BobThePonderer
Joined: Feb 10, 2020
• Posts: 9
February 20th, 2020 at 5:09:04 AM permalink
Quote: gordonm888

Combination math can now quickly explain the difference in the average number of cards drawn with these patterns (for the case of infinite bingo cards).

Thank you.
Your idea to use an infinite population of cards as an explanatory device is very interesting.

Would you please take it a bit further and show the combinatorial math that quickly explains "the difference in the average number of cards drawn with these patterns"?
Wizard

Joined: Oct 14, 2009
• Posts: 25457
February 20th, 2020 at 6:43:22 AM permalink
Yes, I like gordon's infinite card explanation too.
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Wizard

Joined: Oct 14, 2009
• Posts: 25457
February 20th, 2020 at 8:13:50 AM permalink
I've been thinking about this more. Suppose the game is drawn WITH replacement.

What would be the expected number of balls drawn to get both 4 corners and the Diamond pattern (two separate questions).

Four corners: 13.75
Diamond: 12.222222
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
Wizard

Joined: Oct 14, 2009
• Posts: 25457
February 20th, 2020 at 9:26:58 AM permalink
Here are the results of a simulation WITHOUT replacement of the average number of balls needed to call, assuming infinite cards.

Four corners: 12.8289
Diamond: 11.3645
Last edited by: Wizard on Feb 20, 2020
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
BobThePonderer
Joined: Feb 10, 2020
• Posts: 9
February 23rd, 2020 at 7:23:05 AM permalink
An idea:

Following up on Gordon's -

Instead of infinite cards, how about all possible 3 x 3 cards, giving 8! = 40,320 possible cards. Would that help make the combinatorics tractable?

Or, I'm thinking maybe even a brute force spreadsheet listing could show how more columns affect the expected calls. (Should allow exact answers, yes?)

For example:
Let's say for the 1st case (Pattern 1), our 2-mark (single column) Bingo pattern is (using Mike's battleship notation)
A1
A2

Then, for the 2nd case (Pattern 2), also a 2-mark (but, needing two columns)
A1 C1

So, now would it be easy to show that Pattern 1 tends to hit sooner than Pattern 2?
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5751
February 23rd, 2020 at 10:57:16 AM permalink
Quote: BobThePonderer

Instead of infinite cards, how about all possible 3 x 3 cards, giving 8! = 40,320 possible cards. Would that help make the combinatorics tractable?

In a 3 x 3 card, assuming a free space in the center, how are you getting 8! different cards?
Are you saying that each of the numbers 1-8 appear on every card?
In that case, I don't see how A1 A2 is more likely than A1 C1.

Or by "3x3" card, are you limiting it to 1, 2, 3 in column 1, 4 and 5 (and the free space) in column 2, and 6, 7, 8 in column 3?
BobThePonderer
Joined: Feb 10, 2020
• Posts: 9
February 23rd, 2020 at 12:28:33 PM permalink
Yes, thank you, Don, for catching my error.

My idea (an extension of Gordon's idea of infinite cards -or- all-possible cards, I think) of a smaller 3x3 analogue of the regular 5x5 bingo card, numbered as you say, would have only 6 x 2 x 6 = 72 possible cards in the complete set.

So, now that that is corrected, shouldn't having such a small set of possible cards make it easier to show (using combinatorics or brute-force listing) how the two patterns have different expected calls - due to Pattern 1 being confined to one column and Pattern 2 needing two columns?

Thanks
Bob
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 5751
February 23rd, 2020 at 1:35:27 PM permalink
Quote: BobThePonderer

Yes, thank you, Don, for catching my error.

My idea (an extension of Gordon's idea of infinite cards -or- all-possible cards, I think) of a smaller 3x3 analogue of the regular 5x5 bingo card, numbered as you say, would have only 6 x 2 x 6 = 72 possible cards in the complete set.

So, now that that is corrected, shouldn't having such a small set of possible cards make it easier to show (using combinatorics or brute-force listing) how the two patterns have different expected calls - due to Pattern 1 being confined to one column and Pattern 2 needing two columns?

I did a brute force check on all 40,320 permutaions of balls numbered 1-8, assuming all 72 possible cards were in play for each of the games.
Assuming the object of the game was to get either the top left and bottom left corners, or the top left and top right corners, the results were:
15,293 games had the left column corners win
12,187 games had the top row corners win
12,840 games had a left column corners win and a top row corners win at the same time
BobThePonderer
Joined: Feb 10, 2020
• Posts: 9
February 23rd, 2020 at 3:53:01 PM permalink
Quote: ThatDonGuy

I did a brute force check on all 40,320 permutaions of balls numbered 1-8, assuming all 72 possible cards were in play for each of the games.
Assuming the object of the game was to get either the top left and bottom left corners, or the top left and top right corners, the results were:
15,293 games had the left column corners win
12,187 games had the top row corners win
12,840 games had a left column corners win and a top row corners win at the same time

Thanks, Don.

But, I believe your result is for a bingo with EITHER pattern being a winner ...

The original question was concerned about the difference in expected calls for a Pattern 1 game versus the expected calls for a Pattern 2 game.

Can your brute-force attack be applied to each of your patterns separately ... and yield the expected calls for each one?
ThatDonGuy
Joined: Jun 22, 2011