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I am not sure if I got the correct answers to these basic probability questions.

You and I play a die rolling game, with a fair die. The die is equally likely to land on any of its $6$ faces. We take turns rolling the die, as follows.

At each round, the player rolling the die wins (and the game stops) if (s)he rolls either "$3$", "$4$","$5$", or "$6$". Otherwise, the next player has a chance to roll.

  1. Suppose I roll 1st. What is the probability that I win in the 1st round?
  2. Suppose I roll in the 1st round, and we play the game for at most three rounds. What is the probability that
    a. I win?
    b. You win?
    c. No one wins? (i.e., I roll "$1$" or "$2$" in the 3rd round?)

My answers:

  1. $4/6$
  2. a. I think it is $(4/6)^2$ . Can someone explain why it isn't $4/6 + 4/6$ ?
    b. I think this is $4/6$.
    c. I think it is $(2/6)^3$
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  • $\begingroup$ For (1a), your answer is right. For (2a), the correct calculation is $$(4/6) + \left((2/6)(2/6)(4/6)\right)$$. Do you see why? $\endgroup$ – quasi Mar 10 '17 at 9:50
  • $\begingroup$ @quasi: shouldn't that be $\frac{2}{6}\frac{2}{6}$: you have to exclude that you win in the first round in your second case. $\endgroup$ – Student Mar 10 '17 at 9:52
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    $\begingroup$ @Student: Yes, I'll edit my comment -- thanks. $\endgroup$ – quasi Mar 10 '17 at 9:52
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What is the probability that the player who rolls first wins in the first round?

Your answer of $4/6 = 2/3$ is correct.

What is the probability that the player who rolls first wins the game if the game lasts at most three rounds?

The player who rolls first if he or she wins during the first round, second round, or third round. We know that the probability that the player who rolls first has probability $2/3$ of winning in the first round.

For the player who rolls first to win in the second round, both players must roll a 1 or 2 in the first round, then the first player must roll a 3, 4, 5, or 6 in the second round. The probability that this event occurs is $$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{27}$$

For the first player to win in the third round, both players must roll a 1 or 2 for the first two rounds, then the first player must roll a 3, 4, 5, or 6 in the third round. The probability that this occurs is $$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{243}$$

Since these three events are mutually exclusive, the probability that the player who rolls first wins the game is $$\frac{2}{3} + \frac{2}{27} + \frac{2}{243} = \frac{162}{243} + \frac{18}{243} + \frac{2}{243} = \frac{182}{243}$$

The probability that the first player wins cannot be $$\frac{4}{6} + \frac{4}{6} = \frac{8}{6} = \frac{4}{3} > 1$$ since it is impossible for a probability to exceed $1$.

What is the probability that the player who rolls second wins the game if the game lasts at most three rounds.

The player who rolls second can win in the first round if the first player rolls a 1 or 2, then the second player rolls a 3, 4, 5, or 6. The probability that this event occurs is $$\frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{9}$$ For the second player to win in the second round, the first and second players must both roll a 1 or 2 in the first round, the first player must roll a 1 or 2 in the second round, then the second player must roll a 3, 4, 5, or 6 in the second round. The probability that this event occurs is $$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{81}$$ For the second player to win in the third round, both players must roll a 1 or 2 in the first two rounds, the first player must roll a 1 or 2 in the third round, then the second player must roll a 3, 4, 5, or 6 in the third round. The probability that this event occurs is $$\frac{2}{6} \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \cdot \frac{2}{6} \cdot \frac{2}{6} \cdot \frac{4}{6} = \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{2}{729}$$ Since these three events are mutually exclusive, the probability that the second player wins is $$\frac{2}{9} + \frac{2}{81} + \frac{2}{729} = \frac{162}{729} + \frac{18}{729} + \frac{2}{729} = \frac{182}{729}$$

What is the probability that neither player wins a game that lasts at most three rounds?

For this event to occur, both players must roll a 1 or 2 in all three rounds. The probability that this event occurs is $$\left(\frac{2}{6}\right)^6 = \left(\frac{1}{3}\right)^6 = \frac{1}{729}$$

Check: There are three possibilities: The first player wins, the second player wins, or neither player wins. Therefore, the probabilities of these three events should add up to $1$. Observe that $$\frac{182}{243} + \frac{182}{729} + \frac{1}{729} = \frac{546}{729} + \frac{182}{729} + \frac{1}{729} = 1$$

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$\color{red}{\text{REMARK: I wrote this answer supposing that one round is one person roling because }}$ $\color{red}{\text{of the formulation of your second question. If this is not the case, look at N.F. Taussig's}}$ $\color{red}{\text{answer.}}$

First of all: you answered the first question correct!

For your second question: let me denote the persons by person $A$ and person $B$. Suppose person $A$ goes first, then question 2a becomes: 'suppose person $A$ rolls in the first round and at most 3 rounds are played. What is the chance of $A$ winning?'.

Well: either person $A$ wins in the first round, which is a chance of $\frac{4}{6}$. If not, then person $A$ has to win in the third round. Note that this implies you do not win in the first round and that person $B$ does not win in the second round! You not winning in the first round occurs with a chance of $\frac{2}{6}$ and the same chance holds for $B$ not winning round $2$. Now you still have to win in the third round, which occurs (again) with a chance $\frac{4}{6}$.

Conclusion: $A$ wins if $A$ wins round $1$ (game stops) or $A$ does not win round $1$, $B$ does not win round $2$ and $A$ wins round $3$. This gives us the following chance: $$\frac{4}{6} + \frac{2}{6}\frac{2}{6}\frac{4}{6}.$$

Can you apply this reasoning to find the answer to 2b?

To end with some good news: also 2c is correct ;)

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  • $\begingroup$ I think you are misinterpreting the word round. A full round is played when each player has a chance to roll. Therefore, neither player wins if each rolls a 1 or 2 in three successive rounds. $\endgroup$ – N. F. Taussig Mar 10 '17 at 10:27
  • $\begingroup$ @N.F.Taussig: The formulation is not really clear... I mean, the second question states ''suppose I roll in the first round' and 'no one wins (i.e. I role $1$ or $2$ in the third round)'... This suggest that a 'round' is one player roling (well, it does to me). But now you pointed this out, you are probably right... $\endgroup$ – Student Mar 10 '17 at 10:31

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