hi everyone, please help me solve this mathematics probability question:
Mr. A and Mr. B decided to buy lottery one day.
They both spent the same amount of money but in a different way.
Mr. A bought 1 set of 7 numbers randomly generated. (cost: $3.50)
1. X X X X X X X (1-45 random numbers that do not repeat)
Mr. B bought 7 sets of 6 numbers randomly generated (cost: $3.50)
1. X X X X X X (1-45 random numbers that do not repeat)
2. X X X X X X (1-45 random numbers that do not repeat)
3. X X X X X X (1-45 random numbers that do not repeat)
4. X X X X X X (1-45 random numbers that do not repeat)
5. X X X X X X (1-45 random numbers that do not repeat)
6. X X X X X X (1-45 random numbers that do not repeat)
7. X X X X X X (1-45 random numbers that do not repeat)
The result of the lottery draw consists of 6 numbers (1-45).
To win, the winner must have a set with the 6 winning numbers.
Q. Since both Mr. A and Mr. B have spent equal amount of money. Do they have equal chances to win?
If not, which of them have a better probability of winning?
Please explain the full mathematical solution of each senarios with steps.
The denominator in the probability is:
nCr = 45C6 = 45! / 6!*(45-6)! = 8,145,060
Since Mr. A holds 1 of the combinations, his probability of winning is 1 / 8,145,060
Since Mr. B probably holds up to 7 different combinations (unlikely that any of his match each other), then the probability of Mr. B winning is 7 / 8,145,060
Then the P(Mr. A winning) = 7C6 / 45C6 = 7 / 8145060, so they both do in fact have the same probability of winning, if you are willing to "go out on a limb" and say that it's rather unlikely that any of the combinations match....
Mr A has 7 such sets of six (corresponding to the six remaining numbers
when one of the seven is dropped from the set).
If Mr B's numbers are drawn so that no set of six is repeated then he to
has 7 sets of numbers and his chance of winning is identical to Mr A's.
However if a set could be repeated (that is the method of generating
them would allow this to happen), then he has on average slightly less
than 7 sets of numbers, and his chance of winning will be less than Mr A.
(This is the interpretation I would put on this problem, but YMMV).
thank u all for replying.
CaptainBlack: so what u mean is:
1. 1 set of 7 random numbers and 7 sets of 6 random numbers are equal.
2. if not, 1 set of 7 random numbers has a better probability (since/if the 7 sets of 6 random numbers might actually be same).
for example: (7 sets of 6 random numbers)
03 04 07 18 32 40
05 12 22 23 27 44 (possibility of the sets be exactly the same)
04 06 07 10 21 45
20 21 22 23 33 35
17 28 30 31 32 35
05 12 22 23 27 44 (thus, 1 set of 7 random numbers might actually have a slight advantage?)
01 05 06 19 34 37
any one else can confirm?