Probability of rolling a 4 given he won Two players toss one 6-sided die each. The winner is the player with the larger result,
with a tie in case of equal tosses.
(a) What is the probability of player 1 winning if he tosses a 4?
(b) What is the probability that player 1 threw a 4 if he won?
(a) was pretty straight forward since player 2 must roll a 1, 2 or 3 in order for player 1 to win. Which gives use the probability that player 1 wins is 1/2.
(b) I decided to use the formula $$P(A|B) = \frac{P(B|A)\cdot P(B)}{P(A)}$$ which translates to $$P(\text{rolling a 4}|\text{you won}) = \frac{P(\text{you win}|\text{rolling a 4})\cdot P(\text{you win)}}{P(\text{rolling a 4})}.$$ When I plug in; $$P(\text{rolling a 4}|\text{you won}) = \frac{(1/2) \cdot (1/2)}{1/6} = 1.5.$$ Which is not possible. What am I doing wrong?
 A: For part (b): the error is in the formula itself.
The correct formula is:
$$P(A | B) = \frac{P(A \cap B) }{P(B)} = \frac{P(B | A) \cdot P(A)}{P(B)}$$
so your $P(A)$ and $P(B)$ terms are flipped.
However, note that the probability that you win is not $1/2$, because of ties. Ties will occur $1/6$ of the time, so someone will win the other $5/6$ of the time, which means that you will win half of those (i.e. $5/12$ of the time).
A: You need to use Bayes' Theorem here. Here's a good, intuitive way to look at it.
We can break down the event "Player 1 won" into cases:
P(Player 1 rolled a $1$ and won)=$\frac16\times 0 = 0$
P(Player 1 rolled a $2$ and won)=$\frac16\times \frac16 = \frac1{36}$
P(Player 1 rolled a $3$ and won)=$\frac16\times \frac26 = \frac2{36}$
P(Player 1 rolled a $4$ and won)=$\frac16\times \frac36 = \frac3{36}$
P(Player 1 rolled a $5$ and won)=$\frac16\times \frac46 = \frac4{36}$
P(Player 1 rolled a $6$ and won)=$\frac16\times \frac56 = \frac5{36}$
That gives a total probability of $\frac{15}{36}$ that player 1 won. Of that $\frac{15}{36}$, only $\frac3{36}$ comes from the case where player 1 rolled a $4$. Thus the probability that player 1 rolled a $4$, given that player 1 won, is $$\frac{\left(\frac3{36}\right)}{\left(\frac{15}{36}\right)}=\frac3{15}=\frac15$$
Does that all make sense?
