combin function excercise

[Scott]

Hello. I am working on learning how the combin function
works, and how what the results actually means. I
actually got this example from a book but it didn't
explain it very thouroughly. If there are 49 possible
lottery numbers to choose from, and I can choose 6 if I
buy a ticket, then my formula would be =combin(49,6) and
the result is 13,983,816. The book I am working with
leads me to interpret the result as follows: I would need
to buy 13,983,816, choosing 6 numbers on each ticket,in
order to exhaust all the possibile combinations, therefore
eliminating any possibility of losing the lottery.

Where I am getting confused (and it's probably just that
I'm thinking about it backwards), is that if I reduce the
number of choices from 6 down to 5,the resulting number
drops from 13,983,816 to 1,906,884. So, I am interpreting
this as follows: If I reduce the amount of numbers that I
am able to choose per ticket, I would only have to
purchase 1,906,884 tickets in order to exhaust all
possibilities of losing the lottery. This is opposite of
what I was expecting. So I am assuming I am not
understanding the whole thing correctly. Any help is
appreciated.

Furthermore, if I can overcome this obstacle, my next
question is, how to fit into the calculation, if a person
were to purchase 2 tickets. Would this be as easy as
doubling the choices from 6 to 12?

Thanks again

Scott

 

[Tod]

The results are correct. However, Keep a few things in
mind as you interpret these results:

- 13,983,816 is the number of unique combinations of six
tht you can get out of 49 numbers. It doesn't mean that
you would be the only person who'd get that combination.
So don't spend close to $14 mil on a ticket just to find
out that you and five other poeple have to split the
prize.

- Each ticket you purchase has the same odds of winning
at the moment the numbers are drawn. That means if you
buy, one, ten, or a thousand tickets, you do not improve
your odds of winning.

 

[hgrove]

Tod wrote...
...

- Each ticket you purchase has the same odds of winning at the
moment the numbers are drawn. That means if you buy, one,
ten, or a thousand tickets, you do not improve your odds of
winning.

...

Only true if all tickets would have the same numbers. If all ticket
have different sets of 6 numbers, then buying 10 tickets gives 10 time
more chance of winning than buying only one ticket

 

[hgrove]

Scott wrote...
...

Where I am getting confused (and it's probably just that
I'm thinking about it backwards), is that if I reduce the
number of choices from 6 down to 5,the resulting number
drops from 13,983,816 to 1,906,884. So, I am interpreting
this as follows: If I reduce the amount of numbers that I
am able to choose per ticket, I would only have to
purchase 1,906,884 tickets in order to exhaust all
possibilities of losing the lottery. This is opposite of
what I was expecting. So I am assuming I am not
understanding the whole thing correctly. Any help is
appreciated.

Why were you expecting the number of combinations to increase? Follo
your logic: if you reduce the numbers you have to choose to 1, there'
be only 49 possibilities - fewer, not more, than when you choose 6.

On the other hand, if you increase the numbers you have to choose t
48, there are again only 49 combinations. And if you choose all 4
numbers there's only one way to do it.

The number of combinations of k distinct numbers drawn from N possibl
numbers rises as k goes from 0 to N/2, where it peaks, then falls as
goes from N/2 to N. Some simple examples using the 6 letters A-F sho
how this works.

k = 0: nothing
k = 1: A B C D E F
k = 2: AB AC AD AE AF BC BD BE BF CD CE CF DE DF
k = 3: ABC ABD ABE ABF ACD ACE ACF ADE ADF AEF BCD BCE BCF
____ BDE BDF BEF CDE CDF CEF DEF
k = 4: ABCD ABCE ABCF ACDE ACDF ADEF BCDE BCDF BDEF CDEF
k = 5: ABCDE ABCDF ABCEF ABDEF ACDEF BCDEF
k = 1: ABCDEF

Furthermore, if I can overcome this obstacle, my next question
is, how to fit into the calculation, if a person were to purchase 2
tickets. Would this be as easy as doubling the choices from 6 to
12?

Depends on what you mean. If a person buys 2 tickets, they're choosin
two separate and presumably different sets of 6 numbers. This doe
involve choosing 12 numbers overall, but it would have no effect on th
number of combinations

 

[Myrna Larson]

Hi, Harlan:

Thanks for saving my sanity <g>. When I read Tod's reply, I began to wonder
which of us had "lost it".

Myrna Larson