I get
die - coin - coin - coin = 1/3 * 1/2 *1/2 = 1/12
die - die - die - coin = 2/3 * 2/3 * 1/2 = 4/18 = 2/9
so
1/12 + 2/9 = 3/36 + 8/36 = 11/36
Markov plays a game for three turns. On each turn, he either rolls a fair, six sided die or flips a fair coin. If he rolls a 1 or 2 on the die, he will switch to the coin on the next turn, and if he flips a tails on the coin, he will switch to the die on the next turn. If Markov starts by rolling the die, what is the probability that he will flip the coin on the third turn?
What I did
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On the first turn, there is 1/3 chance of going to the coin and 2/3 chance of going to the die.
On the second turn it is the same probability for the die.
The coin has 1/2 and 1/2 for going to the die and staying with the coin.
So is it 1/6 because 1/2*1/3=1/6?
I don't think this is correct, so I need someone to confirm. If this is correct, please type so. If it isn't, please put your solution.
2/9 is not correct.
4 answers
oh
+
die - coin -die - coin = 1/3* 1/2 *1/3 = 1/18
11/36 + 1/18 = 11/36 + 2/36 = 13/36
+
die - coin -die - coin = 1/3* 1/2 *1/3 = 1/18
11/36 + 1/18 = 11/36 + 2/36 = 13/36
Let
D be the outcome of a die at the end of a turn
C be the outcome of a coin at the end of a turn
Make a tree diagram
we start with D
we have either a D or a C at the end of the first toss, two branches
from the D, we have a D or C at the end of the 2nd toss (the beginning of the third toss)
we are now looking at the end of 4 branches
D:DD ---> (2/3)(2/3) = 4/9
D:DC ---> (2/3)(1/3) = 2/9
D:CD ---> (1/3)(1/2) = 1/6
D:CC ---> (1/3)(1/2) = 1/6 , notice that 4/9+2/9+1/6+/6 = 1
we need the cases that end in C
prob = 2/9 +1/6 = 7/18
D be the outcome of a die at the end of a turn
C be the outcome of a coin at the end of a turn
Make a tree diagram
we start with D
we have either a D or a C at the end of the first toss, two branches
from the D, we have a D or C at the end of the 2nd toss (the beginning of the third toss)
we are now looking at the end of 4 branches
D:DD ---> (2/3)(2/3) = 4/9
D:DC ---> (2/3)(1/3) = 2/9
D:CD ---> (1/3)(1/2) = 1/6
D:CC ---> (1/3)(1/2) = 1/6 , notice that 4/9+2/9+1/6+/6 = 1
we need the cases that end in C
prob = 2/9 +1/6 = 7/18
Damon, if we follow your interpretation, the outcomes would be
dddd
dddc --- (2/3)(2/3)(1/3)
ddcd
ddcc --- (2/3)(1/3)(1/2)
dcdd
dcdc -- (1/3)(1/2)(1/3)
dccd
dccc -- (1/3)(1/2)(1/2)
then there would be 4 outcomes that end with a coin, for a total of
4/27 + 1/9 + 1/18 + 1/12 = 43/108
I think you found the coutcomes at the 4th toss
dddd
dddc --- (2/3)(2/3)(1/3)
ddcd
ddcc --- (2/3)(1/3)(1/2)
dcdd
dcdc -- (1/3)(1/2)(1/3)
dccd
dccc -- (1/3)(1/2)(1/2)
then there would be 4 outcomes that end with a coin, for a total of
4/27 + 1/9 + 1/18 + 1/12 = 43/108
I think you found the coutcomes at the 4th toss