![GEK1544 The Mathematics of Games
Suggested Solutions to Tutorial 6
1. In a poker hand of getting 5 consequent cards out of 52 cards, 鍖nd 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 揃 C(4, 3)
ways to form three of a kind. The remaining two cards must be di鍖erent from the three of
a kind, and must be di鍖erent 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 揃 C(4, 3)] 揃 [4 12] 揃 [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 揃 C(4, 2)
ways to a pair. The other pair must be di鍖erent, so there are
12 揃 C(4, 2)
ways. The remaining card must be di鍖erent 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 揃 C(4, 2)] 揃 [12 揃 C(4, 2)] 揃 [4 11]
= 123, 552 .
P (2, 2)
123, 552
= P ( two pairs , no better) = 0.047539.
C(52, 5)](https://image.slidesharecdn.com/s-6-100502121632-phpapp02/85/S-6-1-320.jpg)
![As a pair of 4 has appeared, X = 4 . Suppose X = K, the probability to obtain three
consequent X is
4 3 2
揃 揃
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 揃 揃 揃 = 揃 揃 .
47 46 45 47 46 45
If X = K, then the probability is
3 3 1
揃 揃 .
47 46 45
Hence the total is
40 3 2 3 2 1 40 3 2 1 2 3 41 3 2
揃 揃 + 揃 揃 = 揃 揃 + 揃 揃 = 揃 揃
47 46 45 47 46 45 47 46 45 47 46 45 47 46 45
3. Let {x1 , x2 , 揃揃揃, 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 揃 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
鍖ll 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 = 揃 揃 揃 = xk
and xk+1 , 揃 揃 揃, xn are all distinct. There are n empty space to be 鍖lled with the numbers
{x1 , x2 , 揃 揃 揃, xn } . E.g. in case {4, 4, 3, 2}](https://image.slidesharecdn.com/s-6-100502121632-phpapp02/85/S-6-3-320.jpg)
S 6
- 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, 鍖nd 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 揃 C(4, 3) ways to form three of a kind. The remaining two cards must be di鍖erent from the three of a kind, and must be di鍖erent 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 揃 C(4, 3)] 揃 [4 12] 揃 [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 揃 C(4, 2) ways to a pair. The other pair must be di鍖erent, so there are 12 揃 C(4, 2) ways. The remaining card must be di鍖erent 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 揃 C(4, 2)] 揃 [12 揃 C(4, 2)] 揃 [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 ) = 揃 揃 , 47 46 45 2 45 44 P (J, J, X, J, Y ) = 揃 揃 , 47 46 45 2 45 44 P (J, J, X, Y, J) = 揃 揃 . 47 46 45 In the above, we use the notation (J, J, J, X, Y ) , X = J, Y = J to denote that the following 鍖rst draw is an Jack, second and third draw are other cards di鍖erent from Jacks. X and Y may or may not be the same. Etc. Likewise, 鍖nd 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) = 揃 揃 . 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 461 = 46 . Likewise, the probability that the 鍖fth card is not a Jack 46 45 is 451 = 44 . Hence 45 45 2 45 44 P (J, J, J, X, Y ) = 揃 揃 . 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 472 = 45 . The probability that the fourth 47 47 card is a Jack is 46 . The probability that the 鍖fth card is not a Jack is 451 = 44 . Hence 2 45 45 45 2 44 2 45 44 P (J, J, X, J, Y ) = 揃 揃 = 揃 揃 . 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) = 揃 揃 = 揃 揃 = 揃 揃 . 47 46 45 47 46 45 47 46 45 Likewise, 2 1 P (J, J, J, J, X) = 揃 揃 1, 47 46 45 2 1 2 1 P (J, J, X, J, J) = 揃 揃 = 揃 揃 1, 47 46 45 47 46 2 45 1 2 1 P (J, J, J, X, J) = 揃 揃 = 揃 揃 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 揃 揃 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 揃 揃 揃 = 揃 揃 . 47 46 45 47 46 45 If X = K, then the probability is 3 3 1 揃 揃 . 47 46 45 Hence the total is 40 3 2 3 2 1 40 3 2 1 2 3 41 3 2 揃 揃 + 揃 揃 = 揃 揃 + 揃 揃 = 揃 揃 47 46 45 47 46 45 47 46 45 47 46 45 47 46 45 3. Let {x1 , x2 , 揃揃揃, 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 揃 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 鍖ll 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 = 揃 揃 揃 = xk and xk+1 , 揃 揃 揃, xn are all distinct. There are n empty space to be 鍖lled with the numbers {x1 , x2 , 揃 揃 揃, 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 , 揃 揃 揃, 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 xnk , 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 , 揃 揃 揃, 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)) .