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After I buy one lottery ticket with odds of $14M$ to one, if I buy another ticket with different numbers, does this slash the odds to $7M$ to one? If so, if I double my tickets again to $4$ Tickets, it will halve again to $3.5M$ to one? So $8$ tickets will be $1.75M$ to one?

$16$ Tickets $\rightarrow.875M$ to one

$32 \rightarrow .4475$M to one

$64 \rightarrow 218750$ to one

$128 \rightarrow 109375$ to one

$256 \rightarrow 54587$ to one

Something seems wrong here?

Pad Regan
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  • Your final division by 2 is about a factor 10 too large. – TMM May 19 '17 at 11:40
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    There is nothing wrong here at all as long as you ensure that the tickets are different from each other. Your chance of winning goes up linearly with the number of tickets you buy. Your expected loss also goes up linearly with the number of tickets you buy. – Ross Millikan Jul 14 '17 at 21:03
  • @user1147844 Please do not keep making cosmetic edits to old questions! This is your second change to this old question. – Ethan Bolker Apr 07 '23 at 19:25

2 Answers2

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Nearly but not quite. Start with a simpler case e.g. two dice and you are trying to get at least one 6. What are the odds? A common naive answer is 2/6. However, this is obviously wrong, consider 3, 4, 5, 6, and 7 dice. With 6 dice, getting at least one 6 is likely but certain. With 7 dice, it is not more than certain.

The correct answer can be seen by listing all 36 ways that two dice could fall. You will see that only 11 out of 36 contain a 6 so the odds are only 11/36. The overcounting is due to one of the 36 cases being double 6.

Back to the lottery, this approach is not so practical but fortunately there is an alternative. It is easier to calculate the chances that you don't win with $n$ tickets.

Don't win with 1 ticket: $\frac{13999999}{14000000}$.

Don't win with 2 tickets: $(\frac{13999999}{14000000})^2$.

Don't win with 3 tickets: $(\frac{13999999}{14000000})^3$.

Don't win with n tickets: $(\frac{13999999}{14000000})^n$.

Subtract from 1 to get the odds of winning.

With a very low probability such as $\frac{1}{14000000}$, the first few will be very close to the naive answers. As $n$ gets bigger, the difference will become more obvious.

Thanks to comments by Losvos and TMM: note that I have accidentally assumed that each ticket is an independent random choice as would be typical with dice. If you are careful not to repeat your selections then the naive logic will be correct. Buy all possible 14000000 tickets and you are sure to win. However, it is very likely that you will win less than the cost of the tickets. Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot.

badjohn
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  • This can't be right, what if you buy every ticket? – Bobson Dugnutt May 19 '17 at 11:38
  • @Lovsovs A good point. My answer is based on the dice case. I'll amend it. – badjohn May 19 '17 at 11:39
  • @TMM, Agreed, I have edited it due to Losvos's similar point. – badjohn May 19 '17 at 11:46
  • "Even if the jackpot is higher, due to rollovers, be aware that someone else might win and you will share the jackpot." - What do you mean? If I somehow mamage to buy all tickets for the next draw, am I not guaranteed the sole jackpot? – TMM May 19 '17 at 11:51
  • @TMM It depends on the rules of the specific lottery. Here in the UK at least, it is possible and common that two people buy a ticket with the same numbers. It is quite common that the jackpot is shared by 2 or more people. Quite early on, over 100 people shared the jackpot. I can imagine that many got very excited that they had won over £1000000. They may have calmed down and realised that it might be only £500000 or £333333 but they probably were not expecting just £80000. Nice but but as life changing as a million. I just hope that not too many resigned their jobs. – badjohn May 19 '17 at 12:07
  • Still, if someone buys all tickets, wouldn't he win all those 100 shares of the jackpot? – TMM May 19 '17 at 12:23
  • @TMM If you buy all 14000000 combinations and I buy just one but I am very lucky to get the right one then we share the jackpot. I will be a lot up on the deal and you will probably be a lot down. In the real cases that I have heard of when someone has tried to systematically buy all tickets (because of a rollover), the lottery organization has actually tried to stop them. – badjohn May 19 '17 at 12:58
  • If you have 2 tickets there are not ...999 losers – paparazzo May 19 '17 at 16:04
  • @Paparazzi I am not sure whether you are agreeing or disagreeing with me. My last point was that in the UK lottery, and many others, there are not preprinted tickets with each possible number combination once. Instead, you choose the numbers when you buy your ticket. This means that any number of tickets may be sold. Some combinations may be bought by multiple people and some not at all. So, sharing a prize is possible. Also, no winner is possible and the prize usually rolls over. It is only in this rollover case, that there is any chance that buying all combinations may make sense. – badjohn May 19 '17 at 17:32
  • If I have 2 tickets of 10 then I do not have 9 losers. By your equation could buy all and not be 100%. – paparazzo May 19 '17 at 17:37
  • @Paparazzi Please see the correction at the end of my post. If you are sure that your tickets have different numbers then you have a 2 in 10 chance of winning. My original answer was on the (rather unlikely) assumption that you choose the numbers randomly each time with no memory and there is a chance that you buy the same number twice. – badjohn May 19 '17 at 17:41
  • The question is clear on that matter. – paparazzo May 19 '17 at 17:42
  • @Paparazzi Agreed. I made a mistake and hence the correction that I posted. – badjohn May 19 '17 at 17:43
  • @badjohn OK, we were talking about different lottery systems then. I was indeed referring to a system where each combination is printed once, and you can't choose your combination. – TMM May 19 '17 at 19:15
  • @TMM There's a lottery which prints 14000000 tickets? Where? It's common for much smaller competitions though I find it surprising with such a large number. Of course, if it's a hypothetical lottery then that's fine. My apologies for going down the wrong path. – badjohn May 19 '17 at 20:11
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The odds are in this case given by $$\text{number of tickets you have}:{\text{number of tickets you don't have}}.$$

The important thing is that the denominator is not $\text{total number of tickets}$. So if you have $2^n$ tickets and the total number of tickets is $T,$ the odds of you winning will be $$2^n:T-2^n \iff 1:\frac{T}{2^n}-1,$$

which is different to your calculations. Let's take a look:

$n = 1$ gives odds of $ 1:\frac{T}{2}-1$

$n =2$ gives odds of $1:\frac{T}{4}-1,$

and so on, which grows much slower than your solution.

Edit: Here you can see a graph of your odds of winning as a function of the amount of $n$, the number of times you double your number of tickets. As you can see, it does actually grow exponentially, but it is off to an extremely slow start. At around $n=24$, you have bought all the tickets, so here the odds are $1:0,$ which gives the singularity shown.

So the bottom line is: Don't play the lottery! ;)

Bobson Dugnutt
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