Two players take turns playing a game where $P(win)= 0.28$ for each time a player takes a turn. If A starts, the probability that A wins the game, …

Question

Two players take turns playing a game where P(win) $= 0.28$ for each time a player takes a turn. If A starts, the probability that A wins the game eventually is $0.54$.

A fair coin is tossed to decide who starts the game, with heads letting A start first. What is the probability that the toss was heads, if B wins?

I am assuming that a conditional probability should be used, but an having issues setting the problem up. I can find P(B wins) if A goes first, but I do not understand how to handle the coin toss.

Answer 1

Use Bayes' Theorem:

$$P\left(H|B\right)=\frac{P(B|H)P(H)}{P(B)}$$

In this case, P(B) and P(H), respectively the probability that B wins and the probability that the coin lands Heads are both .5 so they cancel out. Therefore $P(H|B)=P(B|H)$, which according to you is $1-.54 = .46$ but I haven't checked your calculation.

Answer 2

Partial answer:

Call $C$ "A wins" and $D$ "A starts". You are asked for $P(D|C^c)$. Using Bayes' rule:

$$P(D|C^c)=\frac{P(C^c|D)P(D)}{P(C^c|D)P(D)+P(C^c|D^c)P(D^c)}.$$

Now $P(D)=P(D^c)=1/2$, so we can plug that in and then simplify, reducing the problem to

$$\frac{P(C^c|D)}{P(C^c|D)+P(C^c|D^c)}.$$

$P(C^c|D)=1-P(C|D)=1-0.54=0.46$, so that much is given. I do not sufficiently understand the problem statement to be able to calculate $P(C^c|D^c)$. Perhaps you can do that part?

Answer 3

Let $x=$ A wins. $y=$B wins. $z=$ Heads. We have $P(x)=P(y)=P(z)=0.5$. $$\text{ Now } 0.5=P(z)=P(x\cap z)+P(y\cap z)$$ and $$0.54=P(x|z)=P(x\cap z)/P(z)=P(x \cap z)/0.5=2 P(x \cap z)$$ so $$P(x\cap z)=0.27.$$ Therefore $$P(z\cap y)=P(y\cap z)=P(z)-P(x \cap z)=0.5-0.27=0.23.$$ Therefore $$ P(z|y)=P(z \cap y)/P(y)=0.23/0.5=0.46.$$