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# Practice Paper 2 Question 9

Using a fair coin we can generate random integers in $$\{1,2,3,4\}$$ with equal probability by doing:

• $$(a)$$ toss coin, if heads go to $$(b)$$ otherwise go to $$(c)$$
• $$(b)$$ toss coin, if heads output $$1$$ otherwise output $$2$$
• $$(c)$$ toss coin, if heads output $$3$$ otherwise output $$4$$

By altering just one of the lines $$(a),$$ $$(b)$$ or $$(c),$$ we can generate random integers in $$\{1,2,3\}$$ with equal probability. Identify which line and give the new version. Prove that it is correct.

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## Warm-up Questions

1. You roll a die $$3$$ times. What is the probability of getting $$1,2$$ and $$3$$ in any order?
2. A geometric series has terms $$3, 6, 12, 24, 48, \ldots.$$ Find an expression for the $$n^{th}$$ term in the sequence, in terms of $$n.$$

## Hints

• Hint 1
Which of the $$3$$ lines must we definitely change?
• Hint 2
You are allowed to go to another step, as done in step (a).
• Hint 3
... you are also allowed to output a number for heads, and go to another step for tails.
• Hint 4
If you change $$(c)$$ to output 3 for heads and go to a another step for tails, which step should that be?
• Hint 5
Focus on the possible sequences of coin tosses that will output $$1$$ (for now). Notice any pattern?
• Hint 6
Try expressing the probability of each sequence of coin tosses in terms of the length of the sequence.
• Hint 7
Sequences are independent. What can you say about the total probability of outputting $$1?$$
• Hint 8
What about the probabilities of outputting $$2$$ and $$3?$$

## Solution

Since we are not outputting $$4,$$ we must change the second part of $$(c)$$ and we must go to another step instead of outputting $$4.$$ Going to any other step than to (a) would make the probabilities of 1, 2, 3 unequal (this is easy to verify). Hence the only viable option is to go to (a) instead of outputting 4:

$$(c)$$ toss coin, if heads output $$3$$ otherwise go to $$(a)$$

To prove the probabilities are indeed equal, let's consider first the probability $$P(1)$$ of outputting $$1,$$ which we get when we flip $$HH$$ or $$TTHH$$ or $$TTTTHH$$ and so on. The probability of flipping any particular sequence of heads or tails of length $$k$$ is $$\big(\frac{1}{2}\big)^k.$$ Therefore $$P(1)$$ is just the sum of the even powers of $$\frac{1}{2}$$ (infinite geometric series): $P(1) = \sum_{n=1}^{\infty}\frac{1}{4^n} = \frac{1/4}{1-1/4} = \frac{1}{3}.$ Probabilities for $$2$$ and $$3$$ are the same by symmetry of the coin.

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