How to calculate odds in excel (for instance matching pairofdice)

How can one easily calculate the probability of several dice forming a pair
or two pair etc ?

For instance the chance of rolling a 10 on a 10-sided die is 1 in 10 (10%)
but it gets to complicated for me to calculate the chance of say two pair
(any) when rolling 10 dice.

Thanks in advance

Hi

The summary probability of two independet events equals with multiple of
probabilities for those events. I.e. for your example 0.1*0.1=0.01 = 1%


--
Arvi Laanemets
( My real mail address: arvi.laanemets<at>tarkon.ee )


"Diceroller" <Diceroller@discussions.microsoft.com> wrote in message
news:FDA7DB86-8247-4D50-AD45-CE5C8BB531D2@microsoft.com...
> How can one easily calculate the probability of several dice forming a
> pair
> or two pair etc ?
>
> For instance the chance of rolling a 10 on a 10-sided die is 1 in 10 (10%)
> but it gets to complicated for me to calculate the chance of say two pair
> (any) when rolling 10 dice.
>
> Thanks in advance

Not sure if it helps, but you MIGHT want to check out the COMBIN
function?
Arvi Laanemets wrote:
> Hi
>
> The summary probability of two independet events equals with multiple of
> probabilities for those events. I.e. for your example 0.1*0.1=0.01 = 1%
>
>
> --
> Arvi Laanemets
> ( My real mail address: arvi.laanemets<at>tarkon.ee )
>
>
> "Diceroller" <Diceroller@discussions.microsoft.com> wrote in message
> news:FDA7DB86-8247-4D50-AD45-CE5C8BB531D2@microsoft.com...
> > How can one easily calculate the probability of several dice forming a
> > pair
> > or two pair etc ?
> >
> > For instance the chance of rolling a 10 on a 10-sided die is 1 in 10 (10%)
> > but it gets to complicated for me to calculate the chance of say two pair
> > (any) when rolling 10 dice.
> >
> > Thanks in advance

Thanks for the help but how would this work with larger equations ?
Ie having 10 dice and trying to calculate the odds of getting one pair,
which I think is somewhere around 99%)
However I don't know how to 'program' this into Excel.

"Arvi Laanemets" wrote:

> Hi
>
> The summary probability of two independet events equals with multiple of
> probabilities for those events. I.e. for your example 0.1*0.1=0.01 = 1%
>
>
> --
> Arvi Laanemets
> ( My real mail address: arvi.laanemets<at>tarkon.ee )
>
>
> "Diceroller" <Diceroller@discussions.microsoft.com> wrote in message
> news:FDA7DB86-8247-4D50-AD45-CE5C8BB531D2@microsoft.com...
> > How can one easily calculate the probability of several dice forming a
> > pair
> > or two pair etc ?
> >
> > For instance the chance of rolling a 10 on a 10-sided die is 1 in 10 (10%)
> > but it gets to complicated for me to calculate the chance of say two pair
> > (any) when rolling 10 dice.
> >
> > Thanks in advance
>
>
>

Diceroller" <Diceroller@discussions.microsoft.com> wrote in message
news:AFD3FC74-F7F5-480C-A5CD-2D718F5A7323@microsoft.com...
> "Arvi Laanemets" wrote:

>> "Diceroller" <Diceroller@discussions.microsoft.com> wrote in message
>> news:FDA7DB86-8247-4D50-AD45-CE5C8BB531D2@microsoft.com...
>> > How can one easily calculate the probability of several dice forming a
>> > pair
>> > or two pair etc ?
>> >
>> > For instance the chance of rolling a 10 on a 10-sided die is 1 in 10
>> > (10%)
>> > but it gets to complicated for me to calculate the chance of say two
>> > pair
>> > (any) when rolling 10 dice.

>> The summary probability of two independet events equals with multiple of
>> probabilities for those events. I.e. for your example 0.1*0.1=0.01 = 1%

> Thanks for the help but how would this work with larger equations ?
> Ie having 10 dice and trying to calculate the odds of getting one pair,
> which I think is somewhere around 99%)
> However I don't know how to 'program' this into Excel.

The probability of not getting a pair when you roll the second die is 90%
[there are 9 numbers that don't match, and one that does.]
If you don't get a pair with the 2nd, then when you roll the 3rd die, the
further probability then is 80% [there are 8 numbers which don't match, and
2 that do match one or other of the numbers already thrown.]

If you follow this on, the probability of not throwing a pair with 10 dice
is =0.9*0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.1, which is about 0.036%.
You thus have 99.964% probability of throwing at least one pair.
--
David Biddulph

Thank you, this shortens the calculation, yet I can't seem to tweak this for
use with for instance a three of a kind roll ? OR how to chance increases for
a higher pair when you have more dice to throw.

"David Biddulph" wrote:

> Diceroller" <Diceroller@discussions.microsoft.com> wrote in message
> news:AFD3FC74-F7F5-480C-A5CD-2D718F5A7323@microsoft.com...
> > "Arvi Laanemets" wrote:
>
> >> "Diceroller" <Diceroller@discussions.microsoft.com> wrote in message
> >> news:FDA7DB86-8247-4D50-AD45-CE5C8BB531D2@microsoft.com...
> >> > How can one easily calculate the probability of several dice forming a
> >> > pair
> >> > or two pair etc ?
> >> >
> >> > For instance the chance of rolling a 10 on a 10-sided die is 1 in 10
> >> > (10%)
> >> > but it gets to complicated for me to calculate the chance of say two
> >> > pair
> >> > (any) when rolling 10 dice.
>
> >> The summary probability of two independet events equals with multiple of
> >> probabilities for those events. I.e. for your example 0.1*0.1=0.01 = 1%
>
> > Thanks for the help but how would this work with larger equations ?
> > Ie having 10 dice and trying to calculate the odds of getting one pair,
> > which I think is somewhere around 99%)
> > However I don't know how to 'program' this into Excel.
>
> The probability of not getting a pair when you roll the second die is 90%
> [there are 9 numbers that don't match, and one that does.]
> If you don't get a pair with the 2nd, then when you roll the 3rd die, the
> further probability then is 80% [there are 8 numbers which don't match, and
> 2 that do match one or other of the numbers already thrown.]
>
> If you follow this on, the probability of not throwing a pair with 10 dice
> is =0.9*0.8*0.7*0.6*0.5*0.4*0.3*0.2*0.1, which is about 0.036%.
> You thus have 99.964% probability of throwing at least one pair.
> --
> David Biddulph
>
>
>