What is the probability that Fra wins?

Question

Fra and Sam want to play a game. They have two classic coins Head-Tail.

They flip the coins at the same time.

If the result is $HH$, Fra wins. If the result is $HT$ (or $TH$), they flip again and result is again $HT$( or $TH$) Sam wins. In the other cases they continue.

So for example if happens $HT$ and after $HH$, Fra wins. The important for Fra is that $HH$ is an outcome.

The question is: what is the probability that Fra wins?

My work:

There is the outcome $TT$ that is the canceler of the game in the sense that is like they start again from the beginning. So for finishing the game the possible outcomes are:

$HHXX,HTHH,HTHT,HTTH,THHH,THHT,THTH$ where $XX \in \{HH, HT, TH, TT\}$ so Fra wins in $6$ cases on $10$. So the probability is $\frac{3}{5}.$

What do you think about it? Thanks and sorry for my bad English.

Answer 1

Your result is correct, it can be obtained in a cleaner way as follows. Let $p$ be the probability of Fra to win. Fra wins in the following cases:

1) First throw is HH (prob. $1/4$).

2) First throw is TT (prob. $1/4$) and then Fra wins (prob. $p$).

3) First throw is TH or HT (prob. $1/2$) and second throw is HH (prob. $1/4$).

4) First throw is TH or HT (prob. $1/2$), second throw is TT (prob. $1/4$) and then Fra wins (prob. $p$).

Hence: $$ p={1\over4}+{1\over4}p+{1\over8}+{1\over8}p, \quad\text{that is:}\quad p={3\over5}. $$

Answer 2

I did it in a slightly different way and got a result that seems to follow yours.

I first drew a tree (sadly, I don't know any way to draw it with mathjax), by considering TT as a draw (as you told, the game virtually start over from the beginning).

I found this probabilities:

$$\begin{array}{lcr} \text{Throw} & \text{Probability} & \text{Winner} \\ \hline HH & \frac{1}{4} & Fra \\ HTHH & \frac{1}{16} & Fra \\ HTHT & \frac{1}{16} & Sam \\ HTTH & \frac{1}{16} & Sam \\ HTTT & \frac{1}{16} & Draw \\ THHH & \frac{1}{16} & Fra \\ THHT & \frac{1}{16} & Sam \\ THTH & \frac{1}{16} & Sam \\ THTT & \frac{1}{16} & Draw \\ TT & \frac{1}{4} & Draw \end{array} $$ So, $$p(Fra) = \frac{1}{4} + 2\times \frac{1}{16} = \frac{3}{8} $$ $$p(Sam) = 4\times \frac{1}{16} = \frac{1}{4} $$ $$p(Draw) = \frac{1}{4} + 2\times \frac{1}{16} = \frac{3}{8} $$

Now, if we decide not taking care of the throws leading to draws, we get:

$$p(Fra)_{without\_draws} = \frac{p(Fra)}{(1-p(Draw))} = \frac{\frac{3}{8}}{\frac{5}{8}} = \frac {3}{5}$$

Answer 3

It is a Markov chain with $4$ states : $X$ is the start of the game, $HH$ is Fra winning the game, $HT/TH$ is when Sam scores one point, $L$ is when Fra loses the game.

Starting at $X$, we have a probability of $\frac{1}{4}$ that we reach state $HH$, a probability of $\frac{1}{4}$ that we start again ( outcome is $TT$) and a probability of $\frac{1}{2}$ that Sam scores one point. That is

$$p_X=\frac{1}{4}p_{HH}+\frac{1}{2}p_{HT}+\frac{1}{4}p_{X}$$

At states $HT$, we have a probability of $\frac{1}{4}$ that we reach state $HH$, a probability of $\frac{1}{4}$ that we start again ( outcome is $TT$) and a probability of $\frac{1}{2}$ that Sam scores another point. $$p_{HT}=\frac{1}{4}p_{HH}+\frac{1}{2}p_{L}+\frac{1}{4}p_{X}$$

where $p_{XYZ}$ is the probability that FRa wins when he is at state $XYZ$. Obiviously, we have $p_{HH}=1$ and $p_L=0$. Solving the two equations, we have $$p_X=\frac{3}{5}$$

Answer 4

ns=10000

For t=1 to ns

[alfa]

x=rnd(1)

if x<0.25 then f=f+1:goto[beta]

if x>0.75 then goto[alfa]

y=rnd(1)

if y<0.25 then f=f+1:goto[beta]

if y>0.75 then goto[alfa]

[beta]

next t

print f/ns

Simulation with Just Basic; ns is number of simulations, try other values.