Week 2 Problems sol PDF

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Week 2 Problems sol...

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Week 2 Problems P2.1 I toss a coin 99 times. Let

be the number of times in that process that a run of 4 consecutive

tosses contains an equal number of H and T. Find

.

Solution: Either by listing all 16 possible outcomes for four tosses or by Binomial with the probability of equal numbers of H and T in 4 tosses is let

. For

be the random variable that takes the value 1 if the four tosses starting with

the -th comprise two H and two T and variable

otherwise. Then

. The random

is then the total number of occurrences f the required sequences of four

tosses. By additivity of expected value,

P2.2 Let

be the probability of obtaining exactly

Write down the value of

heads and

tails when tossing a coin

times.

.

Stirling's wonderful formula

is good approximation for large . Use that to find an approximate value for form

where

approximation of

and

for

for large

of the

are constants you should determine. How accurate is your ? What does your formula give for

?

Solution: From Binomial,

Now, by Stirling's formula

and

Dividing, we get For Binomial is

.

this approximation gives

whereas the (rounded) true value from

and so the approximation is pretty good.

The approximation for

is

an overflow in my calculator at least.

and trying to do an exact calculation causes

P2.3 Consider a game of tennis where there is no "deuce / advantage'' and the first player to score four points wins the game. Suppose player A wins each point independently with probability and B wins with probability

.

Suppose that player B won the first two points, so now B needs 2 points to win and A still needs 4 points. Find approximately the value of

such that A wins with probability

equation satisfied by this value of

. (You should find an

and then find an approximate solution by plotting the graph

or asking Wolfram Alpha or using any other tool.) Hint: imagine that instead of stopping when one player reaches four points, the players play 5 more points whatever happens. Solution: Considering the hint, suppose the players play 5 more points from the given position. Then at that stage either A will have scored the 4 points he needs or B will have scored the 2 points he needs, but both these things can not have occurred. Whichever player reaches his target is the won who would have won under the "first to four points" rule. So A wins provided he scores at least 4 out of 5 points. So his probability of winning by Binomial is

So we are looking for a solution to Wolfram Alpha provided a solution of

. Replacing the

with an

and typing it in to

.

P2.4 You and I play the following game. Hidden from you, I put a coin in my hand: with probability

it

is a 10 pence coin and otherwise it is a 20 pence coin. You now guess which coin is in my hand: you guess it is 20 pence with probability

and otherwise you guess it is a 10 pence coin.

You get to win the coin if you guess correctly and otherwise win nothing. What (in terms of

and

) is your expected gain in pence from playing this game once with me? Challenge: suppose we are going to play repeatedly and you want to maximise your gain and I wish to minimise my loss. What value of

should I choose and what value of

should you

choose? (This question is somewhat ill-defined, but it does have an interesting possible answer.) (Note: anything labelled "challenge" will not be part of the hand-in.) Solution: Drawing a tree starting with my choice of coin and then considering your choice, there are two routes that lead to your inning something: I select a 10-pence coin (probability ) and you guess it is a 10-pence coin (probability So the probability of this outcome is I select a 20-pence coin (probability So the probability of this outcome is So your expected gain from a game is

).

. ) and you guess it is a 20-pence coin (probability ). .

The challenge: So what should our strategies be? I would like to choose 10 pence to minimise my loss, but if I do it every time, you (once you catch on) will guess 10p every time and so win 10 pence every time. And similarly if you are always guessing the same, I will always choose the other coin and you will win nothing. We could both choose

. That gives you an expected gain of 7.5 pence per game. But

if I am equally likely to have either coin, you should guess "20 pence" all the time and that puts your expected gain at 15p (until I catch on and change my strategy). The magic point is clear if you rewrite the expected gain in two different ways:

So if I choose

your expected gain is

if you choose

your gain is

both settle for this!  

pence irrespective of your strategy. And similarly,

pence irrespective of my strategy. Arguably, we should...