Lottery with repeated numbers
March 25th, 2014 at 5:56:40 PM
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Quick question about the probability of winning a unique lottery scenario. If a lottery number pool consists of the numbers 1-6 and each number is repeated 6 times (a total of 36 numbered balls) what is the probability of winning? There is the similar rule that the order of the draw does not matter. If the number 123 is chosen and the numbers 312 are drawn then 123 is still a winning ticket.
My original analysis leads me to believe that it is best to choose three different numbers. Even though 555 is a possible winning number it is not as likely to be drawn as 246.
I then tried to think of the number of different ways three repeated numbers could be drawn: (6)(6 C 3) = 120
Followed by two of the same number and one other number: (6)(6 C 2)(5)(6 C 1) = 2,700
Followed by three unique numbers: (6)(6 C 1)(5)(6 C 1)(4)(6 C 1) = 25,920
In each of these the C is representing the combination operation. I'm just not sure I've calculated things correctly. Could someone offer some advice?
My original analysis leads me to believe that it is best to choose three different numbers. Even though 555 is a possible winning number it is not as likely to be drawn as 246.
I then tried to think of the number of different ways three repeated numbers could be drawn: (6)(6 C 3) = 120
Followed by two of the same number and one other number: (6)(6 C 2)(5)(6 C 1) = 2,700
Followed by three unique numbers: (6)(6 C 1)(5)(6 C 1)(4)(6 C 1) = 25,920
In each of these the C is representing the combination operation. I'm just not sure I've calculated things correctly. Could someone offer some advice?
March 25th, 2014 at 6:07:48 PM
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It is best to choose three different numbers. Think about it like this.
You need one out of the three numbers for the first ball. That means there is a 18/36 chance of finding a ball you need. The second ball has two possible numbers you can use for a probability of 12/35. The last ball can only be one number which has a 6/34 chance of occurring.
If you pick the same number three times then you have a 6/36, 5/35, 4/34.
These all assume you remove the ball after each pick. If you put them back in then the actual odds change, but the concept is the same.
If order matters then it's still better to pick different numbers if you remove them. If order doesn't matter and you put them back then both choices are the same.
You need one out of the three numbers for the first ball. That means there is a 18/36 chance of finding a ball you need. The second ball has two possible numbers you can use for a probability of 12/35. The last ball can only be one number which has a 6/34 chance of occurring.
If you pick the same number three times then you have a 6/36, 5/35, 4/34.
These all assume you remove the ball after each pick. If you put them back in then the actual odds change, but the concept is the same.
If order matters then it's still better to pick different numbers if you remove them. If order doesn't matter and you put them back then both choices are the same.
March 25th, 2014 at 6:29:47 PM
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They have this in CA.
The interesting thing is that it's a shared prize pool. So, yes, most people pick 3 different numbers. But, if you don't and you win, you will win more (because you share with fewer other winners)
Eg, I am looking at the last two drawings (there are 2/day) and one of them was 6-6-9, and only 38 people won the "any order" prize (called "box") for $152 each. The other number chosen was 4-6-8, and 84 people won $74 each. Even though the 2nd prize pool was about $500 larger, the winners of the 1st prize pool won over twice as much each.
The interesting thing is that it's a shared prize pool. So, yes, most people pick 3 different numbers. But, if you don't and you win, you will win more (because you share with fewer other winners)
Eg, I am looking at the last two drawings (there are 2/day) and one of them was 6-6-9, and only 38 people won the "any order" prize (called "box") for $152 each. The other number chosen was 4-6-8, and 84 people won $74 each. Even though the 2nd prize pool was about $500 larger, the winners of the 1st prize pool won over twice as much each.
March 25th, 2014 at 6:32:18 PM
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I assume that all 36 balls are in the same "bucket", that only three are drawn, and when a ball is drawn, it is not put back.
In that case, since order does not matter, there are 36 C 3 = 7140 different ways to draw balls.
Of these, 6 C 3 = 20 are 111, 20 are 222, and so on; each ticket with the same digit three times has a 20/7140 = 1/357 chance of winning.
For any particular two-and-one combination (e.g. 116, 223), there are 6 balls that match the "single", and 6 C 2 = 15 that match the pair, or 90 combinations of balls that win; your chance of winning is 90/7140 = 3/238, or about 1/79.
For any particular combination of three different digits, there are 6 balls that match the first digit, 6 that match the second digit, and 6 that match the third digit, or 216 winning combinations; your chance of winning is 216/7140 = 18/595, or about 1/33.
Your numbers are slightly off as the order of the drawn numbers seems to matter in the three unique numbers case but not in the two-and-one case. In your unique-numbers case, any digit can be first, any other digit second, and any remaining digit third, so your count includes all six permutations of 123 separately, but in the two-and-one case, 665 is counted just once. Your unique-numbers value should be 6 C 3 (the number of unique three-digit combinations) x 6 x 6 x 6 = 4320. Notice that 120 + 2700 + 4320 = 7140 = 36 C 3.
Also, you're actually solving a different problem; you want to know the best chance of winning with one ticket, as opposed to a particular "class" of tickets. Using a similar methodology to yours, if the lottery consists of a three-digit number from 0 to 999, then, since there are 512 winning numbers that do not have either 0 or 9 as any of its digits and only 488 that do, then you should be more likely to win if you choose a number that does not have a 0 or a 9 - but, of course, all 1000 numbers have a 1/1000 chance of winning.
In that case, since order does not matter, there are 36 C 3 = 7140 different ways to draw balls.
Of these, 6 C 3 = 20 are 111, 20 are 222, and so on; each ticket with the same digit three times has a 20/7140 = 1/357 chance of winning.
For any particular two-and-one combination (e.g. 116, 223), there are 6 balls that match the "single", and 6 C 2 = 15 that match the pair, or 90 combinations of balls that win; your chance of winning is 90/7140 = 3/238, or about 1/79.
For any particular combination of three different digits, there are 6 balls that match the first digit, 6 that match the second digit, and 6 that match the third digit, or 216 winning combinations; your chance of winning is 216/7140 = 18/595, or about 1/33.
Your numbers are slightly off as the order of the drawn numbers seems to matter in the three unique numbers case but not in the two-and-one case. In your unique-numbers case, any digit can be first, any other digit second, and any remaining digit third, so your count includes all six permutations of 123 separately, but in the two-and-one case, 665 is counted just once. Your unique-numbers value should be 6 C 3 (the number of unique three-digit combinations) x 6 x 6 x 6 = 4320. Notice that 120 + 2700 + 4320 = 7140 = 36 C 3.
Also, you're actually solving a different problem; you want to know the best chance of winning with one ticket, as opposed to a particular "class" of tickets. Using a similar methodology to yours, if the lottery consists of a three-digit number from 0 to 999, then, since there are 512 winning numbers that do not have either 0 or 9 as any of its digits and only 488 that do, then you should be more likely to win if you choose a number that does not have a 0 or a 9 - but, of course, all 1000 numbers have a 1/1000 chance of winning.
March 26th, 2014 at 12:31:43 AM
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Quote: nross16Quick question about the probability of winning a unique lottery scenario. If a lottery number pool consists of the numbers 1-6 and each number is repeated 6 times (a total of 36 numbered balls) what is the probability of winning?
Instead of your 1-6 game ("your game"), you can think of the lottery as one with 36 unique balls numbered 1-36 which get to be drawn the usual way (traditional game). To map to your game, the numbers on the traditional balls are meant to be in modulus of 6. Then it is easy to see that each of your ticket are equivalent to some number N traditional tickets (with available numbers 1-36).
The number N is then the mapping of winners between your game and the traditional game. This indeed depends on the pattern chosen on your game
In fact N = C(6,n1) * C(6,n2) * C(6,n3) * C(6,n4) * C(6,n5) * C(6,n6), where nx is the multitude of x chosen on the 1-6 ticket.
If your ticket is 111, then you hold 20 traditional tickets.
If your ticket is 112, then you hold 90 traditional tickets.
If your ticket is 123, then you hold 216 traditional tickets.
