# Is it better to play 1 dollar on 10 lottery draws or 10 dollars on one lottery draw?

If I had 10 dollars to spend on a 1 dollar lottery draw, would I have more chance of winning if I spent all 10 dollars in one draw or bought 1 dollar tickets for 10 separate draws?

Edit: in terms of lottery definition, you pick 6 numbers from a pool of 49 numbers (1-49), that is classed as one lottery ticket. So each 1 dollar represents a selection of 6 numbers. Across multiple tickets you can pick the same numbers as appear on your previous tickets. If you are familiar with Euromillions or UK Lotto, it's that kind of lottery.

http://www.national-lottery.co.uk/player/p/lotterydrawgames/lotto.ftl

Edit 2:

Let me re-phrase the question. The probability of winning the jackpot in the lottery is 1 in 13,983,816.

Would buying 10 tickets for one draw change those odds to 10 in 13,983,816 ? and if so is that better than playing in 10 different draws at 1 in 13,983,816 odds each?

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Please define "more chance of winning". – joriki Mar 19 '12 at 12:38
Both current answers assume a certain type of lottery and aren't applicable to other kinds of lotteries, so you might want to specify what sort of lottery you were thinking of. – joriki Mar 19 '12 at 12:49
It would be better if you just gave me the \$10, and I get us a couple beers. You'd end up with the same loss (\$10), but we'd both have beer. – zzzzBov Mar 19 '12 at 15:50
URL ends with "Lotto For The Loss". Quite apropos. – Austin Mohr Mar 20 '12 at 5:58
I'm always amazed at the irony of the lottery providing state education funding... – tacos_tacos_tacos Nov 29 '12 at 3:10

Your expected gains (or rather losses) are the same for both methods. However, if you get tickets for separate draws, there is an ever so tiny chance that you will win more than once, and correspondingly the chance that you will win (at least once) will be an ever so tiny bit smaller.

As an extreme example of this phenomenon, replace $10$ by the total number of tickets in one draw. Then taking them all in the same lottery ensures a win in that lottery, but taking them in all different lotteries does not ensure any win, but might lead to multiple wins.

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This answer could use a graph to show how the area under both curves is the same but the shape of the multi-drawing cases are increasingly Gaussian. – Ben Jackson Mar 19 '12 at 19:47
Taking them in all different lotteries is about a 63.2% chance of winning at least one. – Joe Z. Dec 24 '12 at 4:43
@Joe Z. - To elaborate on your 63.2%... Chance of winning at least once: $1 - \left(\frac{x - 1}{x} \right )^{n}$ where n is the number of times played and x is the odds. The case referenced here is $n=x$. As x gets large the equation goes to $1-\frac{1}{e}$ or 63.2%. There is a big difference between playing every combination in a single drawing and every combination of as many drawings. I know I would prefer 100% chance of winning once over 63.2% chance of winning infinite times, but lack of capital makes this an impossibility and taxes make it pointless. – J.Money Dec 23 '13 at 20:09

Suppose there are only ten tickets - if you buy them all in one draw you have to win. But if you buy in successive draws you can lose every time.

If there is more than one winning ticket in the first case, you could end up winning twice and dividing the pool between your two winning tickets.

In the second case there is a possibility of winning in multiple draws.

So if you are interested in the maximum return for your stake you will need to factor in the value of the win(s) in each case.

In the first case - buying to tickets in a draw with a single win, the tickets represent mutually exclusive events (probabilities add).

In the second case the outcomes are (in the absence of other information) independent - and the easiest way of calculating the probability of winning at least once is first to calculate the probability of losing each time, and multiply using the rule for independent events. Then you should be able to see how to finish this off.

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The question is poorly formed, and feels more like game theory than math, per se. By any purely objective standard, the two opportunities provide the same expected outcome. However they have different risk profiles, and so either might be preferable depending on the reward function of the person asking.

If you are optimizing to Maximin, for example, it's better to play ten smaller hands (or not to play at all).

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+1 But I would say game theory is a branch of math. Heck, one of the most prominent game theorists is the mathematician John Nash. – Alex Becker Mar 19 '12 at 18:47
I wouldn't argue with that, but game theory involves an element of subjectivity as it allows for multiple optimization functions and the original poster seemed to want an objective answer without further input. – Michael Stern Mar 19 '12 at 20:00

Okay, I'm the guy (above) with all the "historic" posts, but THIS time I feel I actually have the correct answer to the original "Is it better to play 1 dollar on 10 lottery draws or 10 dollars on one lottery draw?"-question!

As I understand how lotteries operate, there actually ARE 2 sets of statistical averages. The "raw, brute #'s" set that we're all familiar with ("1-in-175,000,000," etc.), and then, once again, the long-term "historic" set -- which is where the house's "historic pay-out" table comes in.

Take the following "short-term"-example, which in theory should be extrapolatable into a longer-term set: 1 person buys 1,000 tickets, 17 people buy 100 tickets, 593 people buy 5 tickets, and 8,756 people buy 1 ticket. Out of which group of people is the winning Grand Prize likeliest to come from? Clearly the 1-ticket group!

Now, while it may be true that the single 1,000-ticket buyer has a 1,000x better chance of beating any SINGLE member of the 8,756 1-ticket buyers, it's quite clear that, "overall," one of the 8,756 1-ticket buyers has a better chance of beating the single person holding the 1,000 tickets!

See, if I'm correct, then statisticians apparently forget to factor-in the brute reality that lotteries are COMPETITIONS, which are thus RELATIVE to different #'s of GROUPS of ticket buyers, NOT just a matter of "odds in a vacuum!" Again, if I'm correct here, then the smartest buyers of tickets in the above COMPETITIVE scenario would be the 1-ticket buyers, NOT the single 1,000 tickets buyer!

So, again I ask, since I'm sure major lotteries give each of their Grand Prize winners a form to fill-out which has a question on it to the effect, "How many tickets TOTAL did you buy at the time that you bought your winning ticket?," then it follows that these various lotteries know EXACTLY what the "smartest # of tickets to buy" (hence the "Golden #") is for any one of their particular games!!!

I imagine that, over time, said Golden # is relatively stable (let's guess 7-9 tickets), and this is why I surmise that a statistical mathematician can likely determine what said "historical" Golden # likely is WITHOUT necessarily having to consult the HQ of any given lottery (in order for THEM to state what the Golden # is)!

Please, either show me where I'm wrong here, else "get" how my line of reasoning makes total sense! Thank You.

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The question is not well-formed. To say what is best we need a definition of a loss function! One possible objective is to maximize the probability of winning more than a certain amount, say 1000000$. There is a famous boook studying that question, titled "How to gamble if you must" (That is a very advanced book!). In short, the best strategy for that objective is "Bold play": Put all the money on one ticket. - Perhaps here's another way of putting all this. I always buy just one lottery ticket per game, but then had an insight: Sure, the statistical odds are extreme, but perhaps you should ALSO consider which ticket buyers win the most lotteries. In short, are most Powerball jackpots won by one ticket buyers, or by multiple ticket buyers? In other words, let's say 23% of the time single \$2 ticket buyers win, 39% of the time \$10 ticket buyers (5 tickets) win, 21% of the time \$20 ticket buyers (10 tickets) win, with the other denominations winning the remaining 17% of the time. So, if those stats were true, if I wanted to MAXIMIZE my odds, I'd always buy a \$10 5-line ticket, rather than what I have been doing, buying merely a \$2 (actually \\$3 since I go for the Power Play) 1-line ticket. So how do I find out who wins the most Powerball lotteries, single ticket or multiple ticket buyers? Thanks.

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This is a simple binomial problem. Use the formula: p(X=k) = (n choose k)(prob of success)^k(1-prob of success)^(n-k).

For all ten dollars in 1 draw, n=1, k=1. For ten different draws and exactly 1 success, n=10, k=1.
For ten different draws, and at least 1 success, use 1-p(X=0) where n=10.

You will find it is better to put all 10 dollars in 1 lottery draw.

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Welcome to MSE! It is very helpful to format your question in MathJax as it really improves readability. Regards – Amzoti Jan 22 '13 at 17:15

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