March 1st, 2017 at 11:00:47 AM
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Hello everyone and Wizard,

I've been tasked at work to analyze a new bingo game. I have a couple of questions regarding coverall/blackout bingo (all 24 numbers + 1 free space needs to be dabbed). I would appreciate it if anyone could confirm that the following statements are true.

1) the cumulative distribution function of number of calls needed for coverall/blackout (with only 1 bingo card in play) can be modeled by the probability mass function of the hypergeometric distribution (as it can in Keno).

2) for multiple distinct cards in play, say n, can we use the following formula: 1-(1-p)^n to model the cumulative distribution function of number of calls where p is the probability of a single card achieving coverall?

As for 2, the statement is contradictory to Wizard's website; please refer to the Multi-Player Bingo section (https://wizardofodds.com/games/bingo/probabilities/2/). I agree with Wizard that the formula does not hold when you try to model anything other than coverall/blackout. However, I believe the formula is valid when we are modeling coverall/blackout game. I'd be curious to hear if anyone agrees with me.

I've been tasked at work to analyze a new bingo game. I have a couple of questions regarding coverall/blackout bingo (all 24 numbers + 1 free space needs to be dabbed). I would appreciate it if anyone could confirm that the following statements are true.

1) the cumulative distribution function of number of calls needed for coverall/blackout (with only 1 bingo card in play) can be modeled by the probability mass function of the hypergeometric distribution (as it can in Keno).

2) for multiple distinct cards in play, say n, can we use the following formula: 1-(1-p)^n to model the cumulative distribution function of number of calls where p is the probability of a single card achieving coverall?

As for 2, the statement is contradictory to Wizard's website; please refer to the Multi-Player Bingo section (https://wizardofodds.com/games/bingo/probabilities/2/). I agree with Wizard that the formula does not hold when you try to model anything other than coverall/blackout. However, I believe the formula is valid when we are modeling coverall/blackout game. I'd be curious to hear if anyone agrees with me.

March 1st, 2017 at 6:34:09 PM
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Using the HGD provides a close estimate. For the true probabilities, you need to consider that not all distributions of X calls are equal, and the distribution is something that all cards in play are subject to.

So you have to split the calculation into two parts. Let's say you want the probability of a coverall in X calls (or less) when Y cards are in play:

1) Identify (iterate through) every possible distribution of X calls. For example, with 50 calls, there are 20,176 possible distributions (including distributions where a coverall is not possible). Determine the probability of each distribution.

Then, for each distribution:

2) Determine the probability of achieving a coverall given the current distribution. This will of course be zero when there are fewer than 4 Ns or 5 Bs/Is/Gs/Os.

3) Apply the 1-((1-p)

4) Multiply the result of #3 by the distribution probability of #1, and accumulate (sum) these results for each distribution to determine the overall probability of a bingo within X calls or less for Y cards.

For example, for a coverall in 50 calls, here are the true probabilities (vs. the HGD estimates) for a few card quantities:

1 card = 0.0000047150811362 (vs. the HGD estimate of 0.0000047150811337)

100 cards = 0.00047138 (vs. the HGD estimate of 0.00047140)

500 cards = 0.00235428 (vs. the HGD estimate of 0.00235477)

1000 cards = 0.00470206 (vs. the HGD estimate of 0.00470399)

5000 cards = 0.02325221 (vs. the HGD estimate of 0.02329973)

The above calculations assume the following:

- standard North American bingo cards are used (5 Bs, 5 Is, 4 Ns, 5 Gs, 5 Os)

- the cards in play are random

- the numbers are drawn randomly

So you have to split the calculation into two parts. Let's say you want the probability of a coverall in X calls (or less) when Y cards are in play:

1) Identify (iterate through) every possible distribution of X calls. For example, with 50 calls, there are 20,176 possible distributions (including distributions where a coverall is not possible). Determine the probability of each distribution.

Then, for each distribution:

2) Determine the probability of achieving a coverall given the current distribution. This will of course be zero when there are fewer than 4 Ns or 5 Bs/Is/Gs/Os.

3) Apply the 1-((1-p)

^{Y}) formula here with p = the result of #2 and Y = the number of cards in play.4) Multiply the result of #3 by the distribution probability of #1, and accumulate (sum) these results for each distribution to determine the overall probability of a bingo within X calls or less for Y cards.

For example, for a coverall in 50 calls, here are the true probabilities (vs. the HGD estimates) for a few card quantities:

1 card = 0.0000047150811362 (vs. the HGD estimate of 0.0000047150811337)

100 cards = 0.00047138 (vs. the HGD estimate of 0.00047140)

500 cards = 0.00235428 (vs. the HGD estimate of 0.00235477)

1000 cards = 0.00470206 (vs. the HGD estimate of 0.00470399)

5000 cards = 0.02325221 (vs. the HGD estimate of 0.02329973)

The above calculations assume the following:

- standard North American bingo cards are used (5 Bs, 5 Is, 4 Ns, 5 Gs, 5 Os)

- the cards in play are random

- the numbers are drawn randomly

Last edited by: JB on Mar 2, 2017

March 2nd, 2017 at 8:47:20 PM
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JB's answer sounds right to me. If you know the probability of covering any given pattern in x calls is y for one card, you can't assume is it 1-(1-y)^n for n cards. This is because the cards are correlated. All of them have 5 B's, 5 I's... If there is a smooth distribution of columns called, then all of them are more likely to hit early. I've seen it happen lots of times where it takes a long time for one letter to get called. When it finally does, a ton of people call BINGO.

The easiest way to do multi-card bingo math is by random simulation. I know it isn't as elegant as a closed forum answer, but sometimes the easiest way is ... the easiest way.

The easiest way to do multi-card bingo math is by random simulation. I know it isn't as elegant as a closed forum answer, but sometimes the easiest way is ... the easiest way.

It's not whether you win or lose; it's whether or not you had a good bet.

March 3rd, 2017 at 7:44:22 AM
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Hello Wizard, JB,

Thank you for your responses. Indeed, I had not considered the fact that there could be instances where only 4 B's have been called in a given number of calls and there is no coverall bingo possible. I understand now why the cards are correlated.

One of the variants of the new game we are contemplating does not have the column restrictions, i.e. any numbers between 1 and 75 can go anywhere in the 5x5 matrix. I believe HGD works for this game as it is really Keno but in Bingo format.

Cheers,

Thank you for your responses. Indeed, I had not considered the fact that there could be instances where only 4 B's have been called in a given number of calls and there is no coverall bingo possible. I understand now why the cards are correlated.

One of the variants of the new game we are contemplating does not have the column restrictions, i.e. any numbers between 1 and 75 can go anywhere in the 5x5 matrix. I believe HGD works for this game as it is really Keno but in Bingo format.

Cheers,

October 15th, 2021 at 11:31:54 AM
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JB, could you expound upon why "not all distributions of X calls are equal" and what you mean by distribution? Are you referring to the particular collection of balls drawn? I'm curious why the HGD only gives an estimate when it's 1 card as opposed to the true probability. It seems to me that in a 1 card scenario, the game would be identical to Keno in terms of ball drawing and getting catches. Thanks in advance.

October 15th, 2021 at 2:09:02 PM
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Quote:aspiringmathJB, could you expound upon why "not all distributions of X calls are equal" and what you mean by distribution? Are you referring to the particular collection of balls drawn? I'm curious why the HGD only gives an estimate when it's 1 card as opposed to the true probability. It seems to me that in a 1 card scenario, the game would be identical to Keno in terms of ball drawing and getting catches. Thanks in advance.

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I think an example of what JB was referring to is explained in the above post by Wizard.

So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.

October 26th, 2021 at 8:09:48 AM
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Thank you! It is starting to make sense, and I can even figure out how the number of distributions is 20,176. I am however having difficulty getting 1 card = 0.0000047150811362 (vs. the HGD estimate of 0.0000047150811337).

Since the distributions are not uniform, the probability of a particular distribution cannot be 1/20176. So then I thought we could calculate the probability of any particular distribution (B,I,N,G,O) as comb(15,B)*comb(15,I)*comb(15,N)*comb(15,G)*comb(15,O)/comb(75,50).

Then I was thinking, given the distribution (B,I,N,G,O), the probability of getting a coverall would be:

comb(B,5)*comb(I,5)*comb(N,4)*comb(G,5)*comb(O,5)/(comb(15,4)*comb(15,5)^4).

I multiplied these and summed over all 20176 distributions (B,I,N,G,O) but got the HGD probability again.

It didn't dawn on me initially, but this matches the HGD probability calculated as comb(24,24)*comb(51,26)/comb(75,50).

So great, now I have two ways of thinking about calculating the probability with the HGD but am no closer to calculating the probability of a particular distribution or getting a coverall without the HGD. Any thoughts/tips?

Since the distributions are not uniform, the probability of a particular distribution cannot be 1/20176. So then I thought we could calculate the probability of any particular distribution (B,I,N,G,O) as comb(15,B)*comb(15,I)*comb(15,N)*comb(15,G)*comb(15,O)/comb(75,50).

Then I was thinking, given the distribution (B,I,N,G,O), the probability of getting a coverall would be:

comb(B,5)*comb(I,5)*comb(N,4)*comb(G,5)*comb(O,5)/(comb(15,4)*comb(15,5)^4).

I multiplied these and summed over all 20176 distributions (B,I,N,G,O) but got the HGD probability again.

It didn't dawn on me initially, but this matches the HGD probability calculated as comb(24,24)*comb(51,26)/comb(75,50).

So great, now I have two ways of thinking about calculating the probability with the HGD but am no closer to calculating the probability of a particular distribution or getting a coverall without the HGD. Any thoughts/tips?