Quick probablilty question

[rebling777]

How would you calculate the odds of picking 7 out of 10 or 3 out of 10 football games correctly?

Thanks!

[BruceZ]

How would you calculate the odds of picking 7 out of 10 or 3 out of 10 football games correctly?

You may want the odds for 7 or more and 3 or more, but I'll give you the odds for exactly 7 and 3 first. Then the other odds will be easier to understand. I'll assume your chance of picking correctly is 50%, in other words, you're not a successful handicapper.


The odds for exactly 7 and exactly 3 are the same:

C(10,7)/2^10 = C(10,3)/2^10 = 11.7% or 8.5-1


For the odds of 7 or more, it is easiest to use the BINOMDIST function in Excel:

1 - BINOMDIST(6,10,0.5,TRUE) = 17.2% or 5.8-1

That is 1 minus the odds of 0,1,2,..6. If you don't have Excel, this is the same as:

[ C(10,7) + C(10,8) + C(10,9) + C(10,10) ] / 2^10 = 17.2% or 5.8-1

where we have just added the probabilities for 7,8,9,10.


Same thing for 3 or more:

1 - BINOMDIST(2,10,0.5,TRUE) = 94.5% or 1.06-1

In this case it's easiest to take 1 minus the probability of 0,1,2:

1 - [ C(10,0) + C(10,1) + C(10,2) ] / 2^10 = 94.5% or 1.06-1

[rebling777]

Probably a silly question, but what does C(10,7) mean?

[BruceZ]

C(10,7) is the number of ways to choose 7 things from 10 things, irrespective of order. It is combinations of 10 things taken 7 at a time.

C(10,7) = 10*9*8*...*4/(7*6*5*...*1)

That is, 10 ways to pick the first, 9 ways to pick the second, etc. Note that there are 7 numbers in the numerator, and then we divide by the number of different orders 7! Since 7*6*5*4 cancel out, this is the same as 10*9*8/3*2*1 = C(10,3)

C(n,k) is the number of ways to choose k combinations of things out of n things, irrespective of order. You can use Excel to compute it, the function is called COMBIN.

C(n,k) = n! / [ (n-k)!*k! ] = n*(n-1)*(n-2)*...*(n-k+1)/ [ k*(k-1)*(k-2)*...*1 ]


I'm wondering now if I interpreted this problem right. I'm assuming you are trying to pick all 10 teams, and you want to get the probability of getting 7 right or 3 right. Are you actually talking about football pools where you only pick 3 or 7 teams, and they all have to win? In that case it's just (1/2)^3 or (1/2)^7.

[squiffy]

Bruce, I am sure you have heard this question a million times, but can you recommend a good beginning probability text or one that is focused on poker odds. If I have time, I may take an extension course.

But I can probably struggle through a simple text on my own. I took math through Calculus in high school, but none in college.

[BruceZ]

Petriv's book should be a good start. He shows you how to calculate all the basic hold ‘em probabilities, and he doesn’t assume any prior math knowledge. One drawback is that while he walks through each hold 'em calculation in detail, he doesn't really organize his material around concepts that allow you to confidently make calculations on your own. That is, it's sort of a cookbook for doing hold 'em calculations, rather than a textbook on probability.

If you learn probability concepts well, and they are not that difficult to learn, you can then easily apply those same basic concepts to any form of poker or to any other area very easily. I don’t know if just following Mike Petriv’s calculations by rote will get you all the way there, but it may be a good place to start.

Also, the error I found is rather inexcusable for a book that claims to be instructive. Some of the calculations suffer from the over counting problem which so many people fall victim to, and which I have described on this forum. I will post corrections to errors I find, and this should help make up for this flaw.

You might consider Sklansky's book Getting the Best of It. He covers many of the major probability concepts in general, and he applies them to poker as well as other forms of gambling.

As far as textbooks go, I really like Probability and Statistics by Morris DeGroot. You should have no trouble reading that if you have been through calculus. It is quite clear in its presentation. In addition to probability, it also has a great deal on statistical estimation theory which you are probably don't care about. It has 5 chapters on probability, and 5 on statistics. The book by Feller is only probability, and it is a classic. It has everything in it you need, but it is intended primarily for math majors. It is somewhat terse, and it goes into depth on subjects that you may not be interested in.

[squiffy]

Thank so much!!! By the way, do you think Dusty Baker had anything to do with the Cubs' success this year?

[BruceZ]

do you think Dusty Baker had anything to do with the Cubs' success this year?

That's what the players say, so I believe them. It probably didn't hurt that Pittsburg gave them half their team.

[irchans]

The book Lady Luck (http://www.amazon.com/exec/obidos/tg/detail/-/0486243427/qid=1065692547/sr=1-3/ref=sr_1_3/103-1131092-4329431?v=glance&s=books) is a good introductory book.

[irchans]

Read Bruce's answer first. I will just give a few examples.

If you have three Kings, you must have one of the following:

1) Kc Kd Kh
2) Kc Kd Ks
3) Kc Kh Ks
4) Kd Kh Ks

There are 4 items in this list and 4 = C(4,3). C(4,3) is often spoken "4 choose 3" which is the number of possible ways to choose 3 from any list of 4 items.

If you have a pair of kings, you must have one of the following

1) Kc Kd
2) Kc Kh
3) Kc Ks
4) Kd Kh
5) Kd Ks
6) Kh Ks

There are C(4,2)=6 possiblilities. C(4,2) is the number of ways of choosing 2 items from a list of 4.


If you want to count the number of possible spade flushes, it is simply C(13,5) because you are choosing 5 cards from a list of 13. If you want to count the number of possible 5 card hands, it is C(52,5).


These numbers, sometimes called Binomial Coefficients (http://mathworld.wolfram.com/BinomialCoefficient.html), have a tremendous number of applications. You can calculate them with the formula in Bruce's post or generate them with Pascal's Triangle (http://mathworld.wolfram.com/PascalsTriangle.html). Every number except the 1's in Pascal's triangle is created by adding two numbers from the previous row.

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1


Hope that helps.

[BruceZ]

I just noticed that my conversion of percentage to odds were all off by 1. Here's a corrected version:

The odds for exactly 7 and exactly 3 are the same:

C(10,7)/2^10 = C(10,3)/2^10 = 11.7% or 7.5-1


For the odds of 7 or more, it is easiest to use the BINOMDIST function in Excel:

1 - BINOMDIST(6,10,0.5,TRUE) = 17.2% or 4.8-1

That is 1 minus the odds of 0,1,2,..6. If you don't have Excel, this is the same as:

[ C(10,7) + C(10,8) + C(10,9) + C(10,10) ] / 2^10 = 17.2% or 4.8-1

where we have just added the probabilities for 7,8,9,10.


Same thing for 3 or more:

1 - BINOMDIST(2,10,0.5,TRUE) = 94.5% or 0.06-1

In this case it's easiest to take 1 minus the probability of 0,1,2:

1 - [ C(10,0) + C(10,1) + C(10,2) ] / 2^10 = 94.5% or 0.06-1