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Recently I was having a conversation with a philosophy student on gambling and it intrigued me because of what the person was saying. Before I say what the person said, I remember learning in class that the probability of something occurring does not increase as you use it. For example, if I flipped a fair coin, no matter how many times I flip the coin it will always have a 50% chance of landing heads or tails. However according to the person, he said that the more times I play a lottery machine, my chances of winning will increase as well. Can someone explain to me the difference between my statement and his?

I am thinking that my statement only refers to single events, but his refers to total probability?

Belphegor
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    Playing the lottery $N$ times will leave you more likely to win at least once, compared to playing a single game (but you should just expect to lose more money, overall).The probability of winning any given game is indeed constant, no matter how many times you play. – pjs36 May 06 '15 at 22:49
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    Does that imply that if we could live forever and as we keep playing infinitely, we will reach 100% of winning? – Belphegor May 06 '15 at 22:50
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    No, it just means that if we have a $1%$ chance of winning any individual game, that if we play twice, we have a ${2 \choose 1} \cdot (0.01)^1 \cdot (0.99)^1 \approx 1.98%$ chance of winning exacly once, compared to our $1%$ chance of winning each individual game. Playing the lottery repeatedly is an example of a binomial distribution with an extremely low $p$-value.

    It's just important to remember that, while you're more likely to win at least once, you're also expected to have lost more money than just playing once.

     – pjs36 May 06 '15 at 22:55
  • user3718584 - yes. the probability of winning once will approach 100% ( it will never reach exactly 100%, though). The probability of a total loss will aproach 100% as well, because the lottery ticket costs more than the average gain. – user70160 May 06 '15 at 22:55
  • Related: http://en.m.wikipedia.org/wiki/Gambler%27s_fallacy, searching [math.se] for "gambler's fallacy" yields a lot of posts as well. – 11684 May 07 '15 at 10:35
  • @11684 - interesting - in particular the story about the casino in 1913 referenced in the wikipedia post! http://www.fallacyfiles.org/gamblers.html – user70160 May 08 '15 at 02:18
  • Does this answer your question? If I play the lottery with a specific amount of money, should I put it all on one game or distribute it among several games –  Apr 05 '23 at 00:53

4 Answers4

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Exactly. Try it like this: let's say instead of drawing balls, the "lottery" would consist of a single coin flip. If you play once, you have a 50% chance of winning. If you play twice, your probability of winning at least once is already 75% (there are four equally likely outcomes: First round win, second round win (WW), first round win, second round lose (WL), LW and LL). The more you play, the smaller the percentage of "lose-only" outcomes will get. (The exact probabilities can be derived with the binomial distribution as mentioned in a comment above.)

If you keep playing until eternity, and then look back..., you will see that you have won it about half the time... but lost in total money terms, as the price of the lottery ticket exceeds your average gain (usually drastically).

user70160
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    So if x represents the number of tosses of a fair coin and as x approaches infinity, we can expect the outcome to level out to 50% heads and 50% tails – Belphegor May 06 '15 at 22:53
  • Exactly. That's the Law of Large Numbers. http://en.wikipedia.org/wiki/Law_of_large_numbers – user70160 May 06 '15 at 22:58
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    "If you play once, you have a 50% chance of winning. If you play twice, you have doubled your chance of winning it once." So if I play twice, I have a chance of $2\times 50% = 100%$ of winning? That doesn't seem right. – celtschk May 06 '15 at 23:04
  • @celtschk - yes, that was wrong, sorry, should have been more careful. Let me change it. thanks for the note! – user70160 May 06 '15 at 23:31
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    Why did you revert an edit that fixed misspellings? "Loose" means "not tight". "Lose" is the opposite of winning. – Alec May 07 '15 at 09:11
  • @ Alec - yes, sorry, my mistake, got confused. will change it back. thanks for the hint! – user70160 May 07 '15 at 09:24
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    Nitpicking: your probability of winning once sounds like exactly once, so maybe you should phrase it as your probability of winning (at least) once. – anol May 07 '15 at 12:28
  • @anol you are absolutely right - I will edit right now, thanks! – user70160 May 07 '15 at 13:07
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The outcome of any coin toss is independent of the previous result; be it heads or tails, the coin is "memory-less".

However you have a greater chance of winning the lottery if you have more goes it at.

Similarly if you only flipped a coin once you are going to see only one of either heads or tails, if you flipped it 100 times say, chances are you are going to see both.

Gregg
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  • I don't think this is a very good answer, as it doesn't relate the two setups (coin vs lottery) and the use of the word "however" between them suggests that they behave in different ways. They both behave in the same way: each individual coin toss (respectively lottery ticket) has the same chance of winning as any other coin toss (resp. lottery ticket). But if you play a lot of times, you have more chance of winning on at least one of those plays. – David Richerby May 07 '15 at 07:22
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Just to elaborate a bit on what others have said.

Suppose you flip a coin 3 times and that a winning flip is heads. There are 8 different equally-probable outcomes

(H H H), (H H T), (H T H), (T H H), (H T T), (T T H), (T H T), (T T T)

The probability of never winning (which corresponds to only the outcome (T T T) ) is therefore $1/8$, which is lower than the probability of losing a single coin flip.

Josh Burby
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In the example of a coin toss, each flip is discrete, therefore you have a 50% chance of guessing correctly and a 50% chance of guessing wrongly. The result of the previous flip has no bearing on the result of the subsequent flip, therefore you still have a 50/50 chance.

The same can be said for the lottery; the balls are drawn at random, after fully mixing with each other in the machine, we are led to believe that each ball is of identical mass and size, thus no ball is more likely to be drawn than any other. The result of this, is again, a discrete game.

E.G. Let us say that the outcome of a lottery was thus: 8, 12, 42, 24, 5, 38. The probability of winning the national lottery is around 1 in 14m, thus there is a 1 in 14m chance that the next draw will yield the exact same result (order does not matter). In summary, each lottery draw is discrete, thus you have no more, or less, chance of winning each time you play. You can increase your chances by purchasing multiple tickets, but this will not make you more likely to win any subsequent lotteries.

Hope that helped :-)

HelpingHand
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  • I may be wrong, but is it possible that with "each flip is discrete", you mean "each two flips are independent of each other" (as in http://en.wikipedia.org/wiki/Independence_%28probability_theory%29)? Because if I understand you correctly, that's what you describe with "The result of the previous flip has no bearing on the result of the subsequent flip". Again, I may be wrong, just have not come across the term "discrete" being used in that manner. I see it usually being used to distinguish "discrete random variables" (countable number of possible outcomes) from "continuous random variables". – user70160 May 08 '15 at 02:41