Direct: https://www.youtube.com/watch?v=AHP1T8fTxpQ

He is right that there are more horizontal bingos called in a room full of cards. However, his reasoning is all wrong. He seems to argue that even on a single card, this holds true. To be specific, he says the probability of a horizontal bingo is 3x that of a vertical one. I'm a big fan of Matt Parker and his Stand-Up Maths YouTube channel. However, I have to call him out on this one.

Consider the table he shows at the 23:21 point, that he uses to make is 3x argument.

Before going on, he uses an assumption of no free square, which I shall also, to compare apples to apples.

The 0.00087, which he calls P(V

_{5}), is supposed to be the probability of a vertical bingo with 5 daubs. However, he doesn't take into consideration the daub placement in the table. It is easy to see the probability of getting a vertical bingo in 5 daubs is (4/24)*(3/23)*(2/22)*(1/21) = 0.00009411. Matt is not answering the right question. His figure is the probability the first five numbers drawn will be from the same column.

The correct probability of a horizontal bingo in five daubs is also 0.00009411. Not the 0.04400 , which he calls P(H

_{5}).

I attest that with one card, the probability of a horizontal bingo equals a vertical bingo.

Briefly, the reason you see more horizontal bingos in a multi-player game than vertical is many games it takes a long time for one letter to get called, while the other four letters have been called multiple times each. When that fifth letter is finally called, the whole room erupts in "Bingo!" Those bingos are all going to be horizontal. This explanation is based more on experience playing than actual math.

Again, I do not call out Matt Parker lightly and would love to get a third opinion on this. Thank you.

P.S. What is Karen's opinion this?

Here are my results based on my own simulation, with and without the free square.

Win | No Free Square | Free Square |
---|---|---|

Vertical | 35.89% | 29.10% |

Horizontal | 45.82% | 30.77% |

Diagonal | 18.30% | 40.13% |

The only way I could be in loose agreement with the paper is if diagonal bingos counted as horizontal and no free square.

Quote:avianrandyHow is it possible to have a diagonal bingo without using the free square?

link to original post

No free square means that there is a 5th N number. It's not like the square is a blocker.

Winning Bingo | No Free Square | Free Square |
---|---|---|

B column | 7.19% | 2.89% |

I column | 7.18% | 2.88% |

N column | 7.17% | 17.57% |

G column | 7.18% | 2.88% |

O column | 7.17% | 2.88% |

Row 1 | 9.16% | 2.67% |

Row 2 | 9.17% | 2.69% |

Row 3 | 9.16% | 20.06% |

Row 4 | 9.17% | 2.68% |

Row 5 | 9.16% | 2.67% |

NW to SE diagonal | 9.15% | 20.04% |

SW to NE diagonal | 9.14% | 20.09% |

Sample size (bingos achieved):

No free square: 8,431,726

With free square: 3,545,727

Consider 3 types of bingo squares:

a) a square on a diagonal

b) a square in the N column or on the horizontal row of the free space

c) any other square

What is the mean number of rounds (bingo numbers being called before getting a bingo when your first round outcome is (a) or (b) or (c)? What is the mean number of rounds to bingo when your first round outcome is No Match?

Another way to pose a question is:

Consider a BIngo game with four players: Abel, Betty, Carol and Zach

In the first round, Abel gets an 'a' square, Betty gets a 'b' square and Carol gets a 'c' square and Zach gets no square at all (No Match).

What is the probability of winning for each of these four players?

the article its based on (it's short and accessible, but uses images for the math instead of latex. if that is not a crime, it should be.)

Quote:itsmejeffYoutube recommended this video for me today.

the article its based on (it's short and accessible, but uses images for the math instead of latex. if that is not a crime, it should be.)

link to original post

Thank you for finding that article. I can only groan after reading it. It includes this assumption:

That is simply a lazy assumption. I know sometimes simplifying a problem is called for, but this goes too far.

However, it makes me feel badly for being too harsh on Matt Parker. He simply validated the conclusion of this article. However, that assumption can't be made with only 1,000 cards. I think the correct approach to analyze the "paradox" is a random simulation using actual bingo cards. Of course, that is what I did and it's easy to agree with myself.

If so, the phenomenon makes logical sense to me.

Then if you take to an extreme, you could calculate the hypothetical of every possible BINGO card permutation was playing a game:

1) If the first five numbers called were in the B column, how many BINGO cards would be winners?

2) If the first five numbers went B then I then N then G then O, how many BINGO cards would be winners on the horizontal (and separately on the diagonal)?

Quote:unJonOne question on this topic. What drives this seems to be that when there are multiple simultaneous BINGOs to end a game, it’s more likely that those multiple people have a horizontal BINGO than a vertical BINGO? Is that accurate?

If so, the phenomenon makes logical sense to me.

Then if you take to an extreme, you could calculate the hypothetical of every possible BINGO card permutation was playing a game:

1) If the first five numbers called were in the B column, how many BINGO cards would be winners?

2) If the first five numbers went B then I then N then G then O, how many BINGO cards would be winners on the horizontal (and separately on the diagonal)?

link to original post

Good points!

I have been questioning my own post too and would like to correct myself a bit.

Matt's point is that if there were a vast number of bingo cards in play then as soon as there were a possible bingo, somebody would call it.

I might remind he said that probability of the first possible bingo would be, assuming no free square:

Vertical = 24.8%

Horizontal = 75.2%

Assuming a 5th number is called in a column. The probability any given card would have those five numbers is 1/combin(15,5) = 3003.

Assuming the last column to not have a ball called is finally called, enabling a horizontal bingo. Random simulation shows that when this happens the product of balls called in all five columns is 65.3. That makes the probability any given card will have a bingo is 65.3/15^5 = 1 in 11,629.

However, you can't simply multiply the probability of a horizontal/vertical bingo possibility by the probability any given card will have it, because the numbers are correlated by number of balls called.

After spending more hours on this, I retract my earlier comment that there will be more bingos called when they are horizontal.

Bottom line is I think the reason we see fewer vertical bingos in simulations is the effect of removal. Say B15 is the first ball called. The next ball is less likely to be another B than the other letters.

I believe that the guy screwed up his number of possible faces at 10:59. He should be permuting as order is somewhat important here. Two different cards with the same numbers in different positions will give different results. though half of them would give same results in this case as orientation does not matter.

If it was effect of removal then this phenomenon would show up with a single bingo card playing the game.

Quote:unJonWiz,

If it was effect of removal then this phenomenon would show up with a single bingo card playing the game.

link to original post

Yeah, I was thinking that while exercising this morning. However, I am looking for something in plain English to explain why horizontal bingos are more likely in multi-card games. Any suggestions?

Somewhat relatedly, I’ve played a fair amount of bingo at Plaza. When someone is 1 away (and using the computer, not paper), you hear a ding sound. Sometimes I’ll be 2 away and wonder just far behind I am. It seems to be the answer is “if someone is 1 step closer, they have a massive edge” but I can’t quantify it. Typically there are 35-70 people in the room.

Given bingo game with no free space, only horizontal and vertical bingos are considered.

It takes 5 numbers to make a bingo, so let's look at "groups of five"

This bingo game has combin(75,5) - 17,259,930 different combinations for groups of 5 numbers

Vertical bingos = 5*combin(15,5) = 15,015 combinations

Horizontal bingos = combin(15,1)^5 = 759,375 combinations

Horizontal/Vertical = 50.6 (approx)

Let's just look at a game where 16 numbers have been called: There are 2 columns with 4 numbers,2 columns with 3 numbers, and one column with 2 numbers (4,4,3,3,2).

There are zero groups of 5 vertical numbers and 288 groups of 5 horizontal numbers. With 20 bingo cards (with distinct rows and columns) there is 0 chance of a vertical bingo and a 3.8% chance of a horizontal bingo.

What happens when the 17th number is drawn to a (4,4,3,3,2) distribution ?

P=22/59 (5,4,3,3,2) Vert =1 Horiz = 360

P=24/59 (4,4,4,3,2) Vert =0, Horiz = 384

P=13/59 (4,4,3,3,3) Vert =0, Horiz = 432

So, roughly 0.0025 probability of a vertical (with 20 bingo cards) and 0.051 probability of a horizontal.

***********************************

This approach provides a methodology for calculating the probabilities of vertical and horizontal bingos as a function of the number of numbers that have been called and the number of bingo cards.

If 20 numbers have been called:

List all the possible distributions from (4,4,4,4,4) to (15,5,0,0,0) and their probabilities. Ex:

(4,4,4,4,4) 1024 of the possible 759,375 possible horizontal bingos and 0/15015 possible vertical bingos

(15,5,0,0,0) no possible horizontal bingos and 100% probability of a vertical bingo

It is straightforward (and perhaps tedious) to calculate a probabilistically averaged number of horizontal and vertical bingo combinations for any number of bingo balls(numbers) that have been drawn. (again assuming no free space).

**********************

Edit: I realize the math for going from number of combinations to probabilities is incorrect; but I was in a hurry to write out the ideas.

Let’s make a simpler version of bingo with a two by two card where each column can have 5 numbers.

So B can have 1-5 and I have 6-10.

Let’s say I’m playing by myself with the card:

1 6

2 7

Seems clear that there is an equal chance of winning vertical, horizontal or diagonal. Any pairwise combo (say 1,2 vs 1,6) has the same chance of coming first.

Now let’s say that every single possible card was going to be in play. So the game would end after 2 balls every time.

Seems clear that it’s more likely that the first two balls will be in two different columns than the same column. So there will be more horizontal and more diagonal bingos than vertical bingos. That’s the Wiz effect of removal in action.

To put it in my own words, imagine bingo cards are 2x2 only. There are ten balls in the hopper. 1-5 are assigned to the B column and 6 to 10 to the I.

Since position matters in bingo cards, the number of possible bingo cards are p(5,2)*p(5,2) = 20^2 = 400. Where p(x,y) = number of permutations of y items out of x.

Let's assume all 400 cards are in play.

If the first two balls drawn are from the same column, we will have 2*20=40 winners. 2 for the ways the two winning numbers can be positioned on the card and 20 for the ways the other column can contains p(5,2) = 20 numbers.

If the first two balls drawn are from the different columns, we will have 2*16=32 winners. 2 for the ways the two winning numbers can be positioned on the card (top or bottom row) and 16 for the ways the other two numbers on the card that were not called.

The number of ways the first two balls called can be in the same column is 4 and different is 5.

So total winning cards are:

vertical: 4*20 = 80

horizontal: 5*16 = 80

Hmmm. Am I right or where is our point of departure?

Horizontal = 438317 (58%)

Vertical = 317091 (42%)

Total = 775408

Quote:WizardQuote:unJonOne question on this topic. What drives this seems to be that when there are multiple simultaneous BINGOs to end a game, it’s more likely that those multiple people have a horizontal BINGO than a vertical BINGO? Is that accurate?

If so, the phenomenon makes logical sense to me.

Then if you take to an extreme, you could calculate the hypothetical of every possible BINGO card permutation was playing a game:

1) If the first five numbers called were in the B column, how many BINGO cards would be winners?

2) If the first five numbers went B then I then N then G then O, how many BINGO cards would be winners on the horizontal (and separately on the diagonal)?

link to original post

Good points!

I have been questioning my own post too and would like to correct myself a bit.

Matt's point is that if there were a vast number of bingo cards in play then as soon as there were a possible bingo, somebody would call it.

I might remind he said that probability of the first possible bingo would be, assuming no free square:

Vertical = 24.8%

Horizontal = 75.2%

Assuming a 5th number is called in a column. The probability any given card would have those five numbers is 1/combin(15,5) = 3003.

Assuming the last column to not have a ball called is finally called, enabling a horizontal bingo. Random simulation shows that when this happens the product of balls called in all five columns is 65.3. That makes the probability any given card will have a bingo is 65.3/15^5 = 1 in 11,629.

However, you can't simply multiply the probability of a horizontal/vertical bingo possibility by the probability any given card will have it, because the numbers are correlated by number of balls called.

After spending more hours on this, I retract my earlier comment that there will be more bingos called when they are horizontal.

Bottom line is I think the reason we see fewer vertical bingos in simulations is the effect of removal. Say B15 is the first ball called. The next ball is less likely to be another B than the other letters.

link to original post

Based on your “bottom line” comments, perhaps (some of) you have already realized this… If we compare the math behind bingo and video poker, one could say that vertical bingos resemble flushes and horizontal (including diagonal) bingos resemble straights. (Don’t make the mistake of confusing diagonals with straight flushes, as they are merely a specific case of horizontals.) I thought this analogy might help explain the phenomenon qualitatively…

PS I have no insight ATM regarding the 2X2 query presented.

The numbers in each row add up to more than 100% as if, say, a game ended with a horizontal bingo on one card and a vertical on another (or, for that matter, on the same card), it counted as both.

The number of games played is in millions; for example, there were 2,065,000,000 games with 100 cards played.

# of cards | Millions of games | % Horizontal | % Vertical | % Diagonal |
---|---|---|---|---|

100 | 2,065 | 37.3553 | 29.9806 | 39.4867 |

150 | 1,377 | 37.4182 | 29.013 | 41.1205 |

200 | 1,033 | 37.455 | 28.464 | 42.2644 |

250 | 826 | 37.6588 | 27.9315 | 43.1366 |

300 | 689 | 37.8649 | 27.5951 | 43.7846 |

350 | 590 | 37.9625 | 27.2837 | 44.5281 |

400 | 517 | 37.9083 | 27.1692 | 45.1114 |

450 | 459 | 38.222 | 26.8481 | 45.6383 |

500 | 413 | 38.3511 | 26.701 | 46.0191 |

550 | 376 | 38.446 | 26.6114 | 46.442 |

600 | 345 | 38.6525 | 26.54 | 46.7478 |

650 | 318 | 38.7991 | 26.4472 | 47.0814 |

700 | 295 | 38.8047 | 26.3464 | 47.462 |

750 | 276 | 38.9018 | 26.1855 | 47.879 |

800 | 259 | 39.0807 | 26.1151 | 48.1822 |

850 | 243 | 39.1012 | 26.0761 | 48.4679 |

900 | 230 | 39.5083 | 25.8391 | 48.7861 |

950 | 218 | 39.4986 | 26.0917 | 48.9317 |

1000 | 207 | 39.6749 | 26.0623 | 49.0333 |

I'm already going down the path of pure math with a 3x3 card, but it's getting messy. Even that may not get at the "why" issue.

Quote:WizardOutstanding post above, thank you!

To put it in my own words, imagine bingo cards are 2x2 only. There are ten balls in the hopper. 1-5 are assigned to the B column and 6 to 10 to the I.

Since position matters in bingo cards, the number of possible bingo cards are p(5,2)*p(5,2) = 20^2 = 400. Where p(x,y) = number of permutations of y items out of x.

Let's assume all 400 cards are in play.

If the first two balls drawn are from the same column, we will have 2*20=40 winners. 2 for the ways the two winning numbers can be positioned on the card and 20 for the ways the other column can contains p(5,2) = 20 numbers.

If the first two balls drawn are from the different columns, we will have 2*16=32 winners. 2 for the ways the two winning numbers can be positioned on the card (top or bottom row) and 16 for the ways the other two numbers on the card that were not called.

The number of ways the first two balls called can be in the same column is 4 and different is 5.

So total winning cards are:

vertical: 4*20 = 80

horizontal: 5*16 = 80

Hmmm. Am I right or where is our point of departure?

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I concur. Let me step back and ask what your simulation is counting. If a bingo game ends with multiple winners, then does each winner count towards the total of vertical or horizontal winners?

If not, then you would still get 5/9 of the winners games ending with horizontal bingos and 4/9 ending in vertical bingos on a 2x2 bingo card with 10 numbers.

If so, then I wonder if the two extremes may be equal but not say if some number of cards between 1 card and every card is in play.

Quote:unJon

I concur. Let me step back and ask what your simulation is counting. If a bingo game ends with multiple winners, then does each winner count towards the total of vertical or horizontal winners?

Yes, each winner counts equally.

Quote:If so, then I wonder if the two extremes may be equal but not say if some number of cards between 1 card and every card is in play.

link to original post

I tend to think not, but have no good argument why. I think if we expanded this to a 3x3 card, where there could be 3 to 5 numbers called before a bingo, then we would start to see a bias towards the horizontal bingos. I started down that path in Excel, but probably won't finish.

XXXXO

XXXOX

XXOXX

XOXXX

OXXXX

When the 71st ball is drawn, you will have both a horizontal and a vertical bingo. Somehow this bounding case is part of the same mathematical system as a 5x5 grid with one million players, or a 2x2 grid with ten numbers. And I don't think you can understand this mathematical system by examining its extremums. Its like trying to derive the theory of prime numbers by examining the smallest prime number "2". It's not merely a limiting case, it's a trivial and non-typical case.

I suspect that the ratio of horizontal to vertical bingos is a function of how many rounds it takes (on average) for the first bingo to appear. And examining games with an nxn grid with a zillion players in which it takes only n or n+1 rounds to end is probably not going to reveal the answer. Because the essential factors driving the mathematics are not present.

Previously, I posted a rambling discussion about developing a combinatorial math approach to calculating the liklihood of vertical and horizontal bingos on a single bingo card as a function of the round in the bingo game. I'm still looking at that, and think I may have a chance at the Fields Medal, if only I weren't over 40 years of age.

Bingo Type | Probability First Possible Bingo | Probability Random Card Wins | Product | Ratio |
---|---|---|---|---|

Horizontal | 75.20% | 0.000121 | 0.000091 | 52.3% |

Vertical | 24.80% | 0.000333 | 0.000083 | 47.7% |

Total | 100.00% | 0.000173 | 100.0% |

You can see it shows that 52.34% of bingos will be horizontal. However, by direction simulation, I get 54.15%. I do not understand the reason for the 2% difference.

The 0.000121 figure that a random card wins on a horizontal bingo is based on a mixture of math and 18.313 combinations of daubs on the other four columns, per a simulation.

Edit: Perhaps the difference is my simulation of an overall game used 1,000 cards. I'm going to increase that to see how things change...

After running it, I get these results:

Vertical = 41.9%

Horizontal = 58.1%

I'm not happy these results are so much different that 1,000 cards. I suspect one of the simulations, at least, had a bug.

I reran the simulation for 1,000 cards are got these results:

Vertical = 42.14%

Horizontal = 57.86%

For one card, it still comes out 50/50.

Quote:WizardI made a number of changes to my simple multi-card simulation program other than increasing the number of cards to 10,000.

After running it, I get these results:

Vertical = 41.9%

Horizontal = 58.1%

I'm not happy these results are so much different that 1,000 cards. I suspect one of the simulations, at least, had a bug.

I reran the simulation for 1,000 cards are got these results:

Vertical = 42.14%

Horizontal = 57.86%

For one card, it still comes out 50/50.

link to original post

I’m still confused by this.

What makes logical sense to me is that if a horizontal bingo is about 1.5x as likely then there should also be 1.5x as many horizontal bingo cards winning such that the EV of winning vertically and horizontally is the same when you factor in chance of sharing bingo prize.

But the simulation does not appear to show that?

B | I | N | G | Balls before bingo | Ways | Probability Calls | Winning Card | Product |
---|---|---|---|---|---|---|---|---|

1 | 1 | 1 | 1 | 4 | 1 | 0.04399779 | 0.00000658 | 0.00000029 |

1 | 1 | 1 | 2 | 5 | 4 | 0.08799558 | 0.00001317 | 0.00000116 |

1 | 1 | 1 | 3 | 6 | 4 | 0.03315775 | 0.00001975 | 0.00000065 |

1 | 1 | 2 | 2 | 6 | 6 | 0.08034379 | 0.00002634 | 0.00000212 |

1 | 1 | 1 | 4 | 7 | 4 | 0.01023989 | 0.00002634 | 0.00000027 |

1 | 1 | 2 | 3 | 7 | 12 | 0.07167926 | 0.00003951 | 0.00000283 |

1 | 2 | 2 | 2 | 7 | 4 | 0.03859653 | 0.00005267 | 0.00000203 |

1 | 1 | 2 | 4 | 8 | 12 | 0.02567615 | 0.00005267 | 0.00000135 |

1 | 1 | 3 | 3 | 8 | 6 | 0.01854389 | 0.00005926 | 0.0000011 |

1 | 2 | 2 | 3 | 8 | 12 | 0.05991103 | 0.00007901 | 0.00000473 |

2 | 2 | 2 | 2 | 8 | 1 | 0.00806495 | 0.00010535 | 0.00000085 |

1 | 1 | 3 | 4 | 9 | 12 | 0.01517227 | 0.00007901 | 0.0000012 |

1 | 2 | 2 | 4 | 9 | 12 | 0.02450906 | 0.00010535 | 0.00000258 |

1 | 2 | 3 | 3 | 9 | 12 | 0.03540197 | 0.00011852 | 0.0000042 |

2 | 2 | 2 | 3 | 9 | 4 | 0.0190626 | 0.00015802 | 0.00000301 |

1 | 1 | 4 | 4 | 10 | 6 | 0.00350129 | 0.00010535 | 0.00000037 |

1 | 2 | 3 | 4 | 10 | 24 | 0.03267874 | 0.00015802 | 0.00000516 |

1 | 3 | 3 | 3 | 10 | 4 | 0.0078671 | 0.00017778 | 0.0000014 |

2 | 2 | 2 | 4 | 10 | 4 | 0.00879812 | 0.0002107 | 0.00000185 |

2 | 2 | 3 | 3 | 10 | 6 | 0.0190626 | 0.00023704 | 0.00000452 |

1 | 2 | 4 | 4 | 11 | 12 | 0.00842499 | 0.0002107 | 0.00000178 |

1 | 3 | 3 | 4 | 11 | 12 | 0.01216943 | 0.00023704 | 0.00000288 |

2 | 2 | 3 | 4 | 11 | 12 | 0.01965831 | 0.00031605 | 0.00000621 |

2 | 3 | 3 | 3 | 11 | 4 | 0.00946511 | 0.00035556 | 0.00000337 |

1 | 3 | 4 | 4 | 12 | 12 | 0.00695396 | 0.00031605 | 0.0000022 |

2 | 2 | 4 | 4 | 12 | 6 | 0.00561666 | 0.0004214 | 0.00000237 |

2 | 3 | 3 | 4 | 12 | 12 | 0.0162259 | 0.00047407 | 0.00000769 |

3 | 3 | 3 | 3 | 12 | 1 | 0.00195312 | 0.00053333 | 0.00000104 |

1 | 4 | 4 | 4 | 13 | 4 | 0.00145809 | 0.0004214 | 0.00000061 |

2 | 3 | 4 | 4 | 13 | 12 | 0.01020662 | 0.0006321 | 0.00000645 |

3 | 3 | 3 | 4 | 13 | 4 | 0.0049143 | 0.00071111 | 0.00000349 |

2 | 4 | 4 | 4 | 14 | 4 | 0.0023425 | 0.0008428 | 0.00000197 |

3 | 3 | 4 | 4 | 14 | 6 | 0.00507542 | 0.00094815 | 0.00000481 |

3 | 4 | 4 | 4 | 15 | 4 | 0.00253771 | 0.0012642 | 0.00000321 |

4 | 4 | 4 | 4 | 16 | 1 | 0.00051614 | 0.0016856 | 0.00000087 |

Total | 0.75177861 | 0 | 0.00009064 |

This shows that the probability a random card gets a horizontal bingo as soon as a bingo is possible is 0.00009064. We divide that by 0.75177861, to get the probability of a horizontal bingo given that all five columns were called before 5 balls were drawn in any one column. That conditional probability is 0.000120572.

Let's go back to a table I have tried to construct before showing the overall probability of a vertical and horizontal bingo assuming all possible cards are in play.

Overall | Matt Parker | Prob bingo formed | Product | Ratio |
---|---|---|---|---|

Horizontal | 0.75177861 | 0.00012057 | 0.00009064 | 0.52304015 |

Vertical | 0.24822139 | 0.000333 | 0.00008266 | 0.47695985 |

Total | 1 | 0 | 0.0001733 | 1 |

So, bottom line is a 52.3% chance a bingo is horizontal.

I admit this is not in agreement with my 10,000 card simulation at 58.07%. Why, I am not sure. However, it does agree the results in this post.

I welcome all comments and corrections.

Quote:ThatDonGuyHere is a simulation where various numbers of cards from 100 to 1000 were in play at once, and what percentage of the time the winning card had a horizontal, a vertical, or a diagonal bingo.

The numbers in each row add up to more than 100% as if, say, a game ended with a horizontal bingo on one card and a vertical on another (or, for that matter, on the same card), it counted as both.

The number of games played is in millions; for example, there were 2,065,000,000 games with 100 cards played.

# of cards Millions

of games% Horizontal % Vertical % Diagonal 100 2,065 37.3553 29.9806 39.4867 150 1,377 37.4182 29.013 41.1205 200 1,033 37.455 28.464 42.2644 250 826 37.6588 27.9315 43.1366 300 689 37.8649 27.5951 43.7846 350 590 37.9625 27.2837 44.5281 400 517 37.9083 27.1692 45.1114 450 459 38.222 26.8481 45.6383 500 413 38.3511 26.701 46.0191 550 376 38.446 26.6114 46.442 600 345 38.6525 26.54 46.7478 650 318 38.7991 26.4472 47.0814 700 295 38.8047 26.3464 47.462 750 276 38.9018 26.1855 47.879 800 259 39.0807 26.1151 48.1822 850 243 39.1012 26.0761 48.4679 900 230 39.5083 25.8391 48.7861 950 218 39.4986 26.0917 48.9317 1000 207 39.6749 26.0623 49.0333

link to original post

I'm looking for something to check my own simulation against. So, I ran mine with 1,000 cards, a free square and diagonal wins and get these figures:

Vertical: 29.14%

Horizontal: 30.50%

Diagonal: 40.35%

If I divide your figures, Don, by the sum of the probabilities, 1.147705, we still don't agree. What did you do if two different cards won for the same type of bingo at the same time? Personally, I count all wins.

Quote:mipletKind of off topic, but if cards are created using this paten, only 2971 of the possible 3003 G row combinations of numbers are used.

link to original post

Hmmm. Can you tell us in layman's terms why the patent is blocking certain cards?

Quote:WizardQuote:mipletKind of off topic, but if cards are created using this paten, only 2971 of the possible 3003 G row combinations of numbers are used.

link to original post

Hmmm. Can you tell us in layman's terms why the patent is blocking certain cards?

link to original post

It uses relative Prime numbers. 3001, 2999, 1321, 2971, and 2998 for the rows. And 113 (23 for n) for the order of the rows.

Quote:Wizard

I'm looking for something to check my own simulation against. So, I ran mine with 1,000 cards, a free square and diagonal wins and get these figures:

Vertical: 29.14%

Horizontal: 30.50%

Diagonal: 40.35%

If I divide your figures, Don, by the sum of the probabilities, 1.147705, we still don't agree. What did you do if two different cards won for the same type of bingo at the same time? Personally, I count all wins.

link to original post

I don't. I am trying to count what percentage of the games ended with, say, a horizontal win. It doesn't matter how many cards have the horizontal win at once; it is the number of games that matters.

Quote:ThatDonGuyQuote:Wizard

I'm looking for something to check my own simulation against. So, I ran mine with 1,000 cards, a free square and diagonal wins and get these figures:

Vertical: 29.14%

Horizontal: 30.50%

Diagonal: 40.35%

If I divide your figures, Don, by the sum of the probabilities, 1.147705, we still don't agree. What did you do if two different cards won for the same type of bingo at the same time? Personally, I count all wins.

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I don't. I am trying to count what percentage of the games ended with, say, a horizontal win. It doesn't matter how many cards have the horizontal win at once; it is the number of games that matters.

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If all simultaneous winners chop the prize, then it would make sense to count a fractional win when more than one card wins on a certain ball. This would mean the win counts would sum to the number of games played. With one card in play, it is possible that all three types of bingos hit on one ball. It is imperative to give each type (H, V, D) a fair share of the win when there are simultaneous wins on one card or multiple cards.

With a million cards in play, simultaneous wins on one card is less important. The winning cards will almost never make more than one bingo.

I am writing a generalized simulator that can handle 2x2 to 5x5 cards. I can create 100,000 unique 5x5 bingo cards per second. I will use a single integer bitwise operation to check whether any of the 12 bingos occurred. This will be very fast and allow a lot of pipelining of the card scoring algorithm. I should be able to give Mike some numbers to check against his sims.

Quote:MentalIf all simultaneous winners chop the prize, then it would make sense to count a fractional win when more than one card wins on a certain ball. This would mean the win counts would sum to the number of games played. With one card in play, it is possible that all three types of bingos hit on one ball. It is imperative to give each type (H, V, D) a fair share of the win when there are simultaneous wins on one card or multiple cards.

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I agree that fractional wins make logical sense. I have posted before where I did chop the pot. However, for now, I'm trying to keep it simple and get at how many bingos are horizontal/vertical if there are thousands of cards in play. I'm still looking for a confirmation that my simulator is working correctly, which counts all wins equally.

The net is I get the same chances of the winner being vertical as 0.248221389.

Obviously when there's a vertical winner it's when the first 5 numbers in a column come out, so as you say 1 in 3003 cards win. With a horizontal win, it depends on how many numbers in different columns have come out. So typically there will be multiple winning lines.

Quote:WizardI agree that fractional wins make logical sense. I have posted before where I did chop the pot. However, for now, I'm trying to keep it simple and get at how many bingos are horizontal/vertical if there are thousands of cards in play. I'm still looking for a confirmation that my simulator is working correctly, which counts all wins equally.

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Here you go - a simulation of games with anywhere from 100 to 2000 cards, with the number of horizontal, vertical, and diagonal winning cards per 100 games:

# Cards | Millions of Games | Horizontal Bingos | Vertical Bingos | Diagonal Bingos |
---|---|---|---|---|

100 | 23,087 | 40.0736 | 37.6077 | 42.572 |

200 | 11,544 | 41.1257 | 38.3523 | 46.9409 |

300 | 7,696 | 42.2382 | 39.2854 | 49.9984 |

400 | 5,772 | 43.2486 | 40.3311 | 52.416 |

500 | 4,618 | 44.2259 | 41.3196 | 54.7276 |

600 | 3,848 | 45.2176 | 42.3886 | 56.7342 |

700 | 3,299 | 46.1457 | 43.4043 | 58.7252 |

800 | 2,886 | 47.2285 | 44.4691 | 60.5567 |

900 | 2,566 | 48.2205 | 45.4471 | 62.458 |

1000 | 2,309 | 49.2317 | 46.4605 | 64.1357 |

1100 | 2,099 | 50.2316 | 47.4295 | 65.9372 |

1200 | 1,924 | 51.3179 | 48.4531 | 67.6555 |

1300 | 1,776 | 52.2611 | 49.3414 | 69.3828 |

1400 | 1,650 | 53.2861 | 50.4217 | 71.1467 |

1500 | 1,540 | 54.418 | 51.3734 | 72.8261 |

1600 | 1,443 | 55.578 | 52.3545 | 74.3737 |

1700 | 1,358 | 56.3895 | 53.271 | 76.0393 |

1800 | 1,283 | 57.4175 | 54.1232 | 77.7116 |

1900 | 1,216 | 58.5169 | 55.0495 | 79.5404 |

2000 | 1,155 | 59.5971 | 55.8928 | 81.2197 |

Quote:ThatDonGuy

1000 2,309 49.2317 46.4605 64.1357

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I'm happy to report my own simulation is in close agreement with these figures. Here are my exact results:

Horizontal: 261870

Vertical: 247642

Diagonal: 341765

Also, the rules are such that diagonal bingos do not matter. If a horizontal bingo is possible, it must exist. Therefore, any diagonal will exist with at least one horizontal (minimum number of calls) or many of them.

Quote:itsmejeffHow do you calculate how many cards are needed to simulate the results? It is some number between 2 and all possible cards, but it does not appear to be 1000.

Also, the rules are such that diagonal bingos do not matter. If a horizontal bingo is possible, it must exist. Therefore, any diagonal will exist with at least one horizontal (minimum number of calls) or many of them.

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Each simulation is based on a predetermined number of cards. If you check my post that Wizard responds to, you will see that I have simulations for different numbers of cards from 100 to 2000. The number of cards has a definite effect on the results.

As for your second statement, it is true only if every possible card is in play. If the first four numbers happen to all be on the same diagonal of one card, then it is highly unlikely that any card has the same numbers in the middle row of the card (they can't be on any other row as it would not be a bingo since no N numbers have been called yet).

I don't know if this helps but technically (although some will be mirror images of others) there are 3003 ways to have five numbers in any column, however they can be in a different order, so actually it is 15x14x13x12x11 = 360,360.

The other four columns also have the same perms. So the total number of "unique" tickets are 360360

^{5}(call this N).

Just considering five numbers drawn causing a win.

If there are five numbers in one column then there are 360360

^{4}ways for the other columns and 5x4x3x2x1 ways to organise the five drawn numbers. This is N/3003 - i.e. 1 in 3003 tickets win.

If there are five numbers, one in each column, then

(i) each column has to have the drawn number in that column (and hence any other 14) =14x13x12x11/4/3/2/1. So initially, ignoring the order within each column, this is 1001

^{5}. This is N/((15*24)

^{5}).

(ii) Now while a card might have the winning numbers, they have to appear in one of seven lines (five across or two diagonal).

This means, for each of those lines, the winning number has to be in the correct position and the other four can be in any order.

So the multiplier is (4x3x2x1)

^{5}*7 = 55738368.

This means 1 in 108482 cards win.

The chances of a 5 numbers in a column being called = 1*14/89...11/86 = 1 in 2439.

The chances of 5 numbers, 1 in each column = 1*60/89...15/86 = 1 in 48 (or 1 in 50.77 times more likely)

So N/3003/2439.186 + N/108482/48.2296 cards (about N/7324878+N/5232056) will win on the 5th number. (This agrees with, if you were playing one card, then it has 5 ways to win on a vertical column and 7 ways to win on a horizontal or diagonal one.)

The problem with 9th numbers and beyond is you have to cater for cards with two winning lines.

The only thing I could see is going from 16 numbers (no winner) to 17 numbers is there are 15 numbers to make a horizontal winner and 44 to make a vertical winner.

Let Pv = the probability that a vertical line hits for the first win.

Let Pd = the probability that a diagonal line hits for the first win.

I weight my probabilities for split wins. For example, if I have 1000 cards and there are 10 simultaneous winners, then I divide Nh, Nv, and Nd by 10 and add that to my total winners for each category. If I simulate G games, I divide the totals by G to get a normalized probability. As a result, Ph + Pv + Pd = 1.0.

In order to understand the 'paradox', I examined a very simple bingo game. The simplest bingo game is a square 1x1 with one number in it. Clearly, Ph = Pv = 1/4 and Pd = 1/2. That is, the first number called always results in a H and V win and two D wins on a winning card. This is not a very interesting game.

If I uses a 2x2 bingo card, and I only have two balls per row, I get a more interesting game. Ph = Pv = Pd = 1/3. If I break the probabilities down to the six unique bingo lines:

V0 0.167

V1 0.167

H2 0.167

H3 0.167

D4 0.167

D5 0.167

In this game, the first two numbers always make a bingo on every card. There are only four numbers on a card and only four balls numbered 1-4.

Now, I change my game to allow numbers 1-3 in the first column and 4-6 in the second column. I use 1000 cards and simulate 1,000,000 games. (This takes 4 seconds of CPU time.)

V0 0.200

V1 0.200

H2 0.150

H3 0.150

D4 0.150

D5 0.150

You can prove this is correct because there are 15 equally probable ways for the first two balls to come up, and three of these (12, 13, and 23) make V0 a winner. The cases where one ball is in the range 1,2,3 and the other ball is in the range 4,5,6 are equally likely to make a H or D winner.

When I use 15 balls per column, I get an unexpected result: Ph < Pd

V0 0.243

V1 0.241

H2 0.128

H3 0.128

D4 0.130

D5 0.130

I have not thought this through yet to come up with the mathematical explanation.

What I want to emphasize is that the 'paradox' is easily explained for the 2x2 bingo game with 6 balls, but it is OPPOSITE in direction compared to the bingo paradox in the OP. There are more vertical bingo winners compared to horizontal bingo winners.

My code is very efficient and is parameterized. I can change the size from 2x2 to 5x5 by changing one parameter. Likewise, the number of balls per column is changed by changing another parameter. Here are the results for a 4x4 bingo game with 15 balls per column. Note that the paradox flips to favor H and D winners.

V0 0.085

V1 0.085

V2 0.086

V3 0.085

H4 0.110

H5 0.110

H6 0.109

H7 0.111

D8 0.109

D9 0.110

I am able to use two bitwise operations ((win & board) == win) to determine if a bingo was hit. This makes the program very fast, but it limits me to 64 balls in play to stuff the bits into a 64-bit number. This means that I can simulate a 4x4x15 game, but I need to limit myself to only 12 numbers per column for a 5x5 game.

1 7 15 20 21

4 10 11 19 24

2 6 14 17 23

5 9 13 16 25

3 8 12 18 22

##### ..... ..... ..... ..... V0

..... ##### ..... ..... ..... V1

..... ..... ##### ..... ..... V2

..... ..... ..... ##### ..... V3

..... ..... ..... ..... ##### V4

#.... .#... ....# ....# #.... H5

...#. ....# #.... ...#. ...#. H6

.#... #.... ...#. .#... ..#.. H7

....# ...#. ..#.. #.... ....# H8

..#.. ..#.. .#... ..#.. .#... H9

#.... ....# ...#. #.... .#... Da

..#.. ...#. ...#. ...#. #.... Db

I ran a 5x5 bingo game where there are only 5 numbers per column. Above, I show a random card out of the 1000 cards that I generated. The 12 winning lines are shown above as 25-bit numbers. A # indicates that a bit is set. For the V lines, the 25-bit number for a winner has the first 5 bits set. The H and D lines have one bit set per group of 5 bits.

As I simulate a game, I start with a 25-bit number representing the board. I randomly add bits to the board simulating the random picking of balls from the hopper. I use the logical AND operator between a winning line and the board to see if the board has all of the bits for the winner set. If (H5 AND BOARD == H5), there is a bingo on H5. Below, I show what this process looks like after seven numbers have been called.

#.... .#... ....# ....# #.... H5

#.... .#... .#..# .#..# #.... BOARD

=====================

#.... .#... ....# ....# #.... H5 AND BOARD == H5,

so the horizontal line H5 is a bingo

The probabilities for the 12 lines are very biased against vertical lines. Remember that I am splitting the bingo prize with all bingo cards that have a simultaneous bingo. If the number 1-5 are called first, then every single card will have a V0 bingo. Each card will only get credit for 0.001 prizes. The result is that less than 7% of the prize pool goes to vertical bingos.

V0 0.0123

V1 0.0124

V2 0.0123

V3 0.0123

V4 0.0126

H5 0.1336

H6 0.1333

H7 0.1352

H8 0.1338

H9 0.1345

Da 0.1331

Db 0.1348

If you count raw V, H, and D bingos, then you will get a different result than mine. The 'paradox' is partly the result of the smaller number of combinations that produce a vertical bingo and the large number of cards that share numbers within a vertical column. In my extreme example, all columns on every card require the same 5 numbers to be called.