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A hypothetical example:

You have a 1/1000 chance of being hit by a bus when crossing the street. However, if you perform the action of crossing the street 1000 times, then your chance of being hit by a bus increases to about 60% because every time you do the action, the probability of it happening again increases.

What is the math behind this to support this? Just curious.

Lizzie
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    NO- the probability on each crossing (in this model) does not increase with more tries. The probability of being hit at least once does increase with more tries. – Ross Millikan Sep 17 '14 at 16:49
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    The probability on each crossing remains the same! – Oria Gruber Sep 17 '14 at 16:49
  • I think the question you're really asking is "If I repeat a process with X probability Y times, what is the odds of a success in that trial?" which has an answer in a question I asked, here: http://math.stackexchange.com/questions/6140/help-with-a-specific-limit-left-dfracn-1n-rightn-as-n-rightarrow which I'm surprised I remember exists 4 years after posting it... – corsiKa Sep 17 '14 at 19:46
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    Image you flip a coin - it lands on heads 5 times. Your friend bets a bunch of money it will be tails next time. Is he right? 50% chance. Doesn't matter how many times it just landed on heads! Try: http://en.wikipedia.org/wiki/Gambler's_fallacy – user1833028 Sep 18 '14 at 04:55
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    Actually the probability of being hit by a bus a next time decreases in the event of being hit by a bus, because you're probably either in the hospital or in the graveyard, where there aren't many busses. But it's hard to determine the probability for a "next time" that does not occur. – Marc van Leeuwen Sep 18 '14 at 09:55

4 Answers4

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\begin{align} P(\text{hit by bus in 1000 crossings}) & = 1-P(\text{not hit by bus in 1000 crossing}) \\ & = 1-(999/1000)^{1000} \\ & \approx 0.63 \end{align}

E W H Lee
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    @Lizzie It is important to realize form the above, that if you have crossed the street 999 times earlier (successfully) in your life, the change of being hit by a bus the next time is NOT 63%, it is still 1/1000 – imranfat Sep 17 '14 at 16:56
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    This also assumes that, after being hit by a bus, you continue to try crossing the street, in spite of how injured you are. – Dancrumb Sep 17 '14 at 16:58
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    Close to $1-e^{-1}$ – Henry Sep 17 '14 at 17:16
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    @Dancrumb: Right. In an alternative formulation, that being hit by a bus prevents you making any further attempts, we would conclude that if you perform the action 1000 times, then the first 999 of them necessarily must have been successful, and so the conditional probability of being hit at least once is simply the probability of being hit on the last attempt, 1/1000. This conditional probability therefore is not the quantity that the question is talking about, that increases to over 60% with 1000 attempts :-) Actually the math here makes no assumptions what happens after the first bus. – Steve Jessop Sep 17 '14 at 17:53
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    @Dancrumb It doesn't really assume that. The probability of you being hit by a bus at least once (computed in this answer) is the same regardless of what you do after you are hit once. – Caleb Stanford Sep 17 '14 at 23:28
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    @Henry Is that value the limit of $1-(1-1/n)^n$? – kasperd Sep 18 '14 at 10:00
  • @kasperd: yes, and it is a frequent limit of probability questions such as the secretary problem and the couples sitting in a row problem. – Henry Sep 18 '14 at 10:10
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    @Dancrumb's comment has a lot of votes, it's confusing. I think the answer should be edited to clarify that no such assumption is made. – a06e Sep 18 '14 at 12:51
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    @becko I would agree. However, I think that what is more instructive is that either approach (quit after being hit, or not) leads to the same result. – Dancrumb Sep 18 '14 at 13:04
  • @Dancrumb Agree. – a06e Sep 18 '14 at 13:07
  • FWIW: http://math.stackexchange.com/questions/935486/whats-the-probability-of-a-an-outcome-after-n-trials-if-you-stop-trying-once-y/935860#935860 – Dancrumb Sep 18 '14 at 14:04
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You can approximate this very well using a Poisson distribution: If over a large number $N$ of trials an event occurs an average of $\lambda \ll N$ times, the probability that it occurs $k$ times in a set of $N$ trials is $$P(k) \approx \frac{\lambda^k e^{-\lambda}}{k!}.$$

In our case, over $1000$ trials we expect our event with per-trial probability of $1/1000$ to occur an average of 1 time, so $$P(k) \approx \frac{e^{-1}}{k!}.$$ In particular the probability of the event occurring at least once is $$1 - P(0) \approx 1 - \frac{1}{e} = 0.63212\ldots$$ which is very close to the actual value of $$1 - P(0) = 1 - \left(\frac{999}{1000}\right)^{1000} = 0.63230\ldots$$ as E W H Lee gives in his good answer.

This method shows two additional interesting facts:

  • The approximate probability $1 - \frac{1}{e}$ is actually universal, and not particular to the case $N = 1000$ (in fact, this quantity is the limit of the probability as $N \to +\infty$).
  • Over $N \gg 1$ trials the probability than an event with a per-event probability $1/N$ does not occur is very close to the probability that it occurs exactly once: both of these probabilities are $\approx \frac{1}{e}$.
Travis Willse
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  • "This also assumes that, after being hit by a bus, you continue to try crossing the street, in spite of how injured you are" @dancrumb is that true for both answers? Can you expand on this? – Lizzie Sep 17 '14 at 18:48
  • http://math.stackexchange.com/questions/935486/whats-the-probability-of-a-an-outcome-after-n-trials-if-you-stop-trying-once-y?noredirect=1#comment1930125_935486 – Dancrumb Sep 17 '14 at 20:21
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    @Lizzie That applies to this approximate answer too: For the Poisson distribution to apply, the events must be independent, and this includes not changing behavior after an event. In fact, this method immediately gives you approximate probabilities for being hit by a bus exactly two times ($\frac{1}{2e} \approx 0.184$), exactly three times ($\frac{1}{6e} \approx 0.061$), etc. – Travis Willse Sep 18 '14 at 02:01
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Calling $p$ the probability of being hit on one crossing and $q=1-p$ the probability to be safe in one crossing you can compute $p_{1000}$ (the probability of being hit at least once in $1000$ crossings) like this:

$$p_{1000} = p + qp + qqp +...+q^{999}p$$

This mean: you get hit at the first attempt, or you escape the first and get hit at the second, or you escape the first and the second and get hit at the third...

Now we can solve the geometric series: $$p_{1000} = p \sum_{n=0}^{999} q^n = p \frac{1-q^{1000}}{1-q} = p \frac{1-q^{1000}}{p} =1-q^{1000} \approx 0.63$$

DarioP
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For such questions, with every attempt, your chance to win increases. The chance is never 100% (as far as I know) but generally, you could write the probability out as being: 1-((1–1/x)^n) Where x is your denominator (out of what is your chance i.e 1/1000) and n is the number of tries. So for your example, it would be 1-((1–1/1000)^1000) which equals 1-(0.999^1000), which turns out to be about 0.63230457, or 63.230457% There is a lot of confusion about this topic, as intuitively, you would think that if the odds are 1/1000 playing 1000 times would guarantee a win. I was taught this method, and it has always worked for my purposes.

P.S. Turns out, the lottery is even worse than you thought.

Dan B
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