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Five-Card Straight

```Date: 12/10/2002 at 13:23:01
Subject: Poker probability

From a standard 52-card deck you pull out a 5-6-7-8 (suit is
irrelevent for the problem). You shuffle the remaining 48 cards and
deal 3 more cards. What percentage of the time will you complete the
5-card straight (by receiving a 4 or 9 with any one of the three cards
dealt). My friends and I disagree on the answer. Here is my math:

1st draw = 16.6667%
2nd draw = 17.0213%
3rd draw = 17.3913%

adding the three chances together I get 51.08% you will get a 4 or 9.

My reasoning:

Ignoring the slight increase in percentage when a non-winning card is
dealt, your odds are 8/48 or 1/6, similar to dice. If I select one
number from a die and roll 3 times (3 cards), my odds are 50% to roll
my number.

are!
```

```
Date: 12/10/2002 at 20:20:07
From: Doctor Ian
Subject: Re: Poker probability

Hi David,

The problem with these 'at least one' situations is that you have to
consider _all_ of the possibilities:

p(4,-,-)            '-' = not 4 or 9
+ p(-,4,-)
+ p(-,-,4)
+ p(9,-,-)
+ p(-,9,-)
+ p(-,-,9)
+ p(4,9,-)
+ p(9,4,-)
+ ...

It's a headache, to say the least. Fortunately, there is a trick you
can use that works very nicely:

p(at least once) = 1 - p(never)

Does that make sense?

There are four 4's and four 9's remaining in the deck, which has 48
cards. So the probability of drawing something other than a 4 or a 9
would be

(48 - 8)/48

The probability of doing it again would be

(47 - 8)/47

And again,

(46 - 8)/46

So the probability of missing on all three draws would be

40   39   38
-- * -- * --
48   47   46

5   39   19
=  -- * -- * --
6   47   23

5   13   19
=  -- * -- * --
2   47   23

= 1235/2162

= 0.57

So the probability of hitting at least one would be

p = 1 - 0.57

= 0.43

Does this help?

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 12/12/2002 at 10:16:39
Subject: Thank you (Poker probability)

Dr. Ian -

Thank you very much for explaining the poker probability question to
me. I made a program to simulate and came up with the same answer
before I sent the question, but still couldn't see what I was missing.
Thanks again.

```
Associated Topics:
College Probability
High School Probability

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