Geometric Distributions

[ImbaFeaR]

Two players A and B, toss a fair coin. A, who starts the game. stakes a penny each time he throws the coin; B also stakes a penny at each of his throws. The first player to throw heads wins and gathers all the stakes. Find the probability that A wins the game. Explain why A, although he wins more frequently, loses money. Calculate his expected loss on 100 games
Hint: for |z| < 1, 1 + 2z + 3z^2 +4z^3 + ......= (1-z)^-2
I am able to get the probability of A winning is 2/3 by using sum to infinity formula. But now i am stucked at the remaining part of the question. Help please. Oh and do note that 100 games is not 100 consecutive throws. A game only ends when a player throws head. And then they start all over again, for 100 times

 

[DrPhil]

Two players A and B, toss a fair coin. A, who starts the game. stakes a penny each time he throws the coin; B also stakes a penny at each of his throws. The first player to throw heads wins and gathers all the stakes. Find the probability that A wins the game. Explain why A, although he wins more frequently, loses money. Calculate his expected loss on 100 games
Hint: for |z| < 1, 1 + 2z + 3z^2 +4z^3 + ......= (1-z)^-2
I am able to get the probability of A winning is 2/3 by using sum to infinity formula. But now i am stucked at the remaining part of the question. Help please. Oh and do note that 100 games is not 100 consecutive throws. A game only ends when a player throws head. And then they start all over again, for 100 times

A wins, the combined number of throws N is odd, if B wins N is even. You have correctly found
$\displaystyle P(A) = 2/3, P(B)=1/3$.

The number of pennies is N, but A always puts in the 1st penny.
For odd N, A wins (N-1)/2 pennies, and for even N, A loses N/2 pennies.
To find the expectation value for A's "winnings," you have to sum over all possible outcomes, $\displaystyle \sum[P(N)\times gain(N)]$.

$\displaystyle E[A] = (1/2)(0) - (1/4)(1) + (1/8)(1) - (1/16)(2) + (1/32)(2)- (1/64)(3) + (1/128)(3) + \cdot \cdot \cdot$

........$\displaystyle = -(1/8)(1) - (1/32)(2) - (1/128)(3) + \cdot \cdot \cdot$

........$\displaystyle \displaystyle = -\dfrac{1}{2}\sum_{n=1}^\infty \dfrac{n}{4^n}$

ok from there?

 

[soroban]

Hello, ImbaFeaR!

Two players A and B, toss a fair coin.
A, who starts the game, stakes a dollar each time he throws the coin;
B also stakes a dollar at each of his throws.
The first player to throw heads wins and gathers all the stakes.

(a) Find the probability that A wins the game.
Explain why A, although he wins more frequently, loses money. . ??

(b) Calculate his expected loss on 100 games.

I am able to get the probability of A winning is 2/3 by using sum to infinity formula. . I agree!

I listed a few of the outcomes . . .

$\displaystyle A$ wins on his first throw:-$\displaystyle P(A) = \frac{1}{2}$
He wins $1, but he has staked $1.
. . His gain is $0.

$\displaystyle A$ win on his second throw: .$\displaystyle P(ABA) = \left(\frac{1}{2}\right)^3$
He wins $3, but he has staked $2.
. . His gain is $1.

$\displaystyle A$ wins on his third throw: .$\displaystyle P(ABABA) = \left(\frac{1}{2}\right)^5$
He wins $5, but he has staked $3.
. . His gain is $2.

$\displaystyle A$ wins on his fourth throw: .$\displaystyle P(ABABABA) = \left(\frac{1}{2}\right)^7$
He wins $7, but he has staked $4.
. . His gain is $3.

Hence: .$\displaystyle E \;=\;0\left(\frac{1}{2}\right) + 1\left(\frac{1}{2}\right)^3 + 2\left(\frac{1}{2}\right)^5 + 3\left(\frac{1}{2}\right)^7 + \cdots$


$\displaystyle \begin{array}{ccccccc}\text{We have:} & E &=& 1\left(\frac{1}{2}\right)^3 + 2\left(\frac{1}{2}\right)^5 + 3\left(\frac{1}{2}\right)^7 + \cdots \\ \text{Multiply by }\left(\frac{1}{2}\right)^2 & \frac{1}{4}E &=& \qquad\quad\;\; 1\left(\frac{1}{2}\right)^5 + 2\left(\frac{1}{2}\right)^7 + \cdots \\ \text{Subtract: } & \frac{3}{4}E &=& \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^5 + \left(\frac{1}{2}\right)^7 + \cdots \end{array}$

The right side is a geometric series: first term $\displaystyle a =\frac{1}{8}$, common ratio $\displaystyle r = \frac{1}{4}$
. . Its sum is $\displaystyle \dfrac{\frac{1}{8}}{1-\frac{1}{4}} \,=\,\frac{1}{6}$

Hence: .$\displaystyle \frac{3}{4}E \:=\:\frac{1}{6} \quad\Rightarrow\quad E \:=\:\frac{4}{3}\!\cdot\!\frac{1}{6} \quad\Rightarrow\quad E \:=\:\frac{2}{9}$

$\displaystyle A\text{ can expect to win an average of }\,\$\dfrac{2}{9} \:\approx\:22.2\rlap{/}c\,\text{ per game.}$

 

[DrPhil]

$\displaystyle A\text{ can expect to win an average of }\,\$\dfrac{2}{9} \:\approx\:22.2\rlap{/}c\,\text{ per game.}$

This answer is wrong because it does not include the amounts A loses when B wins. On average, A loses twice as much as he wins, for an average loss of $\displaystyle \frac{2}{9}[/tex] pennies on each game.$\(\displaystyle

[I sent a message to Soroban concerning this question, but received no reply.]\)