Probability Question

I have found this..

((N!)/(N-M)!M!)*(P TO THE POWER OF M)*((1-P) TO THE POWER OF (N-M))

where M is the exact number of successes, N is the number of trials, P is the individual probability.

To give a bit of context, I am trying to calculate when a money back special on a bad beat multiple becomes value when compared to exchange prices.  Its not for arbing purposes but just for my own knowledge.  I need a better mathematician than me to help if you can

Any number  on a dice is 5/1 against  so  10 polls  = 50/1 to do the double  good luck  ....

Zipper, what does that mean? It seems dubious mathematically.

Baz a dice 6 munbers  to  name any  one number is  5/1 against  so you say  number 6  roll the dice  there are  5 numbers  who can beat you  hence  5/1 against   .... ok each throw  is  5/1 against  so  5 times 10  = 50   to get the double

It's all ok until the final part. The chance of the 1st rolls alone both being 6 is 35-1. You have 9 chances of any 2 in a row? In fairness I may be misunderstanding the question.

1st 2 rolls that should be

Baz  i thought you were talking about one dice ... two  dice is  35/1 against each   throw .... god iam hard work .

I think it's one dice thrown 10 times? The 1st 2 individual rolls would be 35-1 to both be 6 etc. You have 9 chances of getting 2 in a row. Feels like a 7/2 shot roughly.

Baz  7/2 to throw a double ////   you work for one of the big  3

I am actually wanting to know the probability of rolling any 6 with 10 rolls.  This is where you move away from basic probability.  If there were two rolls then the probability of rolling a six would be roughly 1/6 + 1/6.  If three rolls then roughly 1/6 1/6 +1/6.  I say roughly because with each roll the odds diminish away from the calulations given here.  If you roll it six times the probability obviously can't be 6* 1/6 = 1 because that states that you will definitely roll a 6 which obviously might not happen.  I can't remember the formula to calculate the probability as its very long, and was wondering if anyone knew it or could find it for me.  The complicated formula above I think is it but was looking for someone to confirm

Prestbury,

Do you mean the probability of rolling 1 or more sixes from 10 rolls of a dice?

Yes equimine, exactly

Just find the probablity of NOT rolling it, then subtract this from 1.  Hence 1-(5^10/6^10)  whatever that works out at.

Gets more complicated though if throwing a particular number more than once is required.  Binomial theorem or something?  Dunno, I only did CSE maths.

Agree with Cosmic.

1 - (5/6)^10

1 -( .83^10) equals .33

Therefore probability of throwing 1 or more sixes from 10 throws is 67%.

Sorry shouldn't try and do two things at once

1 - (5/6)^10

1 - (.83 ^10) equals .16

Therefore probability of 1 or more sixes from 10 throws is 84%

Sorry, I think I have mislead you all a bit, I want it to show me the price to throw one 6 exactly from 10 rolls.  My bad, sorry about that.  Thank you thogh cosmic for that.  Its a long time since I studied probability, but it was at university and I remember a very long equation that hurt my head a little.  Maybe I am wrong about this, but you two sound like you have a good grasp of this.  Any ideas?

You need one 6(1/6) and 9 non-6s (5/6).  There are 10 different ways in which this can be achieved (6 first, 6 second, 6 third, ...)

The equation you need is therefore 10 x 1/6 x (5/6)^9 = 32.3%

Equimine, that is what I had, using 1.2*10 to establish chance of no 6. That gives a 16.15% chance of no 6 being thrown.

I can't fathom out any need to move away from probability at all.

Prestbury,

You need a Binomial Distribution.

If you use, or have access to Excel use the inbuilt one in statistical formulas.

Number would be 1
Trials would be 10
Probability would be 0.167
Use "false" in cumulative

The answer is 32.25%

Basically as follows
Roll die 2 times,there are 6*6 outcomes = 36
5*5 outcomes will not contain a 6, =25
So 11 will contain a 6, thus probability of there being at least one 6 =36/11= 3.27
Roll a die 3 times, there are 6*6*6 outcomes = 216
5*5*5 outcomes will not contain a 6, = 125
So 91 will contain a 6, thus probability ot there being at least one 6 = 216/91 = 2.37
etc etc

I don't know the equation.  Binomial?  Asking for one 6 EXACTLY is different from asking for at least one six since at lest one six will include all the permutations for 2 6's etc..

Since I don't know the equation then we have to do it the long way round.  If any other number apart from 6 = X, then the permutations are:

6xxxxxxxxx
x6xxxxxxxx
etc with the 6 in a new position each time making up 10 permutations.

Each permutation will be 1/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*

Therefore the probabilty of throwing one 6 exactly should be 10 times this value?

Jus' used excel to calculate the answer.  Probability comes out at 0.323 if my reasoning is correct.  Might not be though.  It's astonishingly easy to get confused with maths!

Ah!  My answer is the same as Equimines!  So looks like my reasoning is correct.

Nope you're both wrong I'm afraid.
My answer is the correct one.
Remember we are only looking for at least one 6.

The equation is as follows I believe
Binomial =
Trials C Successes (in this case 10 and 1) * Probability of Success ^ Number of Successes (0.167 ^1) * Probability of Failure ^ Number of Failures (.833^9)

10 * ((.167 * (.833^9)))
= .3225

32.25%

The easiest way to approch these things is always to establish the convers first as I have done,
Then simple deduction gets your answer.

The more recent request is for one 6 exactly.

I don't think so Baz but if it is then that's really easy
Example, roll a die 2 times, = 36 outcomes.
To get only one 6, you must either get in on the first or the second, so only 2 outcomes win.
Therefore probability of there being only one 6 are 36/2 = 18.00

Probabilities have to be less than 1.  You must be referring to odds? Odds of 18 is probability of 0.056. If only thrown twice then it's 2*(1/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6*5/6)= 0.064602233. 
Or you can use binomial which is probably the way I'm doing it but put in its general form.

So your answer is wrong.

Equimine
If you're saying what I think you are, that is if you roll a die 10 times, you will give me odds of over 3 to 1 that I can't get one 6 up, then that's what I call a real yummy value bet.
Just think about your maths a bit.
As I say before the odds of me getting at least one 6 up in 3 rolls is 2.37
You guys are overcomplicating a relatively simple situation.

No that's wrong.  If only thrown twice the permutations are:
XX
6X
X6
66

So we want either 6X or X6 .  Both will be equal probabilities of 5/6*1/6.  So probability = 2*(5/6*1/6) = 0.277777778  I think!

First of all, thanks for all the answers.  One six exactly can be done by going the long way around, yes, but I was looking for the equation as what I need isn't do roll a die.  The binomial equation I think is what I need for at least one six.  If you both (equimine and cosmic) got the same result then that looks good. I'll use that and see what I get.

Cosmic
If you roll a die twice there must be 36 outcomes.
If you want to get one specific number only once, then it has to occur either on the first throw or the second, and not both.
That is 2 outcomes out of 36.
Odds are 18.00.

FAFH,

With all due respect, before you start  lecturing on the intricacies of statistical probability, most people realise it is advisable to know the question you are trying to answer.

Prestbury wrote as follows,
"I want it to show me the price to throw one 6 exactly from 10 rolls"

You are not even correct about the odds of getting at least one six which is 84% (1.19)

FINE AS FROG, I know where you are coming from but this is the crux of what I am asking.  The odds of something happening is simplified in the dice example, but imagine if you change that to a football team who are priced at 5/1 and assume it is a fair price.  Then tell me the probability of them winning at least once if played 10 times and the probability of them winning exactly once if played 10 times.  Maybe I am wrong as I have said, but the method you are using does not give a true probability even though intuitively you expect it to.

The outcomes are:

11
12
13
14
15
16
21
22
23
24
25
26
31
32
33
34
35
36
41
42
43
44
45
46
51
52
53
54
55
56
61
62
63
64
65
66


Not there's 36 if you count them all.  How many have one 6 exactly?  Well count them.  It's 10.  So the probabiity is 10/36 = 5/18 = 0.277777778

The exact same answer I got before!

If you look at the equation I have posted in my second post, this is more like the complexity of equation that I believe I need

good to get the brains working though, eh?

Yes we can just use an equation if we can find the right equation that is.  But it's good to work it out from first principles.  Then you can understand where the equation is coming from with a little bit of thought.

OK gents ..last one for me on this  lets keep it simple  iam a layer your a backer  you come to me  and  say  what price would you give   me  throwing a  6  it could be  a 2 or any  other  number  i say  i will lay you 5/1 each throw  so six it is  your having a pound on  roll it 10 times and no 6  how much  have you lost ...£10  thats right   heres  the other side  you lucky  punter   roll again  and every throw  you get a 6  how much  have i lost  .... thats right  £50  pounds  you dont need  all  this  equation stuff.. got to go now  money to be won

it's 1/6 * (5/6 ^ 9) * 10 since where the six is thrown doesn't matter

imagine if you change that to a football team who are priced at 5/1 and assume it is a fair price.  Then tell me the probability of them winning at least once if played 10 times and the probability of them winning exactly once if played 10 times.

There is an easy way to answer the first part of this, but I don't know how to answer the second part.

So, the probability of getting AT LEAST one 6:
The probability on any one throw of getting a 6 is 1/6 = 0.166..
Therefore the probability of NOT getting a six on a throw is 0.833...
The probability of 10 consecutive non sixes is 08.33^10, which is 0.161.
Therefore there is a 0.84ish probability that you will throw at least one 6 if you make 10 attempts.

Cosmic
I agree with your numbers if you were going to throw the dice 36 times twice.
You would expect to get  10 combos at least with one specific number in it.
But what is being proposed here( I think) is that you just throw the dice twice.
I still believe that in this situation, 36 things can happen, but only 2 will win you money.
That is the odds are 18.00.

zipper, I'm not sure where that is going, although what you have said is accurate

get me a drink, what you have written is what I would think it is as well

I am interested in the binomial chat earlier, but what I will do is ask a man who I know will know for sure and get back to you.  Intuitively I believe this is correct, but I think there is a more complex equation when probability of repeated events is used.

Did anyone do a Maths degree?  Must be a one on general betting?

for a single 6 out of X rolls it is the probablility of

rolling a 6 on the first dice AND not rolling a six on the second AND not rolling a six on the third AND not rolling a six on the fourth AND ... AND not rolling a six on the tenth

OR

not rolling a six on the first AND rolling a six on the second AND not rolling a six on the third AND not rolling a six on the fourth AND ... AND not rolling a six on the tenth

OR

.
.
.

OR

not rolling a six on the first AND not rolling a six on the second AND not rolling a six on the third AND not rolling a six on the fourth AND ... AND rolling a six on the tenth.


10 * [1/6 * (5/6)^9] ~= 0.323

Zola,

Yes that's the Binomial formula I gave Prestbury some time ago.

N * [p * (1-p)^(N-1)]

N = number of trials
p = probability of outcome in single trial

^ which is the equation in the second post with M=1

Prestbury
If you want only one 6 in 10 rolls there can only be one answer.
There are 6*6*6*6*6*6*6*6*6*6 possible outcomes, = 60,466,176
It has to be 5*5*5*5*5*5*5*5*5*10 =19,531,250 for there to be one 6 in only one of the 10 rolls
Therefore, odds are 3.09.
Which, I believe , is the answer given below using the Excel binomial calculator.
Cosmic Horizon
Based on this, your calculation of 10 outcomes out of possible 36 in 2 rolls is also correct, being odds of 3.60. Sorry for doubting you.

Zola,

I have no idea, I think it is a long winded method of calculating the probability of "within a range of successes" (i.e between 1 and 3 sixes from 10 rolls)

Equi, it wasn't a question

Your post slipped in inbetween my posts. N * [p * (1-p)^(N-1)] is the equation in the second post with M=1.

As mentioned earlier Prestbury, I would be wary of just picking equations of tinternet if you are not sure of their proper derivation or applicability. Especially when probability is concerned.

On an unrelated note, I had a nice lunch and pint in The Plough in Prestbury yesterday. Marvelous.

Everybody agrees then at last.
The odds are 3.09.
Bingo.

What Zola's Back Heel said is exactly the same as what I did.  So it's def about 32%

Just to clarify this.  Let's imagine we throw the die (dice) 3 times rather than 10 times.  We want to know the probability it will land on 6 EXACTLY once.  So the 6 will occur either with the 1st throw, 2nd throw or 3rd throw.  Let X = any of the numbers from 1-5.  So we have to have either:

1. 6XX = (1/6)*(5/6)*(5/6) = 25/216
2. X6X = (5/6)*(1/6)*(5/6) = 25/216
3. XX6 = (5/6)*(5/6)*(1/6) =25/216

They each come to the same answer for the same reason as 2*3 = 3*2.

We add all of them up and we get 75/216 or 0.347222222 in decimal.

We can work out the general equation.  1/6 = the probability of throwing a 6 with one throw.  Call that P

So in "1" above this turns from (1/6)*(5/6)*(5/6) to p*(1-P)*(1-P)

The number of times we multiply (1-P) with itself will be the total number of throws minus 1. 

Let the total number of throws = N. 

So it becomes P*(1-P)^(N-1)

"2" and "3" will be exactly the same equation.  So in fact we multiply p*(1-P)^(N-1) by the number of throws

= N*(p*(1-P)^(N-1))

So prestbury road this is the exact same equation as you mentioned in your 2nd post where M = 1.  You pasted in the equation:

((N!)/(N-M)!M!)*(P TO THE POWER OF M)*((1-P) TO THE POWER OF (N-M))

But M=1 so we get ((N!)/(N-1)!*1!)*(P^1)*((1-P)^(N-1))

which is ((N!)/(N-1)!*P*(1-P)^(N-1)

N! means N*(N-1)*(N-2)*(N-3)  . . .etc.  So for example 5! means 5*4*3*2*1

so N!/(N!-1) must equal N since all numbers less than N simply cancel each other out on the top and bottom of the fraction. 

Thus we get N*(P^1)*((1-P)^(N-1))  In other words the equation we worked out from first priciples.


This is something I never knew before!  Could come in useful sometime!

Holy crap! Thats a lot of text

I solved it using excel.

Probability rounded to 4 decimal places:

Sixes    P
1    0.1615
2    0.3230
3    0.2907
4    0.1550
5    0.0543
6    0.0130
7    0.0022
8    0.0002
9    0.0000
10    0.0000

You only need Combin(), the rest is arithmetic.

oops copied & pasted wrong, let me correct that...

Sixes    P
0    0.1615
1    0.3230
2    0.2907
3    0.1550
4    0.0543
5    0.0130
6    0.0022
7    0.0002
8    0.0000
9    0.0000
10    0.0000

FINE AS FROG HAIR Joined: 12 Mar 07
Replies: 653 11 Oct 10 15:12 
Everybody agrees then at last.
The odds are 3.09.
Bingo.


I'll lay at that price?

^
in a 0% commission bet.

Or if somoene else wants to back at 3.10....

.
http://www.sendspace.com/file/ipndmb
for the spreadsheet (link at bottom of page)

Can't say I can make head or tails of the numbers you posted Investor.
It's pretty clear that the odds of only getting one six up in 10 rolls of the dice is 3.09.
Remember we're only talking about one six exactly.

Yes FAFH, I agree (disregarding the rounding error).

1/0.323=3.0959...

Investor/FAFH, you're both agreeing with each other. the 0.323011166 for 1/10 sixes is 3.096 expressed as decimal odds.

But this totally fails to answer the OP anyway, who wanted to know the probability of at least one six, not only one six. In which case you need 1-0.161505... = 0.8384944171

At least 1 is easy to calculate, it's 1-(5/6)^10 because it's the complement of 10 rolls without a six.

If you want to know at least 2, at least 3 etc, then the formula to use is the cumulative binomial distribution, which has been covered elsewhere in this thread but can easily be done in excel with Binomdist() function.

ror
It's a bit confusing what the OP wants.
I thought at first like you say, that is at least one six.
Then it was changed to only one six.
Who knows and who cares by now.
It's all gone beyond meaning in a reality sense.

ror:

prestbury road Joined: 06 Feb 06
Replies: 191 11 Oct 10 12:43 
Sorry, I think I have mislead you all a bit, I want it to show me the price to throw one 6 exactly from 10 rolls.  My bad, sorry about that.

but can easily be done in excel

the death of thought