In: Statistics and Probability
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 nm 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.
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
c)
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!