A game works as follows: each player rolls one dice (let's call it ''n''). After both...

A game works as follows: each player rolls one dice (let's call it ''n''). After both players have made the first roll, they have the choice to leave or not. If a player exits, he automatically looses; if not, he rolls a second time (let's call this second roll "m'') and receives n^m points . The player with the most points wins, in case of a tie no one wins.
a) Give the fundamental space of the two throws.
b) Give the fundamental space of the number of points obtained.
c) Show that a player should automatically withdraw if n = 1.
e) Let n₁ and n₂ be the value of n of the first and second player (respectively). Knowing that n₁ = 2 and n₂ = 3, what is the probability that the first player wins?
d) Knowing that the first player will pull out automatically if n₁ = 1, half the time if n₁ < n₂ and never otherwise, what is the probability that the first player will win?
probability of the first player pulling out?

There is 2 players in the game.

Solutions

Expert Solution

a) The fundamental space of 2 throws={(1,1), (1,2),.....(1,6), (2,1),(2,2),......(2,6),(3,1),(3,2),........(3,6),(4,1),(4,2),....(4,6),(5,1),(5,2),.........(5,6),(6,1),...........(6,6)}

since the possible outcome of one throw=1,2,3,4,5,6

b)space of points={1,2,3,4,5,6,8,9,10,12,15,16,18,20,24,25,30,36} so number of points=18

since n=1,2,3,4,5,6 m=1,2,3,4,5,6

Let the first player gets 1 in first throw.

So, n1=1, n2>=1, m1>=1, m2>=1

Then probability that first player does not loose=P(n1m1>=n2m2)= <<0.5

since n1m1=1,2,3,4,5,6 n2m2=1,2,3,4,5,6,8,9,10,,...........36

since the probability of not loosing of 1st player is very very small that means the probability of loosing is very high. So he should withdraw.

e)probability that first player wins= P(n1m1>n2m2)=P(2m1>3m2)= since m1=1,2,3,4,5,6 and m2=1,2,3,4,5,6

Since we are instructed to answer the first 4 parts I have given the same. For the last part, I am giving hint. Please follow that and compute.

d) Let W= event that 1st player wins, A=event that n1=1, B= event that n1<n2 and n1>=2 and C =the event that n1>=n2 and n1>=2

Then P(W)=P(W|A)P(A) +P(W|B)P(B) +P(W|C)P(C) compute them part by part

here P(W|A)=0 (think the reason, it is very easy)

Probability that 1st player pulls out=P(n1=1)=1/6 since there are 6 possible outcome in a single throw.

Thank you!

orchestra answered 1 year ago

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