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Topic:
What probability a sixnumber lottery ticket has 3 consecutive numbers, the others not?
Replies:
7
Last Post:
Feb 12, 2009 11:18 PM




Re: What probability a sixnumber lottery ticket has 3 consecutive numbers, the others not?
Posted:
Feb 11, 2009 2:23 PM


On Feb 11, 3:52 am, Archie <kohmover...@yahoo.com> wrote: > Say you can choose six numbers out of 59. What is the probability that > three (and only three) will be consecutive? This is complicated by the > fact that numbers appear in order from least to greatest. What if the > six numbers contain 2 sets of 3 consecutive numbers? > > I'm not finding any easy formula for this. > > Thanks!
I must confess that I solve this type of problem by searching for the right physical picture to make the calculations more obvious. If I can't get the right picture, I can't do the problem...
In this case, we start with the total number of differnt Lotto6/59 tickets:
Total = 59! / (6! x 53!) = 45 057 474
Now picture a straigh row of holes, numbered from 1 to 60...Yes, 60... You'll see in a minute, I hope...
Take 6 identical black balls and 54 identical white balls. Yeah, 60 again. Obviously, they'll fill the 60 holes...in a lot of different ways...
You need to consider the various ticket possibilities:
A) Run of three and three separated... eg 3, 6, 12, 13, 14, 30
B) Run of three, a separated run of two, and a single separate eg 3, 4, 9, 10, 11, 49
C) Two distinct runs of three eg 12, 13, 14, 51, 52, 53.
Lets do A)
Glue three of the black balls together in a straight line, and glue a white ball to the end of the line.
Take the remaining three black balls, and glue a white ball to each one.
You now have a fourball glueup, three twoball glueups, and 50 single white balls.
You now count how many ways you can pick these 57 items, and put them down in the numbered holes. Black balls indicate the numbers on the ticket. Glueups go with the whitle ball in the higher number hole. I hope you can see that: there must be three consecutive numbers, with no chance of an addon at the front or back; the extra glueons keep any other runs from occuring; and it's impossible for a black ball to land in Hole #60, but Hole #60 is needed to let a black ball to land in #59.
The answer is 57! / ( 3! x 50! )
Now B)
You need the same threeblack plus white glueup, a twoblackplus white glueup, a blackpluswhite glueup, and the balance of 51 white balls.
These number of ways these 54 items can be arranged in the line of holes is
54! / ( 51! )
C) is left as an exercise for the readers...
I got this strange picture from a keychain thingy I once saw for picking Lotto tickets. A shallow plastic case, with the numbers 1 to 49 printed in black on the clear top, and 6 white balls and 43 black balls inside. Shake it up, and read the numbers with whie balls behind them...
The glue was my idea...



