0
$\begingroup$

Given that there 6 black cards and 1 red card, person A is the first person to draw the card while person B is the second person.The first person to draw a red card wins.Note:The cards are shuffled after being replaced.

If the cards are drawn with replacement, is it possible to work out the probability of person B wining.If so, what is the probability.

What i have tried:

$\lim_{n\to\infty} (\frac {1}{7})^{n}$

This value should diverge right? As the denominator is multiplied by 7 for every power.

However the answer is $\frac {6}{13} $

Can someone explain how to get this answer and where did my understanding go wrong?

4 Answers 4

0

The chance that person A wins on any given draw is $\frac{1}{7}$, and it is the same for person B, and so the chance either A or B loses on one of their draws is $\frac{6}{7}$. Person A can win on the first draw, the third draw... any odd draw. What would the probability be for A to win after 2n-1 draws? (It isn't just $\frac{1}{7^{n}}$). Think about what B does at the even draws too!

Once you've answered this, the chance A wins should be the chance they win on the 1st OR the 3rd OR the 5th OR ... 2n+1 ...

3

With replacement:

A wins on the first turn $\frac 17$

B wins on the second turn (this depends on the game not being over already) $\frac 67\frac 17$

A wins on the 3rd turn $(\frac 67)^2\frac 17$

B wins on the 4th turn $(\frac 67)^3\frac 17$

Any player wins on the n-th turn $(\frac 67)^{n-1}\frac 17$

Chance of A winning is the sum of all the even terms $A=\frac 17 \sum_\limits{n=0}^\infty (\frac 67)^{2n}$

$\frac 17\frac{1}{1-(\frac 67)^2}$

A cute way around simplifying this last fraction (though it is not that hard.)

Chance of B winning: $B=\frac 17 \sum_\limits{n=0}^\infty (\frac 67)^{2n+1} = \frac 67 A$

$A + \frac 67 A = 1$

3

An alternative to Doug's solution, which avoids computing an infinite sum: $$P(B)=\frac{6}{7}\left(\frac{1}{7}\right)+\left(\frac{6}{7}\right)^{2}P(B),$$ since $B$ wins on his or her first draw, and otherwise, the game "resets" if both players do not draw red on their first draw. Simplifying yields the correct answer.

  • 0
    I can't understand $(/frac {6}{7})^2$ that part.Why do have that there? The first $/frac{6}{7}$ is because the probability being dependent on A drawing the black card,the second being that B fails to draw red card...But why is $P(B)$ there? – 2017-03-01
  • 0
    The first term is the probability that player B wins on his/her first draw, which is $P(\text{A draws black})P(\text{B draws red}|\text{A draws black}).$ The second term is the probability that player B wins on a later draw, which is $P(\text{A draws black})P(\text{B draws black}|\text{A draws black})P(\text{B wins}|\text{First two draws were black}).$ Now by replacement, B winning is independent of the first two draws being black, so this last factor is just $P(B).$ – 2017-03-02
0

The probability that player A wins on the first round is $1/7$. The probability that player B wins on the next round is $6/7 \times 1/7$. The probability that player A wins on the next round is $6/7 \times 6/7 \times 1/7$. and so on ... So the probability that player A wins (on any given turn) \begin{eqnarray*} \frac{1}{7} \sum_{i=0}^{\infty} \left(\frac{6 \times 6}{7 \times 7}\right)^i \end{eqnarray*} & the probability that player B wins \begin{eqnarray*} \frac{6 \times 1}{7 \times 7} \sum_{i=0}^{\infty} \left(\frac{6 \times 6}{7 \times 7}\right)^i \end{eqnarray*} Sum the geometric series & we have the probability that A wins is $7/13$ and the probability that B wins is $6/13$.