Pointless Prize 2023 – Bicycle Bingo

[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]

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