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Alice and Bob are playing a game where in each round, each of them rolls a fair die independently. If both roll the same number, then the game is repeated. Otherwise the player with the larger number wins. Let X be the number of rounds until the game is decided.
(a) Determine the probability mass function of X.
(b) Compute E[X].
(c) Compute Probability[ Alice win].
(d) Assume that you get paid 10USD for winning in the first round, 1USD for winning in any other round, and nothing otherwise. Compute your expected winnings.



Answer :

(a) The probability mass function of X will be P(X=n) = 5/(6n)

(b) E[X] =1.2

(c) Probability [ Alice win] will be =0.5

(d) The value of expected winnings will be 4.25

(a) Let 'p' represent the probability of getting the same number on both the dices on a roll of pair of die

And (1-p) = probability of not getting the same number on both the dices on a roll of pair of die

Sample space for the same number appearing on both the dices on a roll of pair of die = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

As there are a total of 36 outcomes on a roll of pair of dice,

p = 6/36 = 1/6

Let the events be defined as:

Dn : Both Alice and Bob roll the same number on the nth roll

Wn : The game is finished in the nth roll (i.e. Alice and Bob roll different numbers (as the numbers are different, one will be larger))

Then,P(W) = P(W) , n = 1,2,3,4....infinity

Wn takes place when event D takes place in the first (n-1) rolls and Event W takes place in the nth roll

[tex]$\mathrm{P}(\mathrm{W})=\sum P\left(W_n\right), \mathrm{n}=1,2,3,4 \ldots .$[/tex]. infinity

And, W_n takes place when

Event D takes place in the first (n-1) rolls and event W takes place in the [tex]$n^{\text {th }}$[/tex] roll

[tex]P\left(W_n\right)=P\left(D_1\right)^* P\left(D_2\right)^* P\left(D_3\right) \ldots . .{ }^* P\left(D_{n-1}\right) P\left(W_n\right)P\left(D_n\right)\\=(p)P\left(W_n\right)\\=(1-p)[/tex]

Implies

[tex]$\begin{aligned}& \left.P\left(W_n\right)=(p)^{n-1}\right)^*(1-p) \\& =\left((1 / 6)^{n-1}\right)^*(1-(1 / 6)) \\& =\left((1 / 6)^{n-1}\right)^*(5 / 6) \\& =5 /\left(6^n\right)\end{aligned}$[/tex]

(b)

[tex]$\begin{aligned}& \mathrm{E}[\mathrm{X}]=\sum n * P(X=n), \mathrm{n}=1,2,3,4 \ldots . . \text { infinity } \\& =\sum n * \frac{5}{6^n} \\& =5 * \sum n * \frac{1}{6^n}\end{aligned}$[/tex]

If [tex]$\mathrm{S}=\sum n * \frac{1}{6^n}$[/tex]

Then S is a sum of infinite aritho-geometric series of the form

[tex]$\sum_{k=1}^{\mathrm{inf}} k * r^k \text {, }$[/tex]

The sum of which is given as [tex]$S=\frac{r}{(1-r)^2}$[/tex] for 0 < r < 1

Here, r=(1 / 6)

[tex]\Rightarrow & S=\frac{\frac{1}{6}}{\left(\frac{5}{6}\right)^2} \\& =\frac{6}{25}[/tex]

E[X] = (5) x (6/25) = (6/5)= 1.2

(c) Since the dies are fair, and the outcomes are independent, the probability of each player winning is equal

Therefore,

P(Alice WIn) = P(Bob WIn) = 0.5

(d) If X represent the round in which the game ends, then the probability of us winning in that round

P(W|X=n) = 0.5*P(X=n)

= (0.5)*(5/(6n))

P(W|X=1) = (0.5)*(5/(61)) = (5/12)

P(W|X=2,3,4...) = 0.5 - P(W|X=1) = (1/2) - (5/12) = (1/12)

Leth the variable W represent the winnings

(W|X=1) = 10

(W|X=2,3,4...) = 1

E[W] = P(W|X=1)*(W|X=1) + P(W|X=2,3,4...)*(W|X=2,3,4...)

= (10 x (5/12)) + (1 x (1/12))

= (50/12) + (1/12)= (51/12)= 4.25

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