Lottery chances What's statistically more likely to win you the UK lottery (6 numbers out of 1-49), 
To play the lottery weekly or to save the money and buy multiple tickets 
all in a oner?
There is a discussion in the office where a group of 6 people want to play weekly.
 A: If you just want to say "I've won the lottery", then all in a oner, provided you buy different numbers for the multiple tickets. 
Suppose there are $Q$ possible outcomes in the lottery. The chance of winning with 1 ticket is $1/Q$. Assuming (reasonably) that the individual drawings are independent, the chance you fail to win $k$ times in a row is $(1-1/Q)^k$. 
If you buy $k$ tickets at one time, the probability that the outcomes is not one of the $k$ tickets you bought is $(Q-k)/Q = 1 - k/Q$. 
Observe that $(1-1/Q)^k > 1 - k/Q$ if $Q > 1$. So the chances of not having a winning ticket at all is greater if you play weekly. In other words, buying multiple tickets ever-so-slightly increase your chances of having a winning ticket. (The increase, however, is very, very small.)

But if you play for the pay-out: Assuming the pay-out is constant across the time-frame, let's call it $P$, buying $k$ tickets at once would give expected payout $kP/Q$: there's only one winning number, and the chance of you hitting it is $k/Q$. 
Whereas if you play weekly, there is the chance that you win multiple times! The chances that you hit the jack pot $j$ times is  $\binom{k}{j}(1-1/Q)^{k-j}(1/Q)^j$. But the payout for that would be $jP$. Adding them all up (from $j = 0$ to $j=k$, with a bit of algebra and the binomial theorem, you see that the total expected payout for playing weekly ends up also exactly as $kP/Q$, for playing $k$ times. 
A: If, in the second scenario, you are planning to buy tickets with distinct number combinations, then the odds are exactly the same.
If you buy multiple tickets with the same numbers on them, your chances of winning go down.  However, your "expected winnings" (in the mathematical sense) don't change.
