While playing a friendly game of Bingo the other day, my mind started thinking about how many possible unique cards could be made.
I know you math whizzes out there can calculate the correct number, but I don’t remember enough math to come up with the correct formula.
For those not familiar with the game, each card has a 5 x 5 grid with numbers in each square, except the center square, which is the “Free Space.” The columns are labeled B, I, N, G and O. The B column has 5 numbers between 1 and 15. The I column has 5 numbers between 16 and 30. The N column has 4 numbers between 31 and 45, plus the Free Space. The G column has 5 numbers between 46 and 60, and the O column has 5 numbers between 61 and 75. No number can appear on the card more than once.
Anyone care to whip out their calculator and come up with the number of possible combinations?
The first square under the B can be any one of 15 numbers. Whatever it is, the second can be any of the remaining 14. Continuing the pattern, the possible combinations under the B would be 15 x 14 x 13 x 12 x 11. Same for I, G, and O, while the possible number of combinations for the N column would be 15 x 14 x 13 x 12. The total number of possibilities is all of these numbers multiplied together, or 15^5 x 14^5 x 13^5 x 12^5 x 11^4.
Off of the top of my head, it would be basic permutation and BCP. Each column except the middle one has 15x14x13x12x11 permutations. The “N” column eliminates one space so it would have 15x14x13x12 permutations yeilding 15^5 x 14^5 x 13^5 x 12^5 x 11^4 possible boards.
Just over 5.5 x 10^26 or 550 sextillion. To put it in perspective,
As of 2010 there were thought to be only 300 sextilion stars in the universe.
The mass of Saturn is 560 sextillion kilograms
550 sextillion molecules is just under 1000 moles
Printing cards once a second, you’d need 1.75×10¹⁷ centuries. If you started at any point in human history, when you were done not only would the Sun be dead but most galaxies as we know them would essentially be ‘dead’ as well and, what’s even more profound, it’s possible no new stars would be forming as the Universe itself winds down for its eventual heat death.
Keep in mind that for a blackout game, the order of the numbers does not matter. For any pulled number set, then number of identical cards that win are:
B,I,G,O: The 5 numbers can be arranged 5! ways each, 120^4
N the numbers are arranged 4! ways or 24.
So divide the above number of cards by 24 * 120^4= 4976640000 (about5 billion) to get the number of unique number combinations for blackout.
I thought there were “only” about 10[sup]22[/sup] stars in the universe (which is 10 sextillion, I think). So, somewhere, maybe in both places above, there may be an error, or at least a discrepancy.
I guess the punch line for the OP is “Don’t try to collect the whole set.” Your house would be *really *full. You couldn’t be on Hoarders because there wouldn’t be room to get the camera crew in there.
By your description, you could have the sequence 9, 14, 3, 12, 2. Or you could have the sequence 12, 14, 9, 2, 3. And by your maths , these would count as different.
This does not happen, at least not on any bingo card I’ve ever seen. Every bingo game I’ve seen has the numbers sorted into order so the only possible sequence is 2, 3, 9, 12, 14.
You have to eliminate the various identical sets of numbers from the equation.
Thus for a column of 5 there are (15x14x13x12x11) / (5x4x3x2x1) = 3003 possibilities.
For a column of 4 there are (15x14x13x12) / (4x3x2x1) = 1365 possibilities
Grand total is 3003 x 3003 x 1365 x 3003 x 3003 = 111,007,923,832,370,565
Per Paper density - Wikipedia, I WAG 10# paper as having a thickness of .0017 inches = 1.7E-3 inches per sheet.
American style bingo cards come in 5.5E23 or 5.5E26 varieties depending on who you believe upthread. (sextillion vs septillion).
So assuming the smaller number, that’s a stack just ~1.5E16 miles tall. If it’s really septillions then your stack is 1.5E19 miles tall.
For English style bingo cards available in a mere 1.1E17 varieties, your stack will be a mere 2.9E9 miles tall.
IOW the short English stack will reach to more or less Pluto.
The taller of the two American stacks will reach 10 billion *times *farther than Pluto. In fact it’ll be about 2.5 million light years tall. Which is far enough to reach the Andromeda Galaxy Andromeda Galaxy - Wikipedia
And yes, I ignored the effect the stack’s gravity will have on itself. Certainly a recangular stack of paper roughly 5 inches by 8 inches by 2.5 million light years will generate enough gravity to tend to collapse into a sphere.
there is definitely a difference between combinations and permutations especially given that the amount of permutations in the b column would be 360360 and combinations would be 3003, i am still not certain if my maths has been correct here thou i got 1930716 x 10 ^9 i have a seen away where it was 3003 x 3003 x 1365 x 3003 3003 also another where it was 3003 ^5 + 3003 ^5 + 1365 ^4 + 3003 ^5 + 3003^5.
i done it 3003 C 5 where n = 3003 an n! or k equals 5 then i took away 3003-1365 = 1638 , where 1638 C 5 from the awnser to get the some 2000 trillion. i am still not certain that it it shouldnt be done as 3003 C 3003 which gives an infinate awnser…:smack:
People in the thread are acting like 552 septillion is a big number. But if you somehow put a different bingo card on each atom, you could fit them all into about 16 cans of beer. Actually, 8 cans of beer would be enough if you eliminate redundancy: each bingo card has a mirror-image twin that wins whenever its twin wins.
Bingo cards are sold in sets of 600. There are, in principle, (552 septillion)^600/600! or almost 10[sup]16000[/sup] possible sets. Now that’s a big number.
