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A lottery game, Big Four, is played by choosing four numbers from 1 to 15 (no repetition of numbers; order of the numbers does not matter). You win the top prize in a particular draw if the four numbers on your ticket are the winning numbers. There is only one top prize in each draw.

You have decided to have the following four numbers: 2 – 7 – 11 – 15 and buy only one lottery ticket of this number combination in each draw for the next 200 draws.

What is the likelihood for you to win the top prizes in the next 200 draws?

My Answer (Or thinking flow)

So we consider the chance of me winning at any draw. It is $\frac{1}{15}^4=\frac{1}{50625}$.

I can win in many different scenarios. Win all 200 draws, win no draws, win the 6th draw etc. Gotta consider all scenarios.

  1. 0 wins. Likelihood $=0$
  2. 1 win. I can win in any of the 200 draws. Likelihood $\frac{1}{50625}*^{200} P_{1}=\frac{1}{50625}*200=\frac{8}{2025}$
  3. 2 wins. I can win in any two of the 200 draws. Likelihood $\frac{1}{50625}*^{200} P_{2}=\frac{1}{50625}*39800=\frac{1592}{2025}$
  4. Continuing.....................Till 200

I believe my answer to be very wrong, as I am supossed to do it with a faster method. My tutor is known for giving not-so straightforward questions, so I'm wondering if I need to consider another way, or I could be wrong. Any alternatives welcome too!

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up vote 3 down vote accepted

If $p$ is the probability of winning on any particular draw, then $1-p$ is the probability of not winning.

The probability of not winning $200$ imes in a row is $(1-p)^{200}$.

So the probability of winning at least once is $1-(1-p)^{200}$.

Note that the probability of winning on any particular draw is given by $$p=\frac{1}{\binom{15}{4}}.$$

Remarks: $1$. In any lottery of this type, some provision has to be made for situations in which there is more than one winning ticket. Ordinarily, the top prize is shared, which may not be quite consistent with your assertion that there is only one top prize.

$2$. In order to find the probability that we win at least once, we first found the probability that we win $0$ times, that is, the probability that the event "win at least once" doesn't happen. This type of strategy is fairly often useful.

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Wow... Thats simple and elegant. Thanks. Is there a way to do it directly? What is wrong with my answer? – Yellow Skies Oct 12 '12 at 14:03
The probability of winning on any draw is not correct, the true value is much higher. The probability of $0$ wins is not $0$, it is $(1-p)^{200}$, where $p$ is as in the answer above. That essentially finishes things. But if we don't do it that way, the probability of exactly $1$ win is $\binom{200}{1}p(1-p)^{199}$. Similarly, the probability of exactly $2$ wins is $\binom{200}{2}p^2(1-p)^{198}$, and so on. Find all of them, add up. That is related to the strategy you used. Lots of work! Fairly often, it is easier to find first the probability a certain thing doesn't happen. – André Nicolas Oct 12 '12 at 14:20

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