70.19 The 'Alias Smith and Jones' Poker ProblemAuthor(s): Nick MacKinnonSource: The Mathematical Gazette, Vol. 70, No. 452 (Jun., 1986), pp. 136-137Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/3615777 .

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THE MATHEMATICAL GAZE' I E

70.19 The 'Alias Smith and Jones' poker problem

One week, a character in the spoof western series 'Alias Smith and Jones' arranged the following bet. He was to be dealt 25 playing cards and he had to arrange them into 5 'pat' hands. In poker a 'pat' hand is one that you do not change and it consists of any one of the following:

a full house (e.g. J J J 2 2) a straight (e.g. 3 4 5 6 7, any suits) a flush (all cards same suit) four of a kind (e.g. 3 3 3 3 J).

For example, if the deal is as on the left below, then it can be arranged into the pat hands on the right

4 234689J 2 3 4 8 J (flush) 232345678A

03 4678 Q 2 3 6 7 A (flush) *2 5 7 A

04678Q(flush)

A 3 24 3 0 4Z) 54 (straight)

5 Z? 64 74 82? 9A (straight)

Now, believe it or not, I dealt those 25 at random. In fact the character in the series claimed that it worked nine times out of ten. Perhaps you and your classes might like to tackle the question of calculating the probability that 25 cards do form 5 pat hands. That seems like a very difficult question, so perhaps the following problems might help you get warmed up.

(1) Arrange the following cards into 5 pat hands:

2 5 6 8 9 J K A

558 10 A 25 9 10JKA

4679QKA

(2) Show that with the 25 cards at the beginning of this article with 24 replaced by K4 you cannot make 5 pat hands.

(3a) Find the probability that 25 cards can be arranged into 5 straights. (3b) Find the probability that 25 cards can be arranged into 5 flushes. (3c) Find the probability that 25 cards can be arranged in 5 straights or

flushes.

136

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(4) Even if you start with 26 cards (and are allowed to throw one away) there are still some deals where Jones would fail. Find such a deal.

(5) I have shown that given 37 cards (being allowed to throw 12 away) you can always get 5 pat hands. I am not proud of this upper bound, as I suspect that 27 cards suffice. Establish upper bounds of your own.

(6) Devise programs for arranging 25 cards into pat hands (and hence for finding the probability of success?).

I know that this is quite a meaty problem but it has plenty of associated questions and I am sure that there is a lot of scope for projects. The editor tells me that he will publish any readable and reasonable success at any related problem.

NICK MACKINNON

Blundell's School, Tiverton, Devon EX16 4DN

70.20 Which is bigger-e" or ie?

The values of ex and 7e are certainly close enough that the answer to the question "which is bigger?" is not self-evident. When I was first asked it (to be done without using a computer, calculator or mathematical tables, of course) it took me many hours of hard labour to derive an answer.

There is a fairly simple way to get there, however, by considering the more general question "which is bigger-ex or xe?" for an unknown x.

Let y = eX/xe.

Ifx=l then ex > 1,xe = 1, so y > 1

If x = ethen y = 1.

If x = 10 then ex > 21 = 1024, xe < 103 = 1000, so y > 1.

The function is continuous, so it must have a minimum between x = 1 and x=10.

Differentiating yxe = ex, we get

dy xe + yexe-l = ex

dx

and, substituting for y,

dy ex+1 xe + = ex.

dx x

(4) Even if you start with 26 cards (and are allowed to throw one away) there are still some deals where Jones would fail. Find such a deal.

(5) I have shown that given 37 cards (being allowed to throw 12 away) you can always get 5 pat hands. I am not proud of this upper bound, as I suspect that 27 cards suffice. Establish upper bounds of your own.

(6) Devise programs for arranging 25 cards into pat hands (and hence for finding the probability of success?).

I know that this is quite a meaty problem but it has plenty of associated questions and I am sure that there is a lot of scope for projects. The editor tells me that he will publish any readable and reasonable success at any related problem.

NICK MACKINNON

Blundell's School, Tiverton, Devon EX16 4DN

70.20 Which is bigger-e" or ie?

The values of ex and 7e are certainly close enough that the answer to the question "which is bigger?" is not self-evident. When I was first asked it (to be done without using a computer, calculator or mathematical tables, of course) it took me many hours of hard labour to derive an answer.

There is a fairly simple way to get there, however, by considering the more general question "which is bigger-ex or xe?" for an unknown x.

Let y = eX/xe.

Ifx=l then ex > 1,xe = 1, so y > 1

If x = ethen y = 1.

If x = 10 then ex > 21 = 1024, xe < 103 = 1000, so y > 1.

The function is continuous, so it must have a minimum between x = 1 and x=10.

Differentiating yxe = ex, we get

dy xe + yexe-l = ex

dx

and, substituting for y,

dy ex+1 xe + = ex.

dx x

137 137 NOTES NOTES

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