Last Updated: Sep 4, 2007. Read the Site FAQ!

When it comes to audience familiarity, the best examples of
random probability can be found in the world of gambling.
Therefore, two excellent examples are the lottery, and the game of
5-card poker. **Warning**: ahoy, there be mathematics here! If
you *hated* math in school, you might have trouble getting
through this page. If you think you won't be able to handle it,
feel free to run like a coward to Page
3, but be aware that by skipping this page, you are proving
that *you don't have what it takes to discuss probability at
even the most basic level*. Keep that in mind if you're a
creationist and you intend to write me an E-mail telling me how
wrong I am.

Lottery games are random (by law), so they are obviously a good
example of random probability. Let's take an example lottery
where you must pick 6 unique numbers in any order from 1 to 49.
Remember that the first step is to *count the number of possible
combinations*: there are 49 choices for the first number. You
can't pick the same number twice, so there are only 48 choices
for the second number, 47 choices for the third number, and so on.
Therefore, the total number of possible combinations is
**49*48*47*46*45*44** (or 49!/43! on your calculator), which
equals approximately **10.07 billion**. However, you can pick
the 6 numbers in any order, and there are **720** possible ways
to arrange 6 numbers (6*5*4*3*2*1, or 6! on your calculator; now
you know what that *x!* button is for). Therefore, the overall
probability is 10.07 billion divided by 720 orders, or 13.98
million (probability math short-hand for this whole process is
"49 choose 6", or 49C6). Therefore, the odds of any given
set of numbers coming up in this type of lottery are roughly **1
in 14 million**. And if you play *two* sets of numbers in
the same lottery draw, then your odds of winning are roughly 2 in
14 million, or 1 in 7 million.

**Note**: an important recurring formula in probability is
"x choose y", and it looks like this:

*x choose y = x!/(x-y)!/y!*

This formula, also expressed as xCy (eg- 49C6) will be used
henceforth in lieu of bulky expressions such as
(49*48*47*46*45)/(6*4*3*2*1). The exclamation mark stands for
"factorial". **Fun Fact**: if you enter "49
choose 6" in Google Search, it will automatically compute the
result for you. Try it!

It must be noted that your odds of winning the lottery do
*not* go up if you have been playing every week for the last
10 years. Did you notice that in the above calculation, no mention
whatsoever was made of previous plays? That is because previous
plays do not factor into the probability equation at all. If your
intuition tells you that your 10 years of prior play should count
for something, remember that if one is to learn about scientific
and mathematical concepts, one must first learn to *listen to the
equations, not your intuition*. Your intuition is the sum total
of your life experience *up to now*, and it is a very
unreliable guide when learning about new and unfamiliar things.

So now you know how to calculate the odds of drawing a
particular number in a lottery. Here's an exercise: if you
understood the preceding, then calculate the odds of any given set
of numbers coming up in a lottery where you pick 4 numbers from 1
to 39. But in this lottery, the order of the numbers *does*
matter, and you *can* pick the same number more than once.
Click here for the answer.

In the game of poker, you have 52 cards: 4 suits (clubs, spades,
hearts, and diamonds), with 13 cards (ranked from ace to king,
although ace can be either low or high) in each suit (no jokers or
other wildcards for now). Remember that the first step is to
calculate the number of possible combinations. You can't draw
the same card twice, and the order doesn't matter, so if you
draw a 5-card hand from a deck of 52 cards, then the math is
similar to the first lottery example: "52 choose 5" =
52C5 = **2598960**.

In order to determine the probability of drawing any
*particular* hand out of those 2598960 possibilities, you must
determine how many different examples of that hand exist. For
example, a royal flush is the five highest cards in any given suit,
from ten to ace, like this example:

Click on the cards to see all **4** royal flushes: one for
each suit. Since there are 4 royal flushes, the odds of a royal
flush are 4 in 2598960, or **1 in 649740**. Another way of
calculating the odds is 20/52*51C4.

For a trickier example, the odds of getting a triple (three of the same number) can be computed by determining how many triples exist in the deck. A triple, also known as "three of a kind", is three cards of the same rank, like this:

Click on any card to see all the combinations available. There
are **13** possible ranks of triple, from ace to king, and for
each rank, there are **4** ways to get a triple out of the four
available suits. Therefore, there are 13*4=**52** possible
triples. But we still have to pick the last two cards, don't
we? Remember that there are 52 cards, we've already used up 3,
and the 4th card of the same rank is off-limits because we
don't want four of a kind, so so there are 48 cards left to
choose from for our 1st unknown. For our 2nd unknown, we don't
want a card of the same rank as the 1st unknown because that would
be a full house (a triple and a double), so that means there are
only 44 cards left to choose from. Therefore, there are 48*44 ways
to pick our two unknowns which can be arranged in two orders (1-2
and 2-1), so the total number of combinations is 48*44/2, or
**1056**. Therefore, there are 52*1056=**54912** different
possible hands containing triples, so your odds of a triple are
54912 out of 2598960, or approximately **1 in 47**.

So now you have an idea of how to calculate poker odds. It's trickier than lottery odds, because when you work with poker odds, the total number of possibilities is just the beginning. Here's an exercise: if you understood the preceding, calculate the odds of drawing a full house. That's a triple and a double, and if you paid attention when we solved the triple, the answer should be easy. Click here for the answer.

If you're feeling confident, you can try two more exercises: first, determine the odds of a straight flush. That's five cards in sequential order, all from the same suit. The lowest card can be an ace (aces can be low or high) but it cannot be a 10, because that would make it a royal flush. Click here for the answer. For a more difficult challenge, try to determine the odds of getting a straight flush if you add two jokers to the deck, and make them both wildcards (a wildcard can be used to fill in for any other card, thus increasing the likelihood of getting a rare hand). Click here for the answer.

Hopefully, we learned how simple probability looks at first, but we also learned how tricky it can get. Think of how easily one could go astray when trying to calculate the odds of a triple. If we forgot that the last two cards could be in any order, we would have forgotten to divide by 2!, and our odds would be off by 100%.

Continue to 3. A Series of Unlikely Events

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