This document provides suggested solutions to problems from Tutorial 6 on mathematics of games. It includes: 1) Calculating the probabilities of getting three of a kind vs two pairs in a 5-card poker hand and showing three of a kind is less likely. 2) Finding probabilities of drawing specific card combinations in a 5-card stud hand where a pair and king are already on the table. 3) Explaining there are n!/k! ways to arrange a set of n numbers where k of the numbers are the same and the rest are distinct.

1. GEK1544 The Mathematics of Games
Suggested Solutions to Tutorial 6
1. In a poker hand of getting 5 consequent cards out of 52 cards, find the following.
P ( three of a kind , no better) & P ( two pairs , no better) .
Conclude that
P ( three of a kind , no better) < P ( two pairs , no better) .
Suggested solution. There are 4 Aces, and we need three of them. The count is C(4,
3). Likewise for a pair of K, Q, J, 10,., 2. We have
13 · C(4, 3)
ways to form three of a kind. The remaining two cards must be different from the three of
a kind, and must be different from each other. the choices are 4 × 12 and 4 × 11 . Finally
the order to obtain the last two cards is not important, so the total count is
[13 · C(4, 3)] · [4 × 12] · [4 × 11]
= 54, 912 .
P (2, 2)
54, 912
=⇒ P ( three of a kind , no better) = ≈ 0.021129.
C(52, 5)
Likewise, there are 4 Aces, and we just need two of them. The count is C(4, 2). Likewise
for a pair of K, Q, J, 10,., 2. We have
13 · C(4, 2)
ways to a pair. The other pair must be different, so there are
12 · C(4, 2)
ways. The remaining card must be different from the pairs – the choices are 4 × 11.
Finally the order to obtain the two pairs is not important, so the total count is
[13 · C(4, 2)] · [12 · C(4, 2)] · [4 × 11]
= 123, 552 .
P (2, 2)
123, 552
=⇒ P ( two pairs , no better) = ≈ 0.047539.
C(52, 5)

2. 2. In a 5 card stud you are dealt a pair of Jacks. On the table a pair of 4 and a K have
already appeared. There are now 52 − 5 = 47 cards left undistributed. It is your turn to
draw 3 consequent cards. Show the following.
2 45 44
P (J, J, J, X, Y ) = · · ,
47 46 45
2 45 44
P (J, J, X, J, Y ) = · · ,
47 46 45
2 45 44
P (J, J, X, Y, J) = · · .
47 46 45
In the above, we use the notation (J, J, J, X, Y ) , X = J, Y = J to denote that the
following first draw is an Jack, second and third draw are other cards different from Jacks.
X and Y may or may not be the same. Etc. Likewise, find
P (J, J, J, J, X) = ? ,
P (J, J, X, J, J) = ? .
P (J, J, J, X, J) = ? .
Finally, show that
41 3 2
P (J, J, X, X, X) = · · .
47 46 45
Suggested Solution. Given a pair of Jacks and 47 cards left with 2 Jacks in left behind
2
cards, the probability that the third card is a Jack is 47 . The probability that the fourth
card is not a Jack is 46−1 = 46 . Likewise, the probability that the fifth card is not a Jack
46
45
is 45−1 = 44 . Hence
45 45
2 45 44
P (J, J, J, X, Y ) = · · .
47 46 45
Next, given a pair of Jacks and 47 cards left with 2 Jacks in left behind cards, the
probability that the third card is not a Jack is 47−2 = 45 . The probability that the fourth
47 47
card is a Jack is 46 . The probability that the fifth card is not a Jack is 45−1 = 44 . Hence
2
45 45
45 2 44 2 45 44
P (J, J, X, J, Y ) = · · = · · .
47 46 45 47 46 45
Similar argument shows that
47 − 2 46 − 2 2 45 44 2 2 45 44
P (J, J, X, Y, J) = · · = · · = · · .
47 46 45 47 46 45 47 46 45
Likewise,
2 1
P (J, J, J, J, X) = · · 1,
47 46
45 2 1 2 1
P (J, J, X, J, J) = · · = · · 1,
47 46 45 47 46
2 45 1 2 1
P (J, J, J, X, J) = · · = · · 1.
47 46 45 47 46

3. As a pair of 4 has appeared, X = 4 . Suppose X = K, the probability to obtain three
consequent X is
4 3 2
· ·
47 46 45
and there are 13 − 1 − 1 − 1 choice of X (X = 4, J, & K), hence the total is
4 3 2 40 3 2
10 · · · = · · .
47 46 45 47 46 45
If X = K, then the probability is
3 3 1
· · .
47 46 45
Hence the total is
40 3 2 3 2 1 40 3 2 1 2 3 41 3 2
· · + · · = · · + · · = · ·
47 46 45 47 46 45 47 46 45 47 46 45 47 46 45
3. Let {x1 , x2 , ···, xn } be a collection of n numbers, among which k of them are the same,
and the remaining n − k numbers are all distinct. (E.g, for the numbers { 4, 4 , 3, 2 } ,
n = 4 and k = 2 .) Show that there are
n!
P (n, n − k) =
k!
ways to arrange the n-numbers in order. E.g. P (4, 2) = 4 · 3 = 12 :
4, 4, 3, 2 2, 3, 4, 4
4, 4, 2, 3 3, 2, 4, 4
4, 3, 4, 2 2, 4, 3, 4
4, 3, 2, 4 4, 2, 3, 4
3, 4, 4, 2 2, 4, 4, 3
3, 4, 2, 4 4, 2, 4, 3
Suggested Solution. Consider n persons and r = n − k seats. The number of ways to
fill the seats in the order important manner is
n! n! n!
P (n, r) = = = .
(n − r) ! [n − (n − k)] ! k!
We relate this to the present problem by the following picture. Assume without loss of
generality that
x1 = x2 = · · · = xk
and xk+1 , · · ·, xn are all distinct. There are n empty space to be filled with the numbers
{x1 , x2 , · · ·, xn } . E.g. in case {4, 4, 3, 2}

4. Next, imagine there are n people standing in the n positives. E.g.
Person Person Person Person
There are r = n − k seats, e.g.
⊥ ⊥
On the seats, we arrange the numbers xk+1 , · · ·, xn . E.g.
2 ⊥ 3 ⊥
Next, the seats are open for the people to sit down. The person sits in seat 1 takes the
number xn−k , and return to his originally standing position. Etc. E.g.
Person (3) Person Person Person (2)
For the k people without seats (hence without numbers), we give each one of the one of
the same numbers
{x1 , x2 , · · ·, xk }.
E.g.
Person (3) Person (4) Person (4) Person (2)
This provides a way to arrange the numbers. E.g.
3, 4, 4, 2.
The procedure described above tells us a way to count the combinations. Therefore, the
answer is P (n, r) = P (n, (n − k)) .