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When playing the lottery you have to pick 6 numbers out of 45 possibilities. Since the order of the numbers don't matter, the number of possible combinations for the jackpot (and hence the 6 correct numbers) is given by:

$ \dbinom{45}{6} = C^6_{45} = \frac{45!}{(45-6)!6!} = 8145060 $

Assuming the numbers are picked at random out of a uniform distribution the chance of winning the jackpot is given by:

$P(win) = \frac{1}{8145060}\approx 0.0000123 \%$

Now I can figure two possible scenarios:

  1. You play every week with the same numbers hoping that one day you'll get the jackpot.
  2. You play every week with a different set of numbers hoping that one day you'll get the jackpot.

Now I was wondering if there is a difference in the chance of winning between the two methods. I was thinking that in the first case the chances of winning are larger because if you stick to your number the alternatives might run out. While if you switch you don't have the other options that become eliminated. But then I started wondering, because in the Monty Hall paradox the chances become larger if you switch.

For the first case I believe the chance of winning in the $n^{th}$ drawing is simply calculated as:

$1-(1-P(win))^n$

But in the second case I wouldn't know how to succeed.

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The right hand of your $\,P(win)\,$ is lacking two zeros between the period and the one...winning the lottery is not that easy! –  DonAntonio Jul 22 '13 at 10:53
1  
If the lottery numbers are independent, then it does not matter. In Monty Hall paradox you can increase your chance because you are given additional information after the first pick (and that is not the case in this setting). –  dtldarek Jul 22 '13 at 10:58
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"the alternatives might run out" is where I think you're a bit mistaken; each lottery draw is independent of the last, so it's perfectly possible to have the same combination two weeks in a row (or more!) –  Tim Jul 22 '13 at 11:01
    
@DonAntonio isn't it missing two zero's becaus I put a % behind it? –  Nick Jul 22 '13 at 13:39
    
I don't think so, @Nick: the lacking zeros are between the period and the $\,1\,$ ...:) –  DonAntonio Jul 22 '13 at 13:49

3 Answers 3

up vote 5 down vote accepted

Assuming the outcome of each lottery raffle is completely random (and thus assuming every result is independent from all the preceeding ones), the chance to win writing down a random number each time is the same, i.e. $\,\frac1{8145060}\,$

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Lets say for argument sake the numbers last week were: 1, 2, 3, 4, 5 and 6. Then the probability of getting 1, 2, 3, 4, 5 and six this week is still $\,\frac1{8145060}\,$.

Assuming the system is totally random then it in not remembering previous outcomes so it makes no difference whether or not you change your numbers.

The Monty Hall is different however:

There are 3 doors one with a prize two with nothing. You pick one door the probability that you have won the prize is $\,\frac1{3}\,$. If the host now were to ask you do you want what is behind the door you picked or what's behind both the other doors then most people would easily choose the other two doors as they are twice as likely to contain the prize.

However the host knows where the prize is, and since there is only one price he can always show you a door with nothing behind it and he asks you do you want to swap with the other still closed door. Should you swap?

Yes because you are still being asked in effect do you want what's behind every door except the one you picked so you are twice as likely to get the prize if you swap.

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Considering your two scenarios, you are just dealing with two aleatory indipendent events.

Aleatory: they are completely casual, in the sense that may arise even when you think they're unlikely to occur;

Indipendent: they don't influence each other in any manner ( $p(E_1) = p(E_1|E_2)$ ). For this case, reason as follows:

If one day you chose the numbers

$7 \quad 5 \quad 18 \quad 32 \quad 44 \quad 38$

with probability $1/{45 \choose 6}$ you win the lottery; suppose that day you lose. The following week you choose another set of numbers, say

$23 \quad 3 \quad 34 \quad 41 \quad 12 \quad 9$

Then you start watching the extraction:

First number: $3$ "Well, fine."

Second number: $9$ "I'm really lucky today!"

Now think, however, that the probability for your other numbers to be drawn is much less than it was before the first extraction:

Third number: $7$ "Who told me to chage the numbers?"

In short, you can change, but you don't have to.

In the Monty Hall Paradox you have $\frac{2}{3}$ probability to win the car changing the door, because the host opens a door with a goat, because he knows where the goats are: if you originally had chosen a goat, changing will make you win; the same as if you had originally chosen the other goat; only in one of three cases (if you had chosen the car) you lose changing the door. For this problem, I would have redirected you to Wikipedia, just if its page wasn't so tremendous!

Finally, remember: each lottery draw is completely indipendent from the last one.

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