The language is bit tough to understand.

D

defeated_soldier

Junior Member
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Apr 15, 2006
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i was reading a math problem whcih involves biased coin tossing
.
where it says ,

"The person who tosses a tail first wins the game...."

These words i dont understand....does it wants to say , if the person tosses and if he gets tail as output he wins the match ?

The language is bit tough to understand.

can you please simplify ?
 
though its not important but in case you need , so i am copying the full problem below...

A and B are playing a game of tosses. they used a biased coin in which the P(head)=1/3 and P(tail) is 2/3. the person who tosses a tail first wins the game. they take turns in tossing the coin.
 
Hello, defeated_soldier!

A and B are playing a game of tosses. they used a biased coin in which the P(head)=1/3 and P(tail) is 2/3.
The person who tosses a tail first wins the game.
They take turns in tossing the coin.
The person who tosses a tail first wins the game.

\(\displaystyle \;\;\)Does it say: if the person tosses and gets tail as output, he wins the match? . . . yes
Click to expand...
I'll bet that the question is: What is the probability that the first player wins the game?

A tree diagram is the neatest approach, but it's hard to type one here.
I'll talk my way through it . . .

Suppose \(\displaystyle A\) goes first.

\(\displaystyle A\) could win on the first toss with probability \(\displaystyle \frac{2}{3}\)

\(\displaystyle A\) could win on his second toss. \(\displaystyle \;\)How can this happen?
\(\displaystyle \;\;A\) gets a Head on his first toss . . . prob \(\displaystyle \frac{1}{3}\)
\(\displaystyle \;\;B\) gets a Head on his first toss . . . prob \(\displaystyle \frac{1}{3}\)
\(\displaystyle \;\;A\) gets a Tail on his second toss . . . prob \(\displaystyle \frac{2}{3}\)
Hence: \(\displaystyle \,P(H,\text{2nd toss})\:=\:\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}\:=\:\frac{2}{3}\left(\frac{1}{9}\right)\)

\(\displaystyle A\) could win on his third toss. \(\displaystyle \;\)How can this happen?
\(\displaystyle \;\;A\) gets a Head on his first toss . . . prob \(\displaystyle \frac{1}{3}\)
\(\displaystyle \;\;B\) gets a Head on his first toss . . . prob \(\displaystyle \frac{1}{3}\)
\(\displaystyle \;\;A\) gets a Head on his second toss . . . prob \(\displaystyle \frac{1}{3}\)
\(\displaystyle \;\;B\) gets a Head on his second toss . . . prob \(\displaystyle \frac{1}{3}\)
\(\displaystyle \;\;A\) getes a Tail on his third toss . . . prob \(\displaystyle \frac{2}{3}\)
Hence" \(\displaystyle \,P(H,\text{ 3rd toss})\:=\:\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{1}{3}\cdot\frac{2}{3}\:=\:\frac{2}{3}\left(\frac{1}{9}\right)^2\)

. . . and so on.


\(\displaystyle A\)'s probability of winning is the sum of these probabilities.

\(\displaystyle \;\;\;P(A)\:=\:\frac{2}{3}\,+\,\frac{2}{3}\left(\frac{1}{9}\right)\,+\,\frac{2}{3}\left(\frac{1}{9}\right)^2\,+\,\frac{2}{3}\left(\frac{1}{9}\right)^3\,+\,\cdots\)

This is a geometric series with first term \(\displaystyle a\,=\,\frac{2}{3}\) and common ratio \(\displaystyle r\,=\,\frac{1}{9}\)

The sum is: \(\displaystyle \,P(A)\;=\;\L\frac{\frac{2}{3}}{1\,-\,\frac{1}{9}}\;=\;\frac{3}{4}\)
 
thanks for the clarification.

The question is :;

If A wins the game in the 3rd round,whats the probability that he was the first person to toss the coin ?
 
Use ‘a’ to mean player a tosses a head likewise for ‘b’ and ‘A’ for tail.
Let W stand for A wins on third toss. AF stands for A tosses first.
A wins on his third tosses in one of two ways: ababA or bababA.
\(\displaystyle \L
\begin{array}{r}
P(ababA) & = & \frac{2}{{3^5 }} \\
P(bababA) & = & \frac{2}{{3^6 }} \\
P(AF|W) & = & \frac{{\frac{2}{{3^5 }}}}{{\frac{2}{{3^5 }} + \frac{2}{{3^6 }}}} \\
\end{array}\)
 
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