cardshark
Joined: Nov 30, 2009
• Posts: 233
March 1st, 2017 at 11:00:47 AM permalink
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.
JB
Joined: Oct 14, 2009
• Posts: 2053
March 1st, 2017 at 6:34:09 PM permalink
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)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
Wizard
Joined: Oct 14, 2009
• Posts: 16193
March 2nd, 2017 at 8:47:20 PM permalink
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.
It's not whether you win or lose; it's whether or not you had a good bet.
cardshark
Joined: Nov 30, 2009