Joanna has a deck of cards. Each card in the deck has a three-digit positive integer on it, and there is exactly one card in the deck for every three-digit positive integer.
Joanna randomly selects a card from the deck of cards. Determine the probability that the sum of the digits of the integer on this card is \(15\).
To begin, we determine the number of cards in the deck. There are \(999\) positive integers less than \(1000\). Of these, \(90\) are two-digit integers and \(9\) are single-digit integers. Therefore, there are \(999-90-9=900\) three-digit positive integers, and so \(900\) cards in the deck.
Next, we determine the digit combinations that sum to \(15\).
Case 1: One of the digits on the card is a \(0\).
Then the other two digits on the card must add to \(15\). There are two
possibilities for the
digits. The three digits must be \(0,6,9\) or \(0,7,8\).
Case 2: One of the digits on the card is a \(1\), but the
number does not contain a
\(0\).
Then the other two digits on the card must add to \(14\). There are three
possibilities for the
digits. The digits must be \(1,5,9\) or
\(1,6,8\) or \(1,7,7\).
Case 3: One of the digits on the card is a \(2\), but the
number does not contain a
\(0\) or \(1\).
Then the other two digits on the card must add to \(13\). There are three
possibilities for the
digits. The digits must be \(2,4,9\) or
\(2,5,8\) or \(2,6,7\).
Case 4: One of the digits on the card is a \(3\), but the
number does not contain a
\(0\), \(1\), or \(2\).
Then the other two digits on the card must add to \(12\). There are four
possibilities for the
digits. The digits must be \(3,3,9\) or
\(3,4,8\) or \(3,5,7\) or \(3,6,6\).
Case 5: One of the digits on the card is a \(4\), but the
number does not contain a
\(0\), \(1\), \(2\), or \(3\).
Then the other two digits on the card must add to \(11\). There are two
possibilities for the
digits. The digits must be \(4,4,7\) or
\(4,5,6\).
Case 6: One of the digits on the card is a \(5\), but the
number does not contain a
\(0\), \(1\), \(2\), \(3\), or \(4\).
Then the other two digits on the card must add to \(10\). There is only one
possibility for the
digits. The digits must be \(5,5,5\).
Now that we know what combinations of digits can be on the cards, we can determine the number of cards that can be created from each combination.
Case (a): One of the digits on the card is \(0\).
Earlier we found that there were two such digit combinations: \(0,6,9\) and
\(0,7,8\). This is a special case since \(0\)
cannot appear in the number as the
hundreds digit for the number to be a three-digit number. For each of
the two digit combinations, the \(0\)
can be placed in two ways, in the tens digit or the units digit. Once
the \(0\) is placed, the other two
numbers can be placed in the remaining two spots in two ways. Thus, each
digit combination can form \(2\times
2=4\) three-digit numbers. Since there are two digit
combinations, there are \(2\times 4=8\)
cards in the deck that contain a \(0\)
whose digits add to \(15\).
Case (b): All three digits on the card are
different and the number does not contain a \(0\).
From the earlier cases, there are eight such combinations: \(1,5,9\), and \(1,6,8\), and \(2,4,9\), and \(2,5,8\), and \(2,6,7\), and \(3,4,8\), and \(3,5,7\), and \(4,5,6\). For each of these combinations,
the hundreds digit can be placed in three ways. For each of these three
choices for hundreds digit, the tens digit can be placed in two ways.
Once the hundreds digit and tens digit are selected, the units digit
must get the third number. So each combination can form \(3\times 2=6\)
different numbers. Since
there are eight digit combinations, there are \(8\times 6=48\) cards in the
deck that
contain three different digits other than \(0\) whose digits add to \(15\).
Case (c): Two of the digits on the card are the
same and the number does not contain a \(0\).
From the earlier cases, there are four such combinations: \(1,7,7\), and \(3,3,9\), and \(3,6,6\), and \(4,4,7\). For each of these combinations,
the unique number can be placed in one of three spots. Once the unique
number is placed the other two numbers must go in the remaining two
spots. So each digit combination can form three different numbers. Since
there are four digit combinations, there are \(4\times 3=12\) cards in the deck
that do
not contain a \(0\) but contain two
digits the same and whose digits add to \(15\).
Case (d): The three digits on the card are the
same.
From the earlier cases we discovered only one such combination: \(5,5,5\). Only
one card can be produced
using the numbers from this combination.
Combining the counts from the above four cases, there are \(8+48+12+1=69\) cards in the deck with a digit sum of \(15\). Therefore, the probability that Joanna selects card whose digits add to 15 is \(\frac{69}{900}=\frac{23}{300}\). This translates to approximately a \(7.7\%\) chance.
A game is considered fair if there is close to a \(50\%\) chance of winning. If Joanna was playing a game where she can win by drawing a card whose digits sum to \(15\), then this game is definitely not fair.
Can you create a game using this specific deck of cards that is reasonably fair and fun to play?