If I buy 2 lottery tickets do I double my chance of winning?
There's a lottery. There are 6 balls chosen randomly from 49 and you have to match all the balls to win.
I buy one ticket. If I buy two tickets with different numbers for the same draw, do I double my chance of winning the jackpot.
What's the correct formula here?
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$\begingroup$Yes, you do double your chance of winning. There are $\binom{49}{6}$ different possible draws. The probability of winning with one ticket is the chance that the 6 drawn balls are one specific combination, i.e. $\frac{1}{\binom{49}{6}}$. With two tickets, there are two winning combinations, so the probability is $\frac{2}{\binom{49}{6}}$
Basically, the reason it is doubled is because winning one ticket and winning the other ticket are mutually exclusive, i.e. you can't win both tickets (note the problem statement says "different tickets"). If there was a chance of winning both, the chance would be less than doubled.
$\endgroup$ 10 $\begingroup$It depends on the lottery.
In your lottery, all bills are unique. So, the odds are indeed doubling as others have pointed out, and if you buy all different tickets, you have a 100% winning chance (though the cost of buying all those tickets is far greater than the lottery prize).
Another lottery:
Suppose that there is a lottery where all tickets are thrown in an pool. There is only one prize. The ticket that is randomly selected gets the prize. There are already 50 tickets sold.
If you buy one ticket, then your chance of winning is $\frac{1}{51}$.
If you buy two tickets, then your chance of winning is $\frac{2}{52}<\frac{2}{51}=2\cdot\frac{1}{51}$.
So your odds are improved, but not doubled.
I guess this is where the confusion on the internet is from.
$\endgroup$ 1 $\begingroup$"If I buy two tickets with different numbers for the same draw, do I double my chance of winning the jackpot."
Does "winning the jackpot" mean exclusively? If so, then the answer is "no" because there is no mathematical formula that can predetermine how many tickets will be sold.
If "winning the jackpot" means "picking the winning numbers," then yes, your odds of winning increase proportionally with the number of unique tickets you buy. So if there are 6 million combinations, your first ticket is 1 in 6 million, and it increases with each ticket you purchase:
1 = 1/6 million 2 = 1/3 million 3 = 1/2 million 4 = 1/1.5 million 5 = 1/1.2 million 6 = 1/1 million
$\endgroup$ 1 $\begingroup$It depends on how you choose the second ticket: If you choose the second ticket independently, you do not double the chance of winning, since there is a non-zero chance of getting a ticket with exactly the same numbers -- after all, in this kind of lottery you can choose the numbers, and you cannot prevent two persons choosing the same numbers.
If you choose the tickets carefully such that you have two different number sequences, you indeed double the chance of winning. There are $\binom{49}{6}$ possible numbers, and by choosing two you have a chance of winning of $\frac{2}{\binom{49}{6}}$ as compared to $\frac{1}{\binom{49}{6}}$.
$\endgroup$ 4 $\begingroup$Well I wonder if I was rich enough to buy 14 million unique lottery tickets and the jackpot is a guaranteed 20 million for matching 6 balls im 100% chance of winning but it could be shared equally between other winners but let's say it's not so I win the 20 million plus all the 5 ball numbers and bonus ball plus all the 4 numbers all the 3 numbers etc is that the case or will it be 20 million and that's it
$\endgroup$ $\begingroup$The expected value of the win is typically doubled because expected value is additive. However, as you mention 6 out of 49: In this type of lottery the prize sum depends on the tickets submitted, not even this is true: If you have one ticket ans are the sole winner, you may win a million and if you are the only winner with two identical tickets each of these may win only half a million, so it may happen that there is no increase at all.
If we model that each ticket has a certain independent winning probability $p$, then two tickets raise your winning probbaility not to $2p$, but to $1-(1-p)^2=2p-p^2$, so slightly less than doubled.
$\endgroup$ 1 $\begingroup$Currently for the mega millions there are 302,575,350 different combinations of numbers. So your probability of winning on your first ticket is 1 / 302,575,350, which equals a percentage of 0.00000033049618880057500%. After you play your first ticket you have 302,575,349 different combinations left and a probability of winning on your second ticket of 1 / 302,575,349, which equals 0.00000033049618989285200%. The latter odds is twice that as the first. So ALMOST YES, mathematically your chances of winning improve by just under 100%. However, you'll notice that your ACTUAL ODDS do not improve that much.
$\endgroup$ $\begingroup$Understanding lottery. Let us say there is a lottery where there is 100 tickets but there will be only 1 prize. This is a game of chance, nothing to do with probability. Buy 1 ticket and your chance of winning the prize is 1 in 100. Buy 2 tickers and your chance of winning is 1 in 99 and not 1 in 50 as most people think.
The rational is as follows: of the 2 tickets 1 has to be a looser but you do not know which. Therefore it is unmistakable that you have a looser. It follows then that you may have the winner, so mentally you discard 1 as a loser and maybe the other is the winner reducing your chance to only 1 in 99 and not 1 in 50 as many go on believing.
Think about it - you have to be bright enough to have posed the question.
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