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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?

Yakk
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Rickie
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    my 2 cent: it is indeed doubled. there is the same chance to win the lottery each ticked. but I guess the confusion is because if you draw dice twice the chance to hit 6 is only $11/36$ which is less than $2*1/6$(which is 2 times the chance to hit 6 with one roll). anyway dice example is not valid here. – d_e Jul 22 '15 at 06:34
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    You double the probability of having your combination drawn. Winning is more complicated, since in standard lotteries the jackpot is fairly often shared. (We are assuming that as in the standard 6/49, that possibly no one wins,) – André Nicolas Jul 22 '15 at 06:38
  • I thought the order of the balls mattered in normal lottery. – Asinomás Jul 22 '15 at 06:44
  • Where is the internet divided? – Asinomás Jul 22 '15 at 06:44
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    Order does not matter in standard 6/49. – André Nicolas Jul 22 '15 at 06:45
  • Hi dREaM. I can find two schools of thought. Those that think it's doubled and those that say there's an increase but it's not doubled and they feel that those who think it is is what makes lotteries successful. :) I'm trying to get a definitive answer with a formula from someone who's far better at maths than I am. – Rickie Jul 22 '15 at 07:00
  • It depends on what happens if two players win, and what happens if nobody does. – RemcoGerlich Jul 22 '15 at 07:39
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    The reason why lotteries are successful is a lot simpler, the prize is less than the cost of tickets they sell. A simple corollary of this is that the expected value of a ticket is less than its price. – Asinomás Jul 22 '15 at 07:49
  • How different is different? All numbers different or at least one number different? – Thomas Weller Jul 22 '15 at 08:11
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    Though it does not answer the question, I have to remark that you double your expected loss by being two tickets (unless the organisers have made a massive mistake). – PJTraill Jul 22 '15 at 11:24
  • Please keep in mind that your Chances are greater but your Odds remain the same. Buying 2 tickets will do nothing to increase your odds. Let's say the odds are 1 in 2 million. Buying 2 tickets means you have 2 tickets that have 1 in 2 million odds. – NotJay Jul 22 '15 at 15:31
  • If you buy them all you are certain to win, so.. – TaW Jul 22 '15 at 16:36
  • @NotJay, he asked what were his odds of winning - not what the odds of each ticket were. So by buying two tickets "he" has doubled his chances/odds of winning. – CramerTV Jul 22 '15 at 19:59
  • @CramerTV The question asks, "If I buy two tickets with different numbers for the same draw, do I double my chance of winning the jackpot." There is a difference between chances and odds and the user did indeed ask about chances, not odds! If you buy 1 ticket, you have 1 chance, if you buy 2 tickets, you have 2 chances. You're not cutting the odds in half, you're doubling your chances. – NotJay Jul 22 '15 at 20:07
  • Think about it this way... if you buy something with a 1 in 10 chance of winning a prize, you have a 10% chance of winning. If you buy 10, you don't have a 100% chance of winning, you have ten 10% chances of winning. – NotJay Jul 22 '15 at 20:11
  • @NotJay, two things, in a lottery, if I have a 'choose without replacement' and I buy 10 tickets with a 1 in 10 chance of winning I do have a 100% chance of winning. Second, odds are x losses to y wins. So in this case I have two tickets - each with 9 to 1 odds. Since I own two the odds are 8 to 2. Converting odds to 'chances' is putting the wins over the total. So we start with 10% chance (1/10) and go to a 20% chance (2/10). This is not a choose with replacement scenario which is what you seem to be thinking. – CramerTV Jul 22 '15 at 20:20
  • The question its self is lacking. The question should be in reference to "two unique tickets" which would lead to a more plausible answer. – Luke Snowden Aug 19 '16 at 19:51
  • Does this answer your question? Odds of winning lottery with multiple unique tickets –  Apr 05 '23 at 00:51
  • Does this answer your question? Probability, lottery. Once a week for 30 years, probability of winning at least once? –  Apr 26 '23 at 03:23
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    Does this answer your question? Probability of winning the lottery the more you play it? –  Jun 06 '23 at 07:26

9 Answers9

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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.

msinghal
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    Which gives $$\frac{1}{{49\choose 6}}=0.00000007151123842...$$ I'd better get a better paid job. – pshmath0 Jul 22 '15 at 06:39
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    @pbs Hey, I think it is at least a better chance than happen to discover the solution to one of the Milennium Problems and win $1M. – wythagoras Jul 22 '15 at 06:42
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    In cases like this you can always logically check your results: If you would buy tickets for every combination, your chance of winning would be 100%, if you bought half of all possible tickets it is 50%, so it is proportional! The difference is when you play the lottery two times in a row! Then your chance is not doubled! (Again do the check, you could play the lottery trillions of times without ever winning 100%) – Falco Jul 22 '15 at 12:54
  • The probability of winning is sufficiently low that playing two times in a row essentially does double your chance of winning. You have to repeat this a very large number of times before the error term reaches anywhere near the order of magnitude of the probability. – R.. GitHub STOP HELPING ICE Jul 22 '15 at 16:24
  • There is no error term. $2/\binom{49}{6}$ is exact. – msinghal Jul 22 '15 at 16:25
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    @MihirSinghal I suspect R.. was referring to the error term when $n / { 49 \choose 6 }$ is used as an estimate for the probability of winning at least once when $n$ separate lotteries are played. (He was responding to Falco's comment). – anovstrup Jul 22 '15 at 19:21
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    @wythagoras OTOH, if you do manage to solve a Millenium Problem, you'll benefit society more than you would by winning the lottery – Max Nanasy Jul 22 '15 at 20:05
  • @MaxNanasy Why not both? – Akiva Weinberger Aug 04 '15 at 05:30
  • @Falco When you play twice in a row instead, the chance is still doubled if you remember to count the situation where you win in both draws as two. The chance of winning the first draw and losing the second, is $\epsilon(1-\epsilon)$, and the chance of losing the first one and winning the second is also $(1-\epsilon)\epsilon$. The chance of winning both draws is $\epsilon^2$. If you just add these three, you get $2\epsilon-\epsilon^2$, so you are right. But if you give the double win a "score" (or weight) of two, the total is $2\epsilon$, so again doubled expectation. – Jeppe Stig Nielsen Apr 16 '18 at 20:56
  • @JeppeStiegNielsen One has to distinguish chance and expected return - when we talk about chance we usually talk about the chance of winning at least one time. (Because in most Scenarios we stop playing as soon as we win) - So the chance is not doubled. And if you are taking about expected ROI, it stays constant, since the expected payout doubles but the investment also doubles. – Falco Apr 18 '18 at 08:50
15

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.

wythagoras
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    It seems quite evident in the question that the winning combination is drawn independently of any tickets sold (and so quite possibly nobody gets the jackpot). So the "another lottery" stuff is just a needless diversion; don't add confusion that is absent in the question. – Marc van Leeuwen Jul 22 '15 at 13:46
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"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

Buck
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  • Even if it does mean “win exclusively”, you double your chance by buying a second distinct ticket, if •(0) the draw is fair, •(1) there is only one winner, •(2) the rest of what happens is unaffected by your behaviour and •(3) the probabilities of other participants buying a ticket identical to your first or second ticket are the same. Under these assumptions, the probability of winning with either ticket is the same, and they are mutually exclusive, so may be added. Even if you drop (0, 2 & 3) but have no information about the true situation, your estimate of your chance should double. – PJTraill Jul 22 '15 at 13:00
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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 $\frac{1}{\binom{49}{6}}$.

Bernhard
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  • The problem states, "If I buy two tickets with different numbers" – msinghal Jul 22 '15 at 06:40
  • Correct. I just wanted to clarify this explicitly, since this apparently causes the confusion in the internet the OP was writing about... – Bernhard Jul 22 '15 at 07:38
  • So let me get this right. If I have a 1 in 14 million of chance of winning the lottery, if I buy a further ticket with a different sequence of numbers to the first one for the same draw my chance of winning is slashed to 1 in 7 million? – Rickie Jul 22 '15 at 08:16
  • Yes, that is correct. – Bernhard Jul 22 '15 at 08:23
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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

Alan Dean
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Here's the "correct formula" as you requested.

If $N$ people play, and the probability of any individual person winning is $p$, then the probability that a given person doesn't win is $1-p$. And the probability that nobody wins is $(1-p)^N$, assuming independence between the various plays. So

$$ P(\text{AT LEAST 1 person wins}) = 1-(1-p)^N $$

If $N$ people play each of $L$ lotteries, then the probability that nobody wins any lotteries is $(1-p)^{LN}$, so

$$ P(\text{at least 1 person wins at least one lottery}) = 1-\color{red}{(1-p)^{LN}} $$

There are two ways to estimate $\color{red}{(1-p)^{LN}}$. You can use

1. $\lim\limits_{n\to \infty }(1-\frac{x}{n})^n=e^{-x}$.

2. or the Binomial Theorem to expand $\color{red}{(1-p)^{LN}}$.

Since $p$ is small, you can approximate $p^{LN} \approx 0 \; \forall \; N>1$ without adding too much error into your calculation. This gives you $\color{red}{(1-p)^{LN} \approx 1-LNp}$. So $$ \begin{align} P(\text{at least 1 person wins at least 1 lottery}) & = 1-\color{red}{(1-p)^{LN}} \\ & \approx 1-\color{red}{[1-LNp]} = LNp. \end{align} $$

But as $N$ and $p$ increase, this approximation worsens.

Thus 1 person buying 1 ticket has a probability $p$ of winning, 2 tickets is $2p$ (100% better than 1 ticket), 3 tickets is $3p$ (50% better than two tickets), etc.

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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.

Hagen von Eitzen
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    It's not independent though, the events are mutually exclusive. You can't win both tickets at once the way the problem is stated. – msinghal Jul 22 '15 at 06:39
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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.

annoying_squid
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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.

user258720Errol
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