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June 3rd, 2007, 01:24 PM  #1 
Newbie Joined: Jun 2007 Posts: 3 Thanks: 0  Probability question....quite hard.
Hi there. Can someone help me with this question???? In the soviet Union bus tickets were numbered 000 000 through 999 999. If the sum of the first three digits equalled the last three the ticket was regarded as lucky. What is the probability of getting a lucky ticket? I would really appreciate some help in understanding how to solve this one. Thanks. 
June 3rd, 2007, 01:59 PM  #2 
Newbie Joined: Jun 2007 Posts: 8 Thanks: 0  Not sure
Do you mean that a ticket is lucky if the sum of the first three digits is equal to the sum of the last three digits, ie 123 501 would be lucky ?

June 3rd, 2007, 02:34 PM  #3 
Newbie Joined: Jun 2007 Posts: 3 Thanks: 0 
Hi, thanks for the reply. I would have to say yes to that question. I dont know if there exists some kind of mathematical equation to figure this out or not.

June 3rd, 2007, 05:43 PM  #4 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
We could break it down into cases based on the sum, although it would be somewhat messy. The total number of tickets is 1,000,000, and we want to count the number of tickets where both sides sum to 0, 1, 2, etc., up to 27. We could even cut the number of cases to search in half by noticing that the same number of tickets sum to n as 27n. This should get you started. 
June 3rd, 2007, 11:43 PM  #5 
Newbie Joined: Jun 2007 Posts: 8 Thanks: 0 
Here is a response to the 'messy' method. I am convinced there is also a simpler one, but haven't found it just yet... First of all we want to find out how may ways there are to make a given number n<28, that is, how many different triplets (x,y,z) exist with : x+y+z=n first of all, if n=0 : x=0, y=0, z=0 one ticket if n=1 : x=0, y=0, z=1 three tickets (either x, y, or z is 1) if n=2: then either two are worth 1, and the other 0; three tickets or one is worth 2 and the others 0 three tickets > six tickets if n=3: three 1's one 2, two 1's one 3, three 0's seven tickets Then keep going until 13 (going to be a lot of fun) by not considering the order in which x, y and z are, and then multiplying when there are possible permutations. For a given n, consider; x=n, y=0, z=0 three tickets x=n1, y=1, z=0 six tickets ( = 3! ) then keep going........ have fun, please let me know when you find the clever method 
June 4th, 2007, 03:53 AM  #6 
Senior Member Joined: Nov 2006 From: I'm a figment of my own imagination :? Posts: 848 Thanks: 0 
Here is a method to simplify your counting. To find the number of solutions to the system x+y+z=n where x, y, z, and n are integers, imagine taking n 's and 2 's. The 's represent one and the 's represent divisions. So the number of 's before the first  represents the value of x, the number of 's between the two 's represents the value of y, and the number of 's after the second  represents the value of z. Clearly these three numbers will add up to n. For example:  represents x=2, y=3, z=1, n=6  represents x=3, y=0, z=3, n=6 Now that we have set up our correspondance, we can evaluate it as a permutation with repetition. Our answer is going to be (n+2)!/(n!2!). Now, because we want the left and right side to both have this sum, we need to square our answer. We also need to, before adding up the answers, double all the ones between 1 and 13 inclusive to account for the ones between 15 and 27. We do not, however, double the answer for 14. This can reduce the work by a lot. 
June 4th, 2007, 07:15 AM  #7 
Newbie Joined: Jun 2007 Posts: 8 Thanks: 0 
Wow, you got me on that one! Not bad ! I guess I should work somewhat more on my math curriculum ! 
June 5th, 2007, 07:48 PM  #8 
Newbie Joined: Jun 2007 Posts: 3 Thanks: 0  Here's the answer!!!!!!
11. NOTE: This is a bonus question and it is quite difficult. You can earn 100% without doing it, but you can get up to 10 points for this question. In the Soviet Union bus tickets were numbered 000 000 through 999 999. When you got on a bus you could buy a ticket from a machine, so the number on it could well be regarded random. If the sum of the first three digits of the number equaled the sum of the last three digits, the ticket was called 'lucky'. Many people used to collect lucky tickets and some believed that lucky tickets should be eaten to activate them. What is the probability of getting a lucky ticket? The first step is to notice that the number of lucky tickets can be represented by a sum of lucky tickets with a given value of the sum of the first (or last) three digits. If we fix the sum to be K and denote NK the number of ways K can be obtained by summing 3 digits, then the total number of lucky tickets is N02+N12+…+N272. Finding NK is somewhat tricky. Let NK(m) be the number of ways K can be obtained by summing m digits. We need NK(3). These can be found recursively. Clearly, NK(1) is 1 for Kÿ9 and 0 for K>9. Then we notice that if NK(m1) is known for any K we can use it to find NK(m): for each possible choice of one digit we have a known number of combinations for the rest m1 digits. NK(m) = NK(m1) + NK1(m1) + … + N0(m1) if Kÿ9 and NK(m) = NK(m1) + NK1(m1) + … + NK9(m1) if K>9. So, after doing this smart procedure up until we know all NK(3), and then summing up the squares, we arrive at 55252 total of lucky tickets. The probability is then 55252 / 1000000 = 0.055252 or slightly over 5.5%. Alternatively, you can write a simple computer program and for each ticket test if it is lucky. 
June 5th, 2007, 08:26 PM  #9 
Member Joined: Dec 2006 From: St. Paul MN USA Posts: 37 Thanks: 0  Re: Here's the answer!!!!!!
Here's a oneline solution in Maple: add(i^2,i=coeffs(expand(add(x^j,j=0..9)^3)))/1.0e6;

June 6th, 2007, 06:32 AM  #10 
Newbie Joined: Jun 2007 Posts: 8 Thanks: 0 
so bigben why did you ask the question in the first place ?


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