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 ofa kind, and must be different from each other. the choices are 4× 12 and 4× 11 . Finallythe order to obtain the last two cards is not important, so the total count is

[13 · C(4, 3)] · [4× 12] · [4× 11]

P (2, 2)= 54, 912 .

=⇒ P ( three of a kind , no better) =54, 912

C(52, 5)≈ 0.021129.

Likewise, there are 4 Aces, and we just need two of them. The count is C(4, 2). Likewisefor 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]

P (2, 2)= 123, 552 .

=⇒ P ( two pairs , no better) =123, 552

C(52, 5)≈ 0.047539.

2. In a 5 card stud you are dealt a pair of Jacks. On the table a pair of 4 and a K havealready appeared. There are now 52− 5 = 47 cards left undistributed. It is your turn todraw 3 consequent cards. Show the following.

P (J, J, J, X, Y ) =2

47· 45

46· 44

45,

P (J, J, X, J, Y ) =2

47· 45

46· 44

45,

P (J, J, X, Y, J) =2

47· 45

46· 44

45.

In the above, we use the notation (J, J, J, X, Y ) , X 6= J, Y 6= J to denote that thefollowing 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

P (J, J, X, X, X) =41

47· 3

46· 2

45.

Suggested Solution. Given a pair of Jacks and 47 cards left with 2 Jacks in left behindcards, the probability that the third card is a Jack is 2

47. The probability that the fourth

card is not a Jack is 46−146

= 4546

. Likewise, the probability that the fifth card is not a Jackis 45−1

45= 44

45. Hence

P (J, J, J, X, Y ) =2

47· 45

46· 44

45.

Next, given a pair of Jacks and 47 cards left with 2 Jacks in left behind cards, theprobability that the third card is not a Jack is 47−2

47= 45

47. The probability that the fourth

card is a Jack is 246

. The probability that the fifth card is not a Jack is 45−145

= 4445

. Hence

P (J, J, X, J, Y ) =45

47· 2

46· 44

45=

2

47· 45

46· 44

45.

Similar argument shows that

P (J, J, X, Y, J) =47− 2

47· 46− 2

46· 2

45=

45

47· 44

46· 2

45=

2

47· 45

46· 44

45.

Likewise,

P (J, J, J, J, X) =2

47· 1

46· 1 ,

P (J, J, X, J, J) =45

47· 2

46· 1

45=

2

47· 1

46· 1 ,

P (J, J, J, X, J) =2

47· 45

46· 1

45=

2

47· 1

46· 1 .

As a pair of 4 has appeared, X 6= 4 . Suppose X 6= K, the probability to obtain threeconsequent X is

4

47· 3

46· 2

45

and there are 13− 1− 1− 1 choice of X (X 6= 4, J, & K), hence the total is

10 · 4

47· 3

46· 2

45=

40

47· 3

46· 2

45.

If X = K, then the probability is3

47· 3

46· 1

45.

Hence the total is

40

47· 3

46· 2

45+

3

47· 2

46· 1

45=

40

47· 3

46· 2

45+

1

47· 2

46· 3

45=

41

47· 3

46· 2

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

P (n, n− k) =n!

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 tofill the seats in the order important manner is

P (n, r) =n !

(n− r) !=

n !

[n− (n− k)] !=

n !

k !.

We relate this to the present problem by the following picture. Assume without loss ofgenerality 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}

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 thenumber 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 ofthe 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, theanswer is P (n, r) = P (n, (n− k)) .