There's an event coming up where $FAVORITE_COMPANY is going to announce a bunch of upcoming products. A fan of that company has 25 anticipated/desired announcements, ranked by likelihood, and wants to arrange them into a Bingo card with the highest possible chance of winning. Optionally, house rules might count "four corners plus center square" as a valid Bingo.

My first thought was to rank squares by how many Bingoes each one is part of:

5 in the center, 4 in the corners, 3 on the remaining diagonal spaces, 2 everywhere else

Then, picturing the center as a free space, I realized that Bingoes crossing the center square are more likely since you only need four spaces for a Bingo:

Same as above, but '2's crossing the center square are changed to '2+'

After this I quickly realized that I was in over my head:

  • The center square isn't the only square that affects probability of Bingoes including it, especially if it's merely the most likely announcement, not a free space; the probabilities of every square in a Bingo must be taken into account when ranking each square within it.
  • If we do count four corners + center as a Bingo, our first five placements can be the five most likely announcements without worrying about dumping the highest-chance items into one row/column/diagonal. Otherwise, or after first five placements in any case, we have to figure out whether to evenly distribute items or front-load the most likely items into a single Bingo.

So, three questions:

  1. What order should I place items into a single card to get the highest probability of a win?

  2. If I were to generate several cards with high probabilities of a win, so that consecutive groups of items are shuffled among similar-probability squares, what would an accurate diagram like my two images above look like?

  3. Approximately how much does the spacing between announcement probabilities affect all this, if at all?

EDIT: For the purpose of this scenario, the order of announcements doesn’t matter. If there are multiple cards, they are all examined once at the end of the event, and everyone with at least one Bingo ties for the win with no tie-breakers.


Some thoughts / too long for a comment

First of all the probability model isn't quite clear.

A natural way to interpret the problem is that each announcement $j$ independently has a probability $p_j$ of being made. The number of announcements actually made can range from $0$ to $25$. Then a bingo card wins if any of its $12$ lines (optionally count "$4$ corners plus center" as a $13$th "line") is good. But in this setting, the order of announcement doesn't matter. In particular your card doesn't have to race against other cards. In this case, the prob of winning can be exactly calculated (via Inclusion-Exclusion), although the sum is tedious.

If you have to race against other cards, then the order of announcement matters, and it is unclear (at least to me) what is the best / most natural way to model that. Maybe randomly permute the $25$ announcements first and then for each one announce it with its probability? Or should the random permutation be biased so that a higher-prob announcement has a higher chance of coming first?

In terms of placement, all versions of the problem seem very hard to solve exactly and in general. It certainly seems like a good heuristic to put the higher-prob announcements at the squares shared by the most lines. OTOH, it might also be a good heuristic to put all the higher-prob announcements on the same line -- in the first model this might be your main chance to win, and in the second model this might be your main chance to win first. This might help break ties among squares that tie according to the first heuristic.

  • $\begingroup$ Thanks for your comment! I would say in this scenario that you’re never racing; if there are different cards, you check all of them one time at the end of the event, and everyone with a Bingo wins, no matter when in the event each square was marked. $\endgroup$ May 10 '19 at 20:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.