# Probability of Lottery Match - Two Entries Compared With Two Draws

If the probability of winning the jackpot in a lottery with a single entry is 1/14,000,000, then the probability of winning the jackpot with two entries is 2/14,000,000 or 1/7,000,000.

If I buy a single entry in two different weeks' draws, what is the probability of winning the jackpot in EITHER draw?

I feel like the answer is still 1/7,000,000, and therefore buying two tickets for one draw gives you the same chance of winning the jackpot as buying a single ticket for each of two draws, but I'm not certain!

• it might depend on if there's any bias in the draws etc. in theory. – user451844 Jul 27 '17 at 12:30

Let $p=\frac{1}{14000000}$ i.e. The probability of winning in any one draw.

The probability of winning in just one of the two draws is $2p(1-p)$ and the probability of winning in either of the two draws, or both, is $$1-(1-p)^2$$

• Why, when showing the probability of winning in one of the draws, do we multiply by (1 - p)? I would have thought the probability would just be 2p? – Philip Stratford Jul 27 '17 at 13:01
• $p(1-p)$ is the probability of winning in the first but not in the second – David Quinn Jul 27 '17 at 13:19

Case 1: 2 tickets, one lottery, one prize, $n$ different tickets equally likely

Events: A:= win with ticket A, B:= win with ticket B, W:= win the prize


$P(A)=P(B)=p=1/n$
$P(W)=P(A\cup B)=P(A)+P(B)-P(A\cap B)=p + p - 0 = 2p$

Notice that $P(A\cap B)=0$ (you can't win with 2 tickets in the same lottery), events $A$ and $B$ are not independent, they are mutually exclusive.

Case 2: 2 lotteries, one ticket in each lottery (2 possible prizes), $n$ tickets equally likely in each lottery

Events: A:= win in lottery 1, B:= win in lottery 2, W:= win in at least 1 lottery,
Z:=win in exactly one lottery


$P(A)=P(B)=p=1/n$
$P(W)=P(A\cup B)=P(A)+P(B)-P(A\cap B)=p + p - p^2 = 2p-p^2$

The events $A$ and $B$ are independent in this case (not in Case 1), therefore $P(A\cap B)=P(A)P(B)=p^2$

If you want the probability of winning in exactly one lottery (event $Z$) but not in both:

$P(Z)=P((A\cap \bar B)\cup(\bar A \cap B))=P(A\cap \bar B)+P(\bar A\cap B)-P(\varnothing)$ $P(Z)=p(1-p)+(1-p)p-0=2p-2p^2.$

Conclusion: the probability of winning in Case 2 (in at least a lottery) is a bit smaller than in Case 1 as $p^2=1/n^2>0$ is a very small positive number ($\rightarrow 0$ as $n\rightarrow \infty$), but you get as a bonus the possibility of winning 2 prizes if you are very lucky! not possible in Case 1 :).