Keno, come all! Come and make some sense out of my madness!

[TheWarmestHole]

Hello one and all - and without further ado, I'll jump right into the peculiar, partly bewildering, occasionally amusing tale of my trip down the rabbit hole of Keno.

For those of you that aren't familiar, Keno is a relatively straightforward lottery game, wherein the player is faced with a board consisting of the numbers 1 to 80, from which 20 are going to get chosen, completely at random, every round, with a new round occurring every 3 and a 1/2.

Being the considerable math whiz that I undoubtedly am, *cough cough* it didn't take me hardly any time at all (nods to those who caught the internal double negative in there ) to surmise that this numbers based bonanza so happened to yield a whopping 1 in 4 chance for any one number to be chosen. (Show my work? With pleasure: 20 random number selections, made from a pool of 80 possible choices... 20 selections out of 80 possibilities.... Reduces to a 1 in 4 chance of any one number being chosen! Woo!! #mathwhiz #idiotsavant)

Now keeping that in mind, here's where I've run into trouble, conceptually...

Let's suppose that you aren't looking for the odds of a single number being selected, but rather what the odds are of one out of two numbers being selected? Okay, now how about 1 out of 4 numbers?

Am I right in making the statement that in the event that you make 4 selections from the possible eighty (Let's say... 15, 30, 52, 69) does it not follow that the odds of having at least 1 of our selections chosen for a given round are better than not. As in, there is a likelier chance than not, that we hit at least one of our picks?

Beyond that, I'd really appreciate any help you can give me where it concerns calculating out what the probability will be of having certain outcomes occur...

In the system that I was using, I would choose an anchor number (any number from 1 to 80, let's suppose we pick 30) and then match it up with a range of numbers, (like 1 to 10, usually).

This means I would end up with 10 wagers made in total, with each one consisting of the number 30 and the number 1, for the first wager, 30 and 2 for the second, 30 and 3 for the third...until finally getting 30 and 10 for the tenth.

This seemed to me to suggest that while the odds of hitting that 'anchor' number, 30, don't change from being 1 in 4, or 25percent, IF we do hit 30, then we'll have winning tickets for however many selections there also happens to be for that round, between 1 and 10.

As far as I can figure, that translates to an average of 2.5 numbers between 1 and 10 being selected for any given round, right? (Or think of it like 5 numbers on average getting picked from 1 to 20... 5 picked, times the four groups of 20 in 80 equals.... Yep, 20! So it seems to be accurate...)

So please, one and all, please help me grasp this - and in return I can promise you one heckuva amusing story that all this connects to from my recent experiences in life! Guaranteed to make you laugh!

 

[Romsek]

So you want the probability that you pick 1 or more of the winning 20 numbers given that you chose \(\displaystyle m\) numbers?


[MATH] P[\text{at least 1 winning number out of $m$ picks, $1\leq m \leq 80$}] = \\ 1-P[\text{no winning numbers out of $m$ picks}] = \\~\\ 1 - \dfrac{\dbinom{60}{m}}{\dbinom{80}{m}}\\~\\ \left( \begin{array}{cc} 1 & 0.25 \\ 2 & 0.439873 \\ 3 & 0.583496 \\ 4 & 0.691679 \\ 5 & 0.772816 \\ 6 & 0.833398 \\ 7 & 0.878426 \\ 8 & 0.911734 \\ 9 & 0.936252 \\ 10 & 0.954209 \\ 11 & 0.967292 \\ 12 & 0.976773 \\ 13 & 0.983604 \\ 14 & 0.988499 \\ 15 & 0.991984 \\ 16 & 0.99445 \\ 17 & 0.996185 \\ 18 & 0.997396 \\ 19 & 0.998236 \\ 20 & 0.998814 \\ 21 & 0.99921 \\ 22 & 0.999477 \\ 23 & 0.999658 \\ 24 & 0.999778 \\ 25 & 0.999857 \\ 26 & 0.999909 \\ 27 & 0.999943 \\ 28 & 0.999964 \\ 29 & 0.999978 \\ 30 & 0.999987 \\ 31 & 0.999992 \\ 32 & 0.999995 \\ 33 & 0.999997 \\ 34 & 0.999998 \\ 35 & 0.999999 \\ 36 & 1. \\ 37 & 1. \\ 38 & 1. \\ 39 & 1. \\ 40 & 1. \\ \end{array} \right) [/MATH]

 

[TheWarmestHole]

So you want the probability that you pick 1 or more of the winning 20 numbers given that you chose \(\displaystyle m\) numbers?


[MATH] P[\text{at least 1 winning number out of $m$ picks, $1\leq m \leq 80$}] = \\ 1-P[\text{no winning numbers out of $m$ picks}] = \\~\\ 1 - \dfrac{\dbinom{60}{m}}{\dbinom{80}{m}}\\~\\ \left( \begin{array}{cc} 1 & 0.25 \\ 2 & 0.439873 \\ 3 & 0.583496 \\ 4 & 0.691679 \\ 5 & 0.772816 \\ 6 & 0.833398 \\ 7 & 0.878426 \\ 8 & 0.911734 \\ 9 & 0.936252 \\ 10 & 0.954209 \\ 11 & 0.967292 \\ 12 & 0.976773 \\ 13 & 0.983604 \\ 14 & 0.988499 \\ 15 & 0.991984 \\ 16 & 0.99445 \\ 17 & 0.996185 \\ 18 & 0.997396 \\ 19 & 0.998236 \\ 20 & 0.998814 \\ 21 & 0.99921 \\ 22 & 0.999477 \\ 23 & 0.999658 \\ 24 & 0.999778 \\ 25 & 0.999857 \\ 26 & 0.999909 \\ 27 & 0.999943 \\ 28 & 0.999964 \\ 29 & 0.999978 \\ 30 & 0.999987 \\ 31 & 0.999992 \\ 32 & 0.999995 \\ 33 & 0.999997 \\ 34 & 0.999998 \\ 35 & 0.999999 \\ 36 & 1. \\ 37 & 1. \\ 38 & 1. \\ 39 & 1. \\ 40 & 1. \\ \end{array} \right) [/MATH]

While I did indeed find this to be interesting, it was not actually what I was hoping to get help with comprehending.

Let me rephrase my query: for a given round of Keno I have chosen to make 40 separate wagers, each attempting to predict only two of the twenty numbers that are to be picked for that round.

Those wagers are as follows: essentially 4 random numbers, paired up with the numbers 1 through 10. For the example I gave initially, those four numbers happened to be 15, 30, 52 and 69. Therefore:

15 & 1
15 & 2
15 & 3
15 & 4
15 & 5
15 & 6
15 & 7
15 & 8
15 & 9
15 & 10

And...

30 & 1
30 & 2
30 & 3
30 & ...
(You get the idea...)
30 & 9
30 & 10

And...

52 & 1... etc...
To 52 & 10

And then the last ten...
69 & 1
69 & 2
(...)
69 & 10

So those form the total number of wagers that I make for a given round.

Given that, I'd like to know:
1) What is the probability of at least one of my four random numbers getting chosen for a specific round.

2) Given that one of the random numbers gets picked that round, are there not, mathematically speaking, going to be an average of 2.5 numbers between 1 and 10 that they would pair up with to form winning wagers? (Obviously you can't have 2 and a half numbers picked out of the numbers 1 through 10, but it's an average...)

3) Anything else of interest that occurs to the replying individual!

Thank you!

 

[JeffM]

I usually ignore questions about gambling systems, but you seem to be sane (all things are relative of course) and to have a sense of humor. So I shall indulge you for at least a while.

I must admit, however, I have not yet made much sense of what you are asking because I am so ignorant that I had to look up keno on Wikipedia, a process that resulted in a detour to learn about the One Thousand Character Classic. With ADD of such magnitude, I am going to need to take this in very small steps.

A couple of things.

Probability theory is not intuitive. Although the rules are few and simple, the results do not always agree with common sense. Moreover, any gambling system must take into account payoffs as well as probabilities.

But just to make sure that I get the basic idea of keno. You buy a ticket that says you win if the single number on that ticket matches one of 20 distinct numbers "selected at random" from the integers 1 through 80. As you say, the probability that your number is a winner, given that it is an integer from 1 through 80, is 1/4, meaning that if you play a very large number of times, you will win around 1/4 of the time. (Fractional probabilities do not guarantee exact results in a finite number of trials. If you buy one thousand tickets, you may not win exactly 250 times, but you should win close to 250 times.)

Am I right that a ticket specifies a single number? If a single ticket specifies more than one number, how many and how is a ticket determined to be a winner?

 

[TheWarmestHole]

I usually ignore questions about gambling systems, but you seem to be sane (all things are relative of course) and to have a sense of humor. So I shall indulge you for at least a while.

I must admit, however, I have not yet made much sense of what you are asking because I am so ignorant that I had to look up keno on Wikipedia, a process that resulted in a detour to learn about the One Thousand Character Classic. With ADD of such magnitude, I am going to need to take this in very small steps.

A couple of things.

Probability theory is not intuitive. Although the rules are few and simple, the results do not always agree with common sense. Moreover, any gambling system must take into account payoffs as well as probabilities.

But just to make sure that I get the basic idea of keno. You buy a ticket that says you win if the single number on that ticket matches one of 20 distinct numbers "selected at random" from the integers 1 through 80. As you say, the probability that your number is a winner, given that it is an integer from 1 through 80, is 1/4, meaning that if you play a very large number of times, you will win around 1/4 of the time. (Fractional probabilities do not guarantee exact results in a finite number of trials. If you buy one thousand tickets, you may not win exactly 250 times, but you should win close to 250 times.)

Am I right that a ticket specifies a single number? If a single ticket specifies more than one number, how many and how is a ticket determined to be a winner?

Well thank you for indulging me, and please allow me to clarify...

You can in fact select anywhere from 1 number to a maximum of 10 numbers, at the gamblers discretion. In the sense that I am interested in, we are making a selection of 2 numbers only.

Ergo, a winning ticket would be one where both of my selected predictions are among the 20 randomly chosen numbers.

My question, as I have since come to understand since I originally made this post, is reductively better posited as follows: if you have 1 in 4 odds of guessing a number correctly, what are the odds of guessing at least one number correctly if four guesses are made? (And by extension, what is the probability formula for arriving at this answer in general terms?)

 

[firemath]

One in four...it doesn't change.

If you have a one in four chance of guessing correctly, then your odds of guessing correctly in four guesses are 1 in 4!

Where is your confusion from?

 

[JeffM]

In some senses, we are talking in circles. You keep talking about probabilities of 1/4 (which are NOT odds).

But in fact, that simple number is greatly confusing the issue. Probability theory demands very clear specifications.

Someone selects "at random" q distinct numbers from a set of n numbers, where 0 < q < n + 1. You select p distinct numbers from the same set of n numbers where 0 < p < n + 1 - q. So if I understand this keno thing at all, you buy a ticket (or place a bet) on a single integer from 1 through 80. The house "randomly" selects 20 integers from 1 to 80 without duplicates. So q = 20, and n = 80. Of course you can make more than one bet (buy more than one ticket). In that case, you can buy say two tickets with the same number or two tickets with different numbers. When I say "distinct" numbers, however, I mean that they are different. With me? You want to know what the probability is that you will have at least r matches given p tickets.

Romsek already gave you the formula for that, namely

[MATH]r = 1 - \dfrac{\dbinom{n - q}{p}}{\dbinom{n}{p}} = 1 - \dfrac{\dfrac{(n - q)!}{p! * (n - p - q)!}}{\dfrac{n!}{p! * (n - p)!}} = 1 - \dfrac{\cancel {p!} * (n - p)! * (n - q)!}{\cancel {p!} * n! * (n - p - q)!} \implies[/MATH]
[MATH]r = 1 - \dfrac{(n - p)! * (n - q)!}{n! * (n - p - q)!}[/MATH]
Then romsek assumed that n = 80 and q = 20. That simplifies the formula to

[MATH]r = 1 - \dfrac{(80 - p)! * 60!}{80! * (60 - p!}.[/MATH]
[MATH]p = 1 \implies r = 1 - \dfrac{79! * 60!}{80! * 59!} = 1 - \dfrac{\cancel {79!} * \cancel {20} * 3 *\cancel {59!}}{\cancel {20} * 4 *\cancel {79!} * 59!} = 1 - \dfrac{3}{4} = \dfrac{1}{4}.[/MATH]
You got the right answer, but for the wrong reason.

[MATH]p = 2 \implies r = 1 - \dfrac{78! * 60!}{80! * 58!} = 1 - \dfrac{60 * 59}{80 * 79} = 1 - \dfrac{177}{316} \approx 0.44.[/MATH]
However, I doubt that number does you any good, because it does not give you the probability of at least breaking even. To calculate that, I need to understand whether you can buy tickets containing more than one number, and what the payoff is when you win versus what you the cost is when you lose.