Math problem if you are WAY too bored

[Justin7]

You have a bingo board, with the numbers 1-25. The center square starts off filled.

What is the chance of getting a 5-in-a-row in 10 draws or fewer?

This problem is directly from PS2's Dragon Quest VIII, although I simplified it.

[Justin7]

Oh - assume each draw of a number is random, and there are no duplicates (i.e. normal Bingo, except only 25 numbers with 1 already filled).

[dwaechte]

Sounds like fun, this will get me going in the morning when I get to work. Just to clarify in the meantime, I understand that to mean that every number drawn will be on your card, it's just whether they form a straight line in the first 10 draws, correct?

[Justin7]

Correct. Every number drawn will fill a spot on your board (normal Bingo doesn't follow that).

For double bonus points: figure out the odds of winning in X draws, where X must be >3 and <11.

[dwaechte]

See, now this is why work sucks and my employers are completely unreasonable. I come into the office and want to do dick all for the first little while and do this problem, but nooo, I have to prepare some worksheets for my boss first. Bullshit.

This is what I came up with:

We first figure out the total number of rows/columns/diagonals, in other words the combos that form a “winning” combination. There are 4 which require only 4 good numbers(plus the freebie in the middle), and 8 which require 5 straight numbers.

Then you multiply all those possibilities by the odds that ONE combination would hit.

Basically it would look like this:

4 draws = 4/(24C4)

5 draws = 4(20)/(24C4) + 8/(24C5)

6 draws = 4(20(19)/(24C4) + 8(19)/(24C5)

7 draws = 4(20)(19)(18)/(24C4) + 8(19)(18)/(24C5)

Etc…

Does that work or am I just an idiot?



I think a better question is what the hell were you doing playing Dragon Quest VIII lol? Those games ain’t gonna cap themselves Justin. Unless…