0
$\begingroup$

P and Q are playing a game in which they toss a coin alternately. The probability that the coin shows a head is 0.2. The game continues until the coin shows head and the person on whose turn a head shows wins the game. Find the probability that P wins the game.

P(head)= 0.2 P(tail)=0.8

In the solution it has been given Since Q starts the game first , the probability (P) is given by

P=(0.8)(0.2)+(0.8)(0.8)(0.8)(0.2)+.......till infinity

Two questions:

How Q started the game first? How this series is formed?

$\endgroup$
  • 3
    $\begingroup$ Whoever starts is an assumption, it can't be determined mathematically. $\endgroup$ – lulu Sep 11 '17 at 16:52
  • $\begingroup$ Who starts should be part of the question. The series is more likely top be $0.8^1 \times 0.2 +0.8^3 \times 0.2 +0.8^5 \times 0.2 +\cdots$ $\endgroup$ – Henry Sep 11 '17 at 16:52
  • $\begingroup$ How? Please explain @Henry $\endgroup$ – Sakuzi Markel Sep 11 '17 at 16:55
  • 1
    $\begingroup$ The second player wins if (a) the first player fails then the second player succeeds, or (b) the first player fails then the second player fails then the first player fails then the second player succeeds, or (c) the first player fails then the second player fails then the first player fails then the second player fails then the first player fails then the second player succeeds, or (d) ... $\endgroup$ – Henry Sep 11 '17 at 17:00
  • $\begingroup$ Given only the information in the first paragraph, you can't determine who went first. Are you certain you copied down the question correctly? $\endgroup$ – Kevin Long Sep 11 '17 at 17:00
1
$\begingroup$

We can actually avoid calculating an infinite sum. Let $p$ be the probability that the first player wins. The probability that they win on the first flip is $0.2$; otherwise, they can only win if the next two flips are tails. This has a probability of $0.8^2$, and in that case the game reverts to its original state. Therefore,

$$p = 0.2 + 0.8^2p,$$

yielding $p = \frac59$. So $P$ has either a $\frac59$ or $\frac49$ chance of winning, depending on who starts.


Here is a diagram that might help:

a possibly helpful diagram

So if we start in the $P$ square, there's a $0.2$ chance of winning immediately. There's also a $0.8^2$ chance of moving to $Q$ then right back to $P$, in which case we are back where we started (and where the chance of winning is therefore the same as before). So the equation I gave above can be thought of as:

$$\underbrace{\text{chance of} \\ \text{eventually winning}}_p = \underbrace{\text{chance of} \\ \text{ immediate win}}_{0.2} + \underbrace{\text{chance of returning} \\ \text{ to starting position}}_{0.8^2} \times \underbrace{\text{chance of} \\ \text{ eventually winning}}_p$$

$\endgroup$
  • $\begingroup$ can you please elaborate more. ? $\endgroup$ – Sakuzi Markel Sep 11 '17 at 17:56
  • $\begingroup$ @SakuziMarkel I added a diagram and some further explanation. Does that help? $\endgroup$ – Théophile Sep 11 '17 at 19:11
1
$\begingroup$

If so inclined, @Theophile can explain the elegant method (+1) of that Answer in detail.

Meanwhile, here is a method using series that gives the same result, and may be easier to understand from first principles.

Let P1 denote the first player. If she is to win, it must be on an odd-numbered trial, so

$$P(\text{P1 Wins}) = P(\text{P1 Wins on 1st}) + P(\text{P1 Wins on 3rd}) + P(\text{P1 Wins on 5th}) + \cdots\\ = .2 + (.8)^2(.2) + (.8)^4(.2) + \cdots = .2[.64^0 + .64 + .64^2 + \cdots] = 5/9.$$

Similarly, let P2 denote the player who goes second. If he is to win, it must be on an even numbered trial, so

$$P(\text{P2 Wins}) = P(\text{P2 Wins on 2nd}) + P(\text{P2 Wins on 4th}) + P(\text{P2 Wins on 6th}) + \cdots\\ = (.8)(.2) + (.8)^3(.2) + (.8)^5(.2) + \cdots = .16[.64^0 + .64 + .64^2 + \cdots] = 4/9.$$

Notice that there is some advantage to being the starting player.


Note: If you don't know about summing geometric series, here is an intuitive method:

$S = \sum_{i=0}^\infty (.64)^i = 1 + .64 + .64^2 + .64^3 + \cdots.\;$ Then $.64S = .64 + .64^2 + .64^3 + \cdots.$

So $S - .64S = .36S = 1$ and $S = 1/.36.$

Also, using software to sum the first 101 terms of this rapidly converging series gives very nearly the correct answer. The following is from R statistical software:

.2*sum(.64^(0:100)); 5/9  # '0:100' is vector of integers 0 thru 100
## 0.5555556     # sum of 101 terms of series
## 0.5555556     # 5/9 to 7 places 
.16*sum(.64^(0:100)); 4/9
## 0.4444444     # sum 101 terms   
## 0.4444444     # 4/9
$\endgroup$
0
$\begingroup$

For P to win, P has to get heads first. If Q starts the game, in order for P to win, Q must get a tails every time Q tosses the coin. Given this, P can win the first time they toss the coin, the second time they toss the coin, the third time they toss the coin and so on. But No matter when P gets a head, Q has to get tails every time before P.

Let's say P wins the first time they toss the coin. But since Q starts the game, Q must get a tails first. For both to happen, the probability is (0.8)(0.2).

Let's say P wins in the second chance they get to toss the coin. But in order for that to happen, Q should get a tails in both the chances they toss the coin and P should get a tails the first time and heads the second time. The probability of that happening is (0.8)(0.8)(0.8)(0.2).

This pattern progresses until such a time that P gets a head. Therefore, the sum of all these probabilities gives us the total probability of P winning the competition.

P(P) = (0.8)(0.2) + (0.8)(0.8)(0.8)(0.2) + ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.