Buying every possible combination of something can be an entertaining thought experiment. With lottery tickets, for example, we might wonder what it would cost to purchase every possible number combination. Of course, buying every combination is usually not feasible in reality. But exploring this hypothetical scenario can reveal some interesting insights.
In this article, we’ll look at a few different products and estimate the cost of buying every possible combination. We’ll start with simpler items like dice and playing cards, and work our way up to more complex combinations like lottery tickets and phone numbers. Our goal is to get a sense of just how astronomically expensive it would be to exhaust all possible permutations. This mental exercise will illustrate the sheer number of options for different combinations.
Dice
Let’s start with a simple six-sided die. There are six possible outcomes for a single roll of one die: 1, 2, 3, 4, 5, or 6. To buy every combination, we would need six dice, one for each number.
Cost to buy one of each die:
| Item | Unit Price |
| 1 six-sided die | $1 |
With six dice at $1 each, the total cost would be $6. Pretty straightforward! Now let’s look at the permutations for rolling two dice.
Possible combinations rolling two dice:
| 1-1 | 1-2 | 1-3 | 1-4 | 1-5 | 1-6 |
| 2-1 | 2-2 | 2-3 | 2-4 | 2-5 | 2-6 |
| 3-1 | 3-2 | 3-3 | 3-4 | 3-5 | 3-6 |
| 4-1 | 4-2 | 4-3 | 4-4 | 4-5 | 4-6 |
| 5-1 | 5-2 | 5-3 | 5-4 | 5-5 | 5-6 |
| 6-1 | 6-2 | 6-3 | 6-4 | 6-5 | 6-6 |
There are 6 x 6 = 36 possible combinations when rolling two dice. To purchase every combination, we would need 36 pairs of dice, bringing the total to 72 dice at $1 each.
Cost to buy all combinations of two dice:
| Item | Quantity | Unit Price | Total Price |
| Six-sided dice | 72 | $1 | $72 |
The price increases more dramatically when we move to three dice. There are 6 x 6 x 6 = 216 possible combinations of three dice. Buying every combo would require 216 sets of three dice, meaning 648 total dice at $1 each:
Cost to buy all combinations of three dice:
| Item | Quantity | Unit Price | Total Price |
| Six-sided dice | 648 | $1 | $648 |
You can see how this gets expensive fast! For four dice, there would be 64 = 1,296 possible combinations. Five dice takes us to 65 = 7,776 combos. And if we want every outcome for rolling six dice, it would require 66 = 46,656 sets of dice for a whopping 279,936 dice total! Just for simplicity’s sake, let’s assume our budget caps out at six dice.
Total cost to buy all combinations of six-sided dice:
| Number of Dice | Possible Combinations | Total Dice Required | Price per Die | Total Price |
| 1 | 6 | 6 | $1 | $6 |
| 2 | 36 | 72 | $1 | $72 |
| 3 | 216 | 648 | $1 | $648 |
| 4 | 1,296 | 5,184 | $1 | $5,184 |
| 5 | 7,776 | 38,880 | $1 | $38,880 |
| 6 | 46,656 | 279,936 | $1 | $279,936 |
| Total | 324,726 | $324,726 |
So for six-sided dice, the total cost to cover every possible outcome through six rolls would be $324,726. That’s already quite pricey for our humble dice game! Now let’s look at a scenario with more variation: a standard 52-card deck.
Playing Cards
A standard 52-card playing card deck contains 52 unique cards consisting of four suits (clubs, diamonds, hearts and spades) with 13 ranks (Ace through 10, then Jack, Queen, King) in each suit. That’s 52 factorial (52!) total combinations, which is an unfathomably huge number:
52! = 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000
Clearly, buying every possible permutation of a 52-card deck is utterly impossible. But just for fun, let’s imagine we could buy all 80 undecillion (that’s 36 zeroes!) unique card orders. A standard bicycle playing card deck costs about $2.
If we multiply $2 by 80 undecillion, the total cost comes out to…
$160,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000
That’s 160 octillion dollars, far more than the entire wealth and resources of our planet! This truly puts into perspective just how many combinations exist for a single deck of cards. Let’s bring it back down and look at a simpler scenario: the potential combinations for pulling 5 cards out of a deck.
Possible 5-card combinations from 52-card deck:
| Number of combinations | 2,598,960 |
If we wanted to buy every possible 5-card hand, we would need 2,598,960 sets of cards. At $2 per deck, that’s:
Cost to buy all 5-card combinations:
| Item | Quantity | Unit Price | Total Price |
| 52-card decks | 2,598,960 | $2 | $5,197,920 |
So buying all possible 5-card poker hands would cost over 5 million dollars! And that’s just for 5 cards out of 52…as we saw above, the complete 52! combinations quickly becomes unfathomable. But this gives us a sense of how numbers expand exponentially with more variation.
Speaking of exponential growth, let’s look next at lottery tickets. Lotteries often have players choose 6 numbers out of 40 or 50 options. We’ll use the examples of Powerball and Mega Millions to explore the potential combinations and costs.
Lottery Tickets
Let’s start with Powerball. Players choose 5 numbers from 1-69, and 1 Powerball number from 1-26. That makes the total possible combinations:
Powerball combinations:
| 5 numbers from pool of 69 | C(69,5) = 11,238,513 |
| 1 number from pool of 26 | 26 |
| Total combinations | 11,238,513 x 26 = 292,201,338 |
Multiplying the number of possible selections reveals there are over 292 million total Powerball combinations. At $2 per ticket, purchasing every combo would cost:
Cost to buy every Powerball combination:
| Item | Quantity | Unit Price | Total Price |
| Powerball tickets | 292,201,338 | $2 | $584,402,676 |
Almost $585 million for every possible Powerball ticket! Now let’s look at the even more complex Mega Millions:
Mega Millions combinations:
| 5 numbers from pool of 70 | C(70,5) = 12,271,512 |
| 1 number from pool of 25 | 25 |
| Total combinations | 12,271,512 x 25 = 306,787,800 |
Cost to buy every Mega Millions combination:
| Item | Quantity | Unit Price | Total Price |
| Mega Millions tickets | 306,787,800 | $2 | $613,575,600 |
Buying every possible Mega Millions ticket would cost over $613 million! And that’s just for the base game – we’re not even including the Megaplier option which further multiples the combinations. While realistically no one could purchase every permutation, it illustrates the immense number of options.
Phone Numbers
Finally, let’s look at phone numbers. We’ll use the North American Numbering Plan’s 10-digit format. The first 3 digits are the area code, so there are 800 possibilities (numbers 555 are reserved). The next 3 digits are the central office code, for which there are also 800 options. The last 4 digits are the station code from 0000 to 9999: 10,000 possibilities.
Possible 10-digit phone number combinations:
| Area code options | 800 |
| Central office code options | 800 |
| Station code options | 10,000 |
| Total combinations | 800 x 800 x 10,000 = 6,400,000,000 |
There are over 6 billion potential phone numbers in the North American numbering plan format. The average monthly cost for a phone line is about $15.
To purchase every possible 10-digit number would therefore cost:
Cost to buy every phone number combination:
| Item | Quantity | Unit Price (monthly) | Total Monthly Price |
| 10-digit phone numbers | 6,400,000,000 | $15 | $96,000,000,000 |
Almost $100 billion per month! And that’s just for one month of service. Clearly buying up every available phone number combination would be astoundingly expensive. While hypothetical, this example illustrates the widespread possibilities opened up by numeric variations.
Conclusion
Buying every possible combination of something is an amusing thought experiment to grasp the exponential growth of variations. As we looked at increasingly complex examples from dice to cards to lottery tickets and phone numbers, the potential combinations and costs spiraled astronomically upwards. While exhausting all combinations is not realistically achievable, this exercise helps drive home the sheer size and scale of possible permutations. It adds perspective on probability and numeric variation. Next time you roll some dice, play the lottery or give out your phone number, consider the immense number of options made possible through different combinations.
