 15 words each for each column.

I believe this is analogous to the ball drawing problem:

There are 75 balls in a bag, 25 are white and 50 are rainbow. In drawing 14 balls, what are the odds that 2 drawn are white. Or, if you are CBGAnimal, only 1 is white.

I know how to do this problem with all one color or the other, but I’m not sure how to think about the general problem: what are the odds of drawing x white balls if you draw z balls from the bag?

[Steve O]

@bikingjenn 223513 wrote:

I just love the math/numbers/stats geeking out conversation. Thinking someone on this string has access to some stats software and knows how to model the problem? I am stinking at this game. Only five of the called items have appeared on my card.

Actually 5 out of 15 is probably the most common expected value at this point in the game, since each card has 1/3 of the items. The further you get away from 5, the less likely that outcome is.

[karenbikes2@gmail.com]

1/31/2023 Flag (military)
[ATTACH=CONFIG]29991[/ATTACH]
This flies outside the Bundeswehr-German Armed Force Command in Reston (11150 Sunset Hills Rd). In the photo just to the left of the bottom of the flagpole you can see a section of the Berlin Wall which stands just outside the building entrance.

[LhasaCM]

@Steve O 223525 wrote:

I believe this is analogous to the ball drawing problem:

There are 75 balls in a bag, 25 are white and 50 are rainbow. In drawing 14 balls, what are the odds that 2 drawn are white. Or, if you are CBGAnimal, only 1 is white.

I know how to do this problem with all one color or the other, but I’m not sure how to think about the general problem: what are the odds of drawing x white balls if you draw z balls from the bag?

Yeah – the binomial distribution idea (I used the coin flip analogy because I always think back to Rosencrantz and Guildenstern are Dead when thinking about outlier probabilities, so it’s my default) also is the ball drawing problem with replacement.

Without replacement, the ball drawing problem outlined above (where each draw is pass/fail) follows a hypergeometric distribution, which decreases the odds slightly for the outlier events (8.2% for 0, 1, or 2 “successful” draws out of 14, with 6.5% for exactly 2).

The forum software isn’t the greatest for showing math equations so I’m not going to try to type it since it is a bit complicated, but the general formula for the probability of “k” successful draws out of “n” total draws is easily seen on Wikipedia (https://en.wikipedia.org/wiki/Hypergeometric_distribution).

Basically – the numerator is the number of different possible combinations of having 2 successful draws out of 14 given that there are 25 “successes” out of the 75 total balls, with the denominator being the number of different possible combinations of drawing 14 balls out of 75.

[LhasaCM]

@bikingjenn 223513 wrote:

I just love the math/numbers/stats geeking out conversation. Thinking someone on this string has access to some stats software and knows how to model the problem? I am stinking at this game. Only five of the called items have appeared on my card.

Sent from my iPhone using Tapatalk

As Steve said – 5/15 is the likeliest outcome at this point of the game.

Where modeling the problem becomes trickier is translating it to BINGO (and more computationally intensive and not something to just casually do in a few minutes over lunch), where it’s not just how many “successes” that matter, but the odds of getting them in the right locations relative to each other, which obviously takes longer unless you’re REALLY lucky. Hence why 15 is the expected value for getting 5 marks on a card but it’s 41 for getting a BINGO in the normal game (and I think that model was done *with* a free space)

[Author]

Replies