Puzzles
11 December
Noel has a large pile of cards. Half of them are red, the other half are black. Noel splits the cards into two piles: pile A and pile B.
Two thirds of the cards in pile A are red. Noel then moves 108 red cards from pile A to pile B. After this move, two thirds of the cards in pile B are red.
How many cards did Noel start with?
Note: There was a mistake in the original version of today's puzzle. The number 21 has been replaced with 108.
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Let's say there were originally \(t\) cards of each colour, and \(2a\) red cards and \(a\) black cards in pile A.
After the move, there were \(t-2a+108\) red cards and \(t-a\) black cards in pile B. Two thirds of the card in pile B are red, so:
$$t-2a+108=2(t-a)$$
$$t=108$$
Therefore there were 108×2=216 cards.
Interestingly, it appears that this answer is independent of \(a\): any number of cards could be put in each pile and this situation would still work. However, only one situation could have happened
unless you allow there to at some points have been a negative number of cards in each pile.
5 December
Carol rolled a large handful of six-sided dice. The total of all the numbers Carol got was 521. After some calculating, Carol worked out that the probability that of her total being 521
was the same as the probability that her total being 200. How many dice did Carol roll?
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The totals that are equally likely add up to 7 times the number of dice.
This can be seen by using the fact that the opposite sides of a dice add up to 7: for each way of making a given total with \(n\) dice, there is a way of making \(7n\) minus that total
by looking at the dice from below. Therefore \(T\) and \(7n-T\) are equally likely. (This also holds true (but is harder to explain) if you rearrange the faces of the dice so that the opposite
faces no longer add to 7.)
Therefore today's number is \((521+200)/7\), which is 103.
Bending a straw
Two points along a drinking straw are picked at random. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?
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A triangle will be made if none of the segments of straw is longer than the other two added together. This is the same as requiring that each segment must be less than half the straw.
Let the length of the straw be 1 unit. Call the points \(x\) and \(y\). A triangle is made if either:
- \(x\lt y\), \(x\lt\tfrac12\), \(y-x\lt\tfrac12\), \(1-y\lt\tfrac12\); or
- \(y\lt x\), \(y\lt\tfrac12\), \(x-y\lt\tfrac12\), \(1-x\lt\tfrac12\).
For the second condition, the allowable region is shown below.
This region covers \(\tfrac18\) of the whole square. By switching \(x\) and \(y\) it can be seen that the first condition's region is the same size as the second's, plus they don't overlap. Therefore the probability of making a triangle is \(\tfrac18+\tfrac18=\tfrac14\).
Extension
One point along a drinking straw is picked, then a coin is flipped. If the coin shows heads, a second point above the first is chosen; If tails, a second point below the first is chosen. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?
The sixth cent
You toss 6 fair coins, and I toss 5 fair coins. What is the probability that you get more heads than I do?
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As You have one more coin than me, if you don't throw more heads than me, then you must throw more tails than me.
This means that the probability of you throwing more heads then me plus the probability of you throwing more tails then me is equal to one.
By symmetry, the probabilities for heads and tails must be equal and so the probability that you throw more heads is \(\frac{1}{2}\)
Extension
You toss \(n\) fair coins, and I toss \(m\) fair coins. What is the probability that you get more heads than I do?
Marbles
A bag contains \(m\) blue and \(n\) yellow marbles. One marble is selected at random from the bag and its colour is noted. It is then returned to the bag along with \(k\) other marbles of the same colour. A second marble is now selected at random from the bag. What is the probability that the second marble is blue?
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Let \(M_1\) be colour of the first marble and \(M_2\) be the colour of the second marble
$$\mathbb{P}(M_2=\mathrm{blue})=\mathbb{P}(M_2=\mathrm{blue}|M_1=\mathrm{blue})\mathbb{P}(M_1=\mathrm{blue})\\+\mathbb{P}(M_2=\mathrm{blue}|M_1=\mathrm{yellow})\mathbb{P}(M_1=\mathrm{yellow})\\
=\left(\frac{m+k}{m+n+k}\right)\left(\frac{m}{m+n}\right)+\left(\frac{m}{m+n+k}\right)\left(\frac{n}{m+n}\right)\\
=\frac{m(m+k)+mn}{(m+n)(m+n+k)}\\
=\frac{m(m+n+k)}{(m+n)(m+n+k)}\\
=\frac{m}{m+n}
$$
Extension
A bag contains \(m\) blue and \(n\) yellow marbles. One marble is selected at random from the bag and its colour is noted. It is then returned to the bag along with \(k\) other marbles of the same colour and \(l\) marbles of the other colour. A second marble is now selected at random from the bag. What is the probability that the second marble is blue?
Fair dice
Timothy and Urban are playing a game with two six-sided dice. The dice are unusual: Rather than bearing a number, each face is painted either red or blue.
The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they're different. Their chances of winning are equal.
The first die has 5 red faces and 1 blue face. What are the colours on the second die?
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Let \(A\) and \(B\) be the outcomes of the two dice. Let \(p\) be the probability that the second die lands on red. The probability of the dice being the same is:
$$\frac{1}{2}=\mathbb{P}(A=r)\mathbb{P}(B=r)+\mathbb{P}(A=b)\mathbb{P}(B=b)\\
=\frac{5}{6}p+\frac{1}{6}(1-p)\\
=\frac{1}{6}+\frac{4}{6}p
$$
This means that:
$$\frac{4}{6}p=\frac{1}{2}-\frac{1}{6}\\
=\frac{1}{3}\\
p=\frac{\frac{1}{3}}{\frac{4}{6}}=\frac{1}{2}$$
Extension
If the first die has \(n\) red faces and \(6-n\) blue faces, what colours are on the second die?
The blue-eyed sisters
If you happen to meet two of the Jones sister (two sisters chosen at random from all the Jones sisters), it is exactly an even-money bet that both will be blue-eyed. What is your best guess of the total number of Jones sisters?
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If there are \(n\) sisters and \(k\) of these sisters have blue eyes, then the probability of the two sisters having blue eyes is:
$$\frac{\left(\begin{array}{c}k\\2\end{array}\right)}{\left(\begin{array}{c}n\\2\end{array}\right)}
\\
=\frac{k(k-1)}{n(n-1)}=\frac{1}{2}
$$
This means that:
$$2k(k-1)=n(n-1)$$
The smallest integer solution of this is when there are 4 sisters, 3 of whom have blue eyes.
The next smallest integer solution is when there are 21 sisters, 15 of whom have blue eyes. There are unlikely to be as many as 21 sisters, so 4 sisters are the most likely.
Extension
What is the next integer solution after 21 sisters?
Equal opportunity
Can two (six-sided) dice be weighted so that the probability of each of the numbers 2, 3, ..., 12 is the same?
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Let \(p_1\), \(p_2\), ..., \(p_6\) be the probabilities of getting 1 to 6 on one die and \(q_1\), ..., \(q_6\) on the other. The probability of getting a total of 2 is \(p_1q_1\) and the probabilty of getting a total of 12 is \(p_6q_6\). Therefore \(p_1q_1=p_6q_6\).
If \(p_1\geq p_6\) then \(q_1\leq q_6\) (and vice-versa) as otherwise the above equality could not hole. Therefore:
$$(p_1-p_6)(q_1-q_6)\leq 0$$
$$p_1q_1-p_6q_1-p_1q_6+p_6q_6\leq 0$$
$$p_1q_1+q_6p_6\leq p_1q_6+p_6q_1$$
The probability of rolling a total of 7 is \(p_1q_6+p_2q_5+...+p_6q_1\). This is larger than \(p_1q_6+p_6q_1\), which is larger than (or equal to) \(p_1q_1+q_6p_6\), which is larger than \(p_1q_1\).
Therefore the probability of rolling a 7 is larger than the probability of rolling a two, so it is not possible.
Extension
Can two \(n\)-sided dice be weighted so that the probability of each of the numbers 2, 3, ..., 2\(n\) is the same?
Can a \(n\)-sided die and a \(m\)-sided die be weighted so that the probability of each of the numbers 2, 3, ..., \(n+m\) is the same?
![](javascript:showlimage("straws_1.png"))