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2011-02-01T00:00:01
<mdoxml version="1.0"><p>We are often blasted with opportunities to "take a chance" to try to win an attractive prize for what seems like very little money and with an apparently good chance of striking lucky (how wrong can you be?). Below are some examples of opportunities to win a holiday for two at a well known resort. It will only cost you the price of a bottle of water to take part.</p>
<p class="c1">If you were extravagant enough to "try your luck", which of the following offers the best chance of you winning and how do you know?<br></br>
<mdo:image alt="Image of the options to win prizes" height="372" src="probs4.gif" width="795"></mdo:image><br></br>
I think I would prefer the water - especially on a hot summer's day! But you might not agree.</p>
<br></br>
<p><a href="http://nrich.maths.org/7997">Click here for a poster of this problem</a>.</p>
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<span class="editorial">Rajeev from Fair Field School sent us the
following solution:</span><br></br>
I found that your best chance of winning was the option of tossing
the coin and the chance of winning was $\frac{1}{4096}$.<br></br>
Here is how I worked it out:<br></br>
<ol>
<li>You win first prize if you can toss a fair coin and get 12
heads in a row. With one coin toss, you get half a chance, with 2
coin tosses you get $\frac{1}{4}$ and with 3 you get $\frac{1}{8}$
so with 12 coin tosses you get$(\frac{1}{2})^{12}$ which is
$\frac{1}{4096}$</li>
<li>Throw 5 fair dice and you get 5 sixes and you win the first
prize. With 1 fair die your chance of getting a six is
$\frac{1}{6}$ and with 2 its $\frac{1}{36}$ so with 5 fair dice its
$(\frac{1}{6})^5$ which is $\frac{1}{7776}$</li>
<li>Choose the top 4 from 10 famous pictures and put them in the
right order to win. $\frac{1}{5040}$</li>
<li>Our resident Gardener has listed her seven favourite plants in
order. If you can match the order you win. With 2 there are 2 ways
of ordering, with 3 it is 6 and with 4 it is 24 and so with 7 it is
$7\times6\times5\times4\times3\times2\times1)=5040$ So the
probability of selecting the correct ordering is
$\frac{1}{5040}$</li>
<li>You toss four ten-sided dice and win the first prize if you can
get 4 sixes. With one die it's $\frac{1}{10}$ and with 2 dice it's
$\frac{1}{100}$ and with 4 dice it is $\frac{1}{10000}$</li>
</ol>
<div><span class="editorial">Christian from Takeley explains how to
calculate the probability for number 3:</span></div>
<div>The game which involves the ten famous pictures. You work out
the probability of putting the right picture first, which is one
tenth. You then work out the probability of getting the right
picture and putting it into the second place, which is one ninth.
You continue in this way until you have all four individual
probabilities. You then multiply them all together: one tenth, one
ninth, one eight and one seventh, to get your final answer. The
probability is $\frac{1}{5040}$, which by coincidence is the same
probability as in the gardener game.</div>
<br></br>
<br></br>
<div><span class="editorial">Well done too to Patrick from
Woodbridge School and Yash from Natomas Charter School who sent in
largely complete solutions to this problem. Yash
commented:</span></div>
<div>Since all the numerators are the same, the greatest
probability would be have to be the least denominator. Thus you'd
be best off tossing a coin 12 times, but it's probably not
as much fun.</div>
<br></br>
<div><span class="editorial">And Patrick pointed out:</span></div>
<div>Thus, the best deal seems to be the twelve heads in a row -
which is so unlikely that it would be better to buy that bottle of
water instead!</div>
<br></br>
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<h3>Why do this problem?</h3>
<a href="http://nrich.maths.org/public/viewer.php?obj_id=920&part=">This
problem</a> offers a great opportunity for talking about chance.
Students can use their intuition to rank the options in order, and
then model the situation and manipulate the resulting fractions.
Finally, there is a chance to discuss whether the models used are
appropriate and how this might affect their answer.<br></br>
<h3>Possible approach</h3>
<div>Present the problem, and explain that students are being asked
to try their luck without having access to anything more
sophisticated than paper and pencil. Allow some time for pairs or
small groups to discuss the different options, making sure they
understand what they mean. Each group should come up with which
option they think gives them the best chance or the worst chance of
winning, with some justification for why they believe it.</div>
<br></br>
<div>Now give the groups time to write down the calculation they
would use to represent each situation, encouraging use of fractions
rather than decimals for easier working without calculators.</div>
<br></br>
<div>Check that groups have modelled each situation appropriately.
This is a good opportunity for some discussion about the more
subjective options - those which involve a ranking based on
preference - and how the probability would change given more
information about the people who put them in order.</div>
<br></br>
<div>Once everyone has a model for each option, groups can start to
compare different options. Some comparisons are easier than others,
so encourage use of prime factorisation to compare the size of the
denominators. Laws of indices can be useful to compare for example
$(\frac{1}{2})^{12}$ with $(\frac{1}{10})^4$.</div>
<div>Once comparisons have been made, students should compare their
answers with their intuition about which was best, and consider
whether they would try to win the holiday or keep their money!
Another discussion point could be how much the prize should be
worth in order for the organisers to make money from the
enterprise.</div>
<br></br>
<h3>Key questions</h3>
<div>Which game do you think is easiest to win?</div>
How can we compare two fractions with different denominators?<br></br>
<h3>Possible extension</h3>
<div>Come up with other scenarios with similar probabilities.</div>
<div><a href="http://nrich.maths.org/public/viewer.php?obj_id=4334&part=">The
Better Bet</a> is another problem about choosing which game to play
in order to maximise winnings. </div>
<br></br>
<h3>Possible support</h3>
Allow some calculator use so the focus is on the probability
calculations without the comparisons of fractions getting in the
way. <br></br>
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<p>The answers are not necessarily all different!</p>
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<mdoxml version="1.0">A full solution was received from Andrei of School 205 Bucharest. Joseph Yu from:HKMA David li Kwok Po College and La Viseis of the Preah Sisovath also sent solutions for the "best bet".
<br></br><br></br>The probability of the "toss coin" is: 1 / (2 to the power 12) = 1 / 4096
<br></br><br></br>The probability of the "Famous snaps" 1 / (10 x 9 x 8 x 7) = 1 / 5040
<br></br><br></br>The probability of the "Throw five dice" 1 / (6 to the power 5) = 1 / 7776
<br></br><br></br>The probaility of the " ten side dice" 1 / (10 to the power 4) = 1 / 10000
<br></br><br></br>The probaility of the "Gardener" 1 / (7 x 6 x 5 x 4 x 3 x 2 x 1) = 1 / 5040
<br></br><br></br>So "Toss coin" has the highest chance to win!
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Chances are
Which of these games would you play to give yourself the best
possible chance of winning a prize?
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