Lottery Question (Plz respond quickly!)

Match them to what?

I do not understand the rules :/

Match the balls to the chosen numbers. For example, If the Superball number chosen is 25, and you pick 23, you don't win a prize unless you matched at least two white balls.

As best as I can understand the rules:

Your chances of matching 2 white balls are: 20C2 =190, or 1 chance in 190.

Your chances of matching the blue Superball are: 10C1 =10, or 1 chance in 10.

Your chances of matching all 3 balls are =190 x 10 =1,900. or 1 chance in 1,900.

How many balls do you get for your ticket ?

You have still not explained how trhe game is played to us!!

To answer the Guest's question, you pick three white balls and one blue Superball.

To answer Melody's question, you pick three numbers from 1 to 20. Once you pick a number from there, it can not be chosen again. You also pick a number from 21 to 30. You pick these numbers at random. ( I never played lottery before, so I'm doing my best.) Again, three numbers from 1 to 20 and one number from 21 to 30 is chosen. These numbers are the winning numbers. If it happens that you match at least two of the numbers from 1 to 20 with the winning numbers, or you match the Superball number, you win a prize. ( The order doesn't matter)

Yep seems like a normal lottery to me, Except that the super balls would normally be numbered from 1 to 10 instead of 21 to 30.

In lotteries with super balls, the super balls are a separate set and a different color, and the ball is drawn separately from the other balls – in other words, these are independent events.  If they were not independent then a super ball may not be drawn at all or three super balls may be drawn and none of the others.

To find the over all probability

First, calculate the probability of matching 2 of three numbers drawn from a pool of 20 numbered balls.

1)       Number of ways to select 3 balls from 20 

                20! / (17! * 3!)  =  nCr(20,3) = 1140

2)       Number of ways select to 2 balls from 3 balls

                (3!) / ((2!)*(3-2)!) = 3

3)       Number of ways to select 2 matching balls and one none matching ball in a draw of 3                      from 20 balls

                (20-3)! / (((20-3) - (3-2))! * (3-2)!) = 17

4)       Number of ways of matching 2 balls in a draw of 3 from 20 balls

                 3*17= 51

5)       Probability of matching 2 balls in a draw of 3 from 20 balls

           51/1140 = 0.04473684 or 1 in 22.353

Second, calculate the probability of matching 1 of 10 numbers drawn

This is easy it is just 1 in 10.

Winning requires a match on the first OR second draw. (You could match both, but that is irrelevant here.)

The probability of winning is the same as the rate of winning.  On average, a win will occur 1 in every 22.353 draws for the first set, or 1 in every 10 draws for the second set.

To find the over all probability of having a win, average the two rates.

The word “rate” offers a clue for the proper type of average to use. When averaging rates, use the harmonic average.

H = 2/ (1/10 + 1/22.353) = 13.8182

This equates to each of the draw events having a (1/13.813) probability of matching; but remember there are 2 events, so multiply this by 2.  (1/13.813) * 2 =  0.1447

Over all probability 14.47%

Thanks Ginger, I'm going to try answering independently of you and  see if we get the same answer :))

To answer Melody's question, you pick three numbers from 1 to 20. Once you pick a number from there, it can not be chosen again. You also pick a number from 21 to 30. You pick these numbers at random. ( I never played lottery before, so I'm doing my best.) Again, three numbers from 1 to 20 and one number from 21 to 30 is chosen. These numbers are the winning numbers. If it happens that you match at least two of the numbers from 1 to 20 with the winning numbers, or you match the Superball number, you win a prize. ( The order doesn't matter)

ok, I am going to pretend that the winning numbers are chosen first.  So there are 3 low winning numbers but you only need two of them OR one high winning number and that is the one you need. This will make no difference to the outcome.

Plus there are 2 independent events happening here.

The chance of you getting the high winning number is 1 in 10 that is the easy bit

The number of ways that 3 numbers can be chosen from 20 is 20C3 = 1140   just like Ginger said.

The number of ways that I can get all three numbers is 1

I can get 3 pairs of 2 numbers , just as Ginger said.

The number of ways i can get any one of these pairs and not the third winning number is  17

17*3= 51

So I think there are 51 ways of chosing 2 winning small numbers and not not getting the third.

So that makes a total of 52 ways of getting a winning set of smal numbers.

So I think that the chance of getting 2 or 3 small numbers + getting the big one is

\(\frac{52}{1140}\times\frac{1}{10}=\frac{52}{11400}\)

Which means that the chance of winning is  

\(\frac{52}{1140}+\frac{1}{10}-\frac{52}{11400}\\ =\frac{520}{11400}+\frac{1140}{11400}-\frac{52}{11400}\\ =\frac{520+1140-52}{11400}\\ =\frac{1608}{11400}\\ =\frac{67}{475}\\ \approx 0.141 \)

Ginger's answer and mine are only slightly different. I think it is just rounding error.

If so mine is closer because i didn't round anything until the very end.  

Hi Melody:

Actually the difference in our solutions is significant; it’s not from rounding errors.

The main reason is 

\(\dfrac{52}{1140}+\dfrac{1}{10}\underbrace {- \dfrac{52}{11400}}_{\text{This Term}}\\\)

^{(My computer is not rendering any Latex. I will correct this display, if needed)}

This subtracts 10% of the (2 of 3) ball wins.

I had initially thought of this but dismissed it because of the either/or nature of the win.

On reflection, I can see it does count the win twice when both the super ball and the white balls are a match. It should count one or the other but not both. The problem, now, is by subtracting 52/11400, it subtracts both wins.  So, what should the value be?

My quick answer is half that (26/11400).The reasoning is by subtracting half, then one or the other is subtracted but not both.

What thinkest thee, Dame Melody?

More comments:

I agree with the 52/1140, I neglected to include the 3 of 3 win in my equation, because I was patting my chimp head for figuring out Lancelot Link’s convoluted formula for counting a  Z subset of  matches of R in a  N choose R combination.  

Here are the adjusted numbers in the same format
6)  Add (1 / 1140) to (51 / 1140) giving (52 / 1140) for when all three numbers match.

Calculate number of times both the super ball and two of the three balls match
(52 / 1140)  * (1 / 10) = 52/11400      <---  ½ of this = (26 / 11400)

Subtract half of matches when both the super ball and two of the three balls match (This counts one or the other but not both.)

Calculate the harmonic average

H = 3 / ((52 / 1140) + (1 / 10) - (26 / 11400))  = 20.930232558

Equalized probability of each event (1 / 20.9023) * 3 (for the three events)
= (3 / 20.9023)
= (43 / 300)
= 0.143333…

Note: doing it this way is much easier (This is Melody’s method)

 ((52 / 1140) + (1 / 10) - (26 / 11400)) = (0.143333… )

Over all probability 14.33%

(Only the final calculation is rounded)

This question is interesting, but it seems very advanced.
Do you understand this math, Victoria?  

Comments and criticisms from mathematicians (and articulate trolls) are welcome.

GA

I'm with Melody.

For the 20 ball first part, the number of three ball combinations is 20C3 = 1140.

The number of ways in which we miss with all three numbers is 17C3 = 680.

The number of ways in which we have 1 hit and three misses is 3*17C2 = 408.

The number of ways in which we have 2 hits and 1 miss is 3C2*17 = 51.

The number of ways in which we have three hits is 1.

(Check that all possibilities have been covered, 680 + 408 + 51 + 1 = 1140.)

That gets us the probabilities,

zero hits 680/1140 = 0.59649 (5 dp),

1 hit 408/ 1140       = 0.35789 (5 dp),

2 hits 51/1140        = 0.04474 (5 dp),

3 hits 1/1140          = 0.00088 (5 dp).

An alternative method is to calculate the probabilities directly.

Using H for a hit and M for a miss, the probability of three misses MMM is (17/20)*(16/19)*(15/18) = 0.59649.

The probability of the sequence HMM is (3/20)*(17/19)*(16/18) = 0.119298, and since both of the other 1 hit sequences MHM and MMH have the same probability, the probability of 1 hit will be 3*0.119298 = 0.35789 (5 dp).

The probabilty of the two hit sequence HHM is (3/20)*(2/19)*(17/18) = 0.014912, and since both of the other two hit sequences HMH and MHH have the same probability, the probability of two hits will be 3*0.014912 = 0.04474 (5 dp).

The probability of three hits will be (3/20)*(2/19)*(1/18) = 0.00088 (5 dp).

That means that the probability of at least two hits (two matches) is 0.04474 + 0.00088 = 0.04562.

The probability of not winning that way will be 0.59649 + 0.35789 = 0.95438.

The probability of winning with the superball is 0.1 and not winning with the superball is 0.9.

There are four possible outcomes,

(1) not getting at least two matches and not picking the superball : probability 0.95438* 0.9 = 0.85894,

(2) getting at least two matches, but not the superball : probability                    0.04562*0.9 = 0.04106,

(3) not getting at least two matches but do pick the superball : probability        0.95438*0.1  = 0.09544,

(4) getting at least two matches and picking the superball : probability              0.04562*0.1 = 0.00456,

(the probabilities summing  to 1, as they should).

That means that the probability of winning is 0.04106 + 0.09544 + 0.00456 (or 1 - 0.85894) = 0.14106.

Tiggsy.

Thanks, Melody and Tiggsy......I'm never good at these probability things and your explanations were easy to see......

Tiggsy, your presentation is amazing. This clearly indicates all the mathematical relations, including the 10% “adjustment” for double counts.

Here’s a condensed presentation

\(\text{Prb = }\left(\dfrac{\binom{3}{2}\binom{17}{1} + \binom{3}{3}}{\binom{20}{3}} * \dfrac{9}{10}\right) + \dfrac{1}{10} \approx 0.1411 \text{ or } 14.11%\)

Melody’s Venn diagram also makes clear why the adjustment only affects the intersection of the two events and not the events separately.

At first blink, this question seems straightforward, and relatively simple to solve. However, it has subtleties that require the mathematician to have practiced skills in combinatorics and statistics to fish out the nuances that affect the final counts. Melody and Tiggsy obviously have these skills. 

Someone previously asked this question on here:

https://web2.0calc.com/questions/1-nbsp-in-the-smallstate-lottery-three-white-balls-are-drawn-at-random-from-twenty-balls-numbered-1-through-20-and-a-blue-superball-is

Post #5 has an incidental question that I can answer:  It’s only a swear word when you are trying to count the bleedin’ things!       

GA

In the Small State Lottery,
three white balls are drawn at random from twenty balls labled 1-20 and
a blue Superball is drawn from ten balls labeled 21-30.
To win a prize you must match at least two of the white balls or match the blue Superball.
If you buy a ticket what is the probability that you would win a prize?

Let W be the probability of choosing 2 white balls out of 20 or 3 white balls out of 20 .

Let \(\overline{\mathbf{W}}\)be not W.

Let B be the probability of choosing one blue ball out of 10.

Let \(\overline{\mathbf{B}}\) be not B

\(\begin{array}{|rcll|} \hline W &=& \dfrac{\binom{3}{2}\binom{17}{1} + \binom{3}{3}\binom{17}{0} }{\binom{20}{3}} = \dfrac{13}{285} \\\\ \overline{W} &=& 1 - W = 1-\dfrac{13}{285} = \dfrac{272}{285} \\\\ B &=& \dfrac{\binom{1}{1}\binom{9}{0} }{\binom{10}{1}} = \dfrac{1}{10} \\\\ \overline{B} &=& 1 - B = 1 - \dfrac{1}{10} = \dfrac{9}{10} \\ \hline \end{array} \)

\(\begin{array}{|c|c|c|c|} \hline & \mathbf{B} & \mathbf{\overline{B}} \\ & \dfrac{1}{10} & \dfrac{9}{10} \\ \hline \mathbf{W} & W\cap B & W\cap \overline{B} \\ \dfrac{13}{285} & \dfrac{13}{285} \times \dfrac{1}{10} & \dfrac{13}{285} \times \dfrac{9}{10} \\ \hline \mathbf{\overline{W}} & \overline{W}\cap B & \overline{W}\cap \overline{B} \\ \dfrac{272}{285} & \dfrac{272}{285} \times \dfrac{1}{10} & \dfrac{272}{285} \times \dfrac{9}{10} \\ \hline \end{array} \)

The probability of not a prize is
\(\overline{W}\cap \overline{B} = \dfrac{272}{285} \times \dfrac{9}{10} \\\)

The probability of a prize is
\(\begin{array}{|rcll|} \hline && 1- \overline{W}\cap \overline{B} \\ &=& 1- \left(\frac{272}{285} \times \frac{9}{10} \right) \\ &=& \frac{285\cdot 10- 9\cdot 272}{285\cdot 10} \\ &=& \frac{402}{2850} \\ &=& \frac{67\cdot 6}{475\cdot 6} \\ &=& \frac{67}{475} \\ &=& 0.14105263158 \\ \hline \end{array} \)