by Brian Tomasik
Written 17, 18, 20, 22 June 2003
This page explains a probability question that I invented and tried to solve on my own. I never figured out the general solution, so this piece is incomplete. Moreover, the existing content of this piece may contain errors.
Imagine a bingo board with five spaces across, five spaces down, and no free spaces. A bingo requires five spaces in a row horizontally, vertically, or diagonally. Each of the 25 squares has one unique letter from A to Y. If a different letter from A to Y is called by the bingo-game host each time without repetition, what is the chance of getting a bingo after 5, 6, 7, etc. letters are called? How much does the probability improve each time the number of letters called increases? Is there a general formula for the probability depending on the number of letters called?
If only five letters are called
I began by thinking about the probability of a bingo in just one row of the board. The chance that the first letter called will be in any particular square is 1/25, and since it doesn’t matter which square of the row is chosen first, the probability that any of the squares will be chosen is 5/25. Once the first square is chosen, there remain only 24 squares that can be picked, and if the first square selected was one of the five in that particular row, there are only four possible squares left in that row, making the probability that the next letter is also on a square in that row 4/24. If the second square chosen is also in the row, the probability that the third letter will fall in the row is 3/23, for the fourth letter, it’s 2/22, and for the fifth letter, it’s 1/21. Thus, the probability for a bingo in one row after five letters are called is (5/25)(4/24)(3/23)(2/22)(1/21) = 1/53130 = 1.882175795 · 10-5.
There are twelve ways to get a bingo, with 5 rows across, 5 rows down, and 2 diagonal rows. Therefore, twelve times the above number should be approximately the probability for a bingo any of the twelve ways, although it isn’t exact because, for example, if there were 100,000 ways to get a bingo and the above number were multiplied by 100,000, the result would exceed one. 12(1/53130) = 2/8855 = 2.258610954 · 10-4.
I next thought about the problem in terms of the probability of getting a bingo depending on the square chosen. Since there are only five letters called, each square chosen must be in a single row for a success. Once any square, except the nine on diagonal lines, is selected, which has a probability of 16/25, there remain only eight squares that could be called next that would be in the same row as the first. Once one of these is selected, there are only three other squares in the row to be filled, and subsequently, there will be only two and then only one. Thus, the probability for five in a row beginning with one of the non-diagonal squares and with only five letters called is (16/25)(8/24)(3/23)(2/22)(1/21).
If the square chosen is on the lines of one of the diagonals, which has a 9/25 chance of occurring, there are twelve other squares that, if chosen, would be in the same row. However, once one of these twelve spaces is selected, the rest of the squares must be in one certain row, so the number of possible squares after two letters have been chosen is again three, then two, then one, making the probability of five in a row with only five letters called and beginning with a diagonal square (9/25)(12/24)(3/23)(2/22)(1/21). Adding this to the other probability should be the chance of getting five in a row with five letters called, regardless of the location of the first square: (16/25)(8/24)(3/23)(2/22)(1/21) + (9/25)(12/24)(3/23)(2/22)(1/21) = 59/265650 = 2.220967438 · 10--4, which is quite close, although slightly lower (as it probably should be) than the probability obtained by the first method.
Because they only differ by the numerators of the first and second fractions in each term, I attempted to simplify the two fraction products used above into one term by finding the average of the eight and twelve in the numerators of the second fraction. Sixteen times the numerator is eight, and nine times it is twelve, out of a total of 25 times: (16 · 8 + 9 · 12)/25 = 236/25 = 9.44. Since it is guaranteed that one of the squares on the entire board will be chosen, the first fraction in the product will always be 25/25. Substituting 9.44 for both eight and twelve into one term: (25/25)(9.44/24)(3/23)(2/22)(1/21) = 2.220967438 · 10-4, which is exactly the same answer as above.
If six letters are called
As always, the chance that any square will be chosen is 25/25. I began by investigating the possibilities if the second square chosen is not in line with the first, which has a probability of 14.56/24, or one minus the probability that it will be in the same row: 1-(9.44/24) = 14.56/24. If this situation occurs, either the remaining four squares must fall into a row with the first one chosen, the probability of which is (25/25)(14.56/24)(9.44/23)(3/22)(2/21)(1/20), or the remaining four squares will be part of a row created by the second, rather than the first, square chosen, which has a probability of (25/25)(14.56/24)(9.44/23)(3/22)(2/21)(1/20); the chance of either of these two possibilities, therefore, is the same.
Those situations, however, only show the result if the “extra” (sixth) square comes first or second. Once two squares chosen lie in the same row, there is no possibility that a third square outside of that row could successfully form its own row since only six letters are called. Therefore, there is only (??) way the third square called could be the extra square: (25/25)(9.44/24)(20/23)(3/22)(2/21)(1/20). (The fraction for the extra square was 20/23 because there were only three possible squares out of 23 total squares that would be within the row at that point.) Following the same procedure, the probability if the extra square is fourth is (25/25)(9.44/24)(3/23)(20/22)(2/21)(1/20), if the extra square is fifth, (25/25)(9.44/24)(3/23)(2/22)(20/21)(1/20), and if the extra square is sixth, (25/25)(9.44/24)(3/23)(2/22)(1/21)(20/20). The sum of all of these terms (counting the first and second ones separately, even though they are the same result, because they are obtained two different ways), is approximately 0.0012117598, which, assuming the two probabilities to be correct, is about 5.456000001 times better than the probability when only five letters are called.
My original writeup included speculations about a more general formula, but the details were messy and probably wrong. The formula did work to reproduce the above results for 5 or 6 total letters called.