9
$\begingroup$

Three boys play a game as follows. They put three white balls and a red ball in a box. Andy, Bruce, and Charles, in this order, each choose a ball at random from the box, without replacement. Whoever gets the red ball wins. If none of the three draws the red ball, nobody wins. Which one of the three boys has the largest probability of winning?

Since the balls aren't being replaced, I thought Charles should have the highest probability, but it seems that they all have equal chances of winning. How is this possible?

$\endgroup$
  • 1
    $\begingroup$ Do not identify yourself with Charles, but with the red ball: "what are my chances to be taken out at the third draw?.. Smaller or larger than the chances of one of my white colleagues?" $\endgroup$ – drhab Jun 26 '17 at 11:00
  • 5
    $\begingroup$ Drawing in sequence without replacing is equivalent to drawing at the same time. $\endgroup$ – MattPutnam Jun 26 '17 at 15:51
  • 1
    $\begingroup$ It might be notable that, while their probabilities are equal, there could be an implicit perceived bias- if Andy pulls the red ball and you're playing in the open, the game probably ends before Charles or Bruce draw. As such, Charles may only draw when there's a 1/2 shot of pulling the red ball (so 50% of the times he draws he wins), but he only has a 1/2 shot of drawing at all. $\endgroup$ – Delioth Jun 26 '17 at 20:54
  • $\begingroup$ If one of the answers "worked" for you then be so kind to accept it. $\endgroup$ – drhab Jun 27 '17 at 9:04
  • $\begingroup$ It depends on whether you draw blind or whether each person reveals the color he's drawn. Not really; but obviously knowledge of the remaining balls changes the odds at that point. You have to set the draw order prior to any draws to make them all equal. $\endgroup$ – Carl Witthoft Jun 27 '17 at 11:24
16
$\begingroup$

A winning: $$P(R_A)=\frac{1}{4}.$$ B winning: $$P(W_A)P(R_B|W_A)=\frac34 \cdot \frac13 = \frac14.$$ C winning: $$P(W_A)P(W_B|W_A)P(R_C|W_A\cap W_B)=\frac34 \cdot \frac23 \cdot \frac12=\frac14.$$

$\endgroup$
  • 4
    $\begingroup$ The method is okay of course, but your notation is sloppy. Usually $P(W\mid W)=1$ and $W\cap W=W$. On that part you are not setting a good example for students. $\endgroup$ – drhab Jun 27 '17 at 9:01
  • $\begingroup$ That's the way I like it (+1). $\endgroup$ – drhab Jun 28 '17 at 10:06
13
$\begingroup$

Suppose that instead there are a black, a green, a blue and a red ball in the box. Three balls are taken out one by one without replacement.

Has the red ball a smaller or larger chance to be taken out at the third draw than e.g. the green ball? No, so the probability that it will be taken out at the third draw is $\frac14$.

The same reasoning tells us that this will also be true for the first or second draw.

$\endgroup$
  • $\begingroup$ Beautiful. On paper, you made the problem more complex, but now it's immediately understood intuitively. $\endgroup$ – Alexander Engelhardt Jun 27 '17 at 10:53
  • 1
    $\begingroup$ @AlexxHardt Glad you like it. It was indeed my intention to make the OP understand intuitively. In my optics that has greater value then just being familiar with some mathematical technique. $\endgroup$ – drhab Jun 27 '17 at 10:56
9
$\begingroup$

There are 4 possibilities in which the balls can be drawn (from first to last):

R W W W, W R W W, W W R W, W W W R

Each one is equally as likely. So whether you are first or third you have a 1/4 chance of drawing the Red ball.

$\endgroup$
6
$\begingroup$

Let $X_i$ denote the event that a red ball is drawn at $i$th draw and $X_i^{C}$ denotes the event that a white ball is drawn at $i$th draw.

$P(X_1) = 1/4 = 1-P(X_1^{C})$

$P(X_2 | X_1) = 0$ and $P(X_2|X_1^{C}) = 1/3$ implies,

$$P(X_2) = P(X_2|X_1)P(X_1) + P(X_2|X_1^{C})P(X_1^{C}) = 0 + (1/3)\cdot(3/4) = 1/4$$

Similarly, $P(X_3) = P(X_3|X_1^C,X_2^C)P(X_2^C|X_1^C)P(X_1^C) = (1/2)\cdot(2/3)\cdot(3/4) = 1/4$

Note that other terms in the RHS of the first equality above will be zero (because in all those terms the conditioning will be on the red ball already drawn in either of the first or second draw). For example, $P(X_3|X_2,X_1^C)P(X_2|X_1^C)P(X_1^C) = 0 \cdot (1/3) \cdot (3/4) = 0$

So, the unconditional probabilities of winning are same.

$\endgroup$
5
$\begingroup$

This is equivalent to arranging the four balls in one of $4! = 24$ permutations, and then giving the first ball to Andy, the second to Bruce, and the third to Charles. There is no reason for the red ball to show up more often in one of the four locations than the other, since all permutations are equally likely.

To see why, note that each person chooses completely randomly, so there is no reason to favor one permutation over another.

$\endgroup$
0
$\begingroup$

P(A) = probablilty of andy winning p(B) = probability of Bruce winning P(C) = probability of charles winning

P(A) ^ P(B) ^ P(C) = 0 (mutually exclusive)

P(A)= 1C1/4 = 1/4 P(B) = 2C1/8 = 1/4 P(C)= 3C1/12 = 1/4

Hence all are equal

$\endgroup$
0
$\begingroup$

Probability of Andy Winning = Drawing one red ball out of 4 total balls and is therefore $$P_{A}=\frac{1}{4}$$

Probability of Bruce Winning = Probability of Andy NOT winning and Bruce Winning and is therefore $$P_{B}=\frac{3}{4}.\frac{1}{3} = \frac{1}{4}$$

Lastly,

Probability of Charles Winning = Probability of both Andy and Bruce Losing and Charles winning and is therefore $$P_{C}=\frac{3}{4}.\frac{2}{3}.\frac{1}{2}=\frac{1}{4}$$

$\endgroup$
0
$\begingroup$

Here is some detail on the comment by MattPutnam on the equivalence with simultaneous drawings. I assume that in the process you describe, everyone looks at their ball (and shows it) when they draw it.

Now, consider the same process but the first ones to draw hide their balls then everyone reveals it at the same time. The probabilities should be the same as in the previous process, because the only different thing is when the balls are revealed, not their (blind) choice by the boys.

One step further, we can now consider the experiment when everyone chooses a ball at the same time, that is, one ball is attributed to each one the boys simultaneously, then they reveal them at the same time. Here intuitively none of the boys can be favoured by the process, but the probabilities should be the same as in the previous case as well.

$\endgroup$
0
$\begingroup$

Let me first show why your intuition goes wrong. For this, consider a slight variation of the game: Again, Andy, Bruce and Charles draw a ball in turn, but this time with putting the ball back. The first one who draws the red ball wins. If neither draws the red ball, nobody wins.

Now who has the best chances at this game? Well, obviously Andy: If he draws a red ball, he already has won, while Bruce can only win if he draws a red ball after Andy drew a white ball, and Charles can only win if both Andy and Bruce drew a white ball, and he then drew a red one.

Now, how to fix this first-mover advantage? Well, obviously we have to increase the probability of the others to win to make up for it. One obvious way to do so is if there are less white balls in the box for the later players. So just remove a white ball after each draw.

However, if Andy draws the red ball, he has won anyway, so it doesn't matter whether and what we remove afterwards, so we can instead just say Andy keeps the ball he has drawn. If the ball was white, it improves the chances of Bruce, and if it was red, it doesn't matter anyway. For the same reason, Bruce can just keep his ball, and of course for Charles it doesn't matter anyway since he's the last. But that now is exactly the game you described.

So your mistake was to see the advantage of having less white balls for the later players, but you didn't take into account the first-mover's advantage that Bruce can only win if Andy hasn't already, and Charles only if neither Andy nor Bruce has.

However so far this only shows that there are two effects working into opposite directions (thus showing that your intuition was based on incomplete analysis of the situation), but it doesn't yet show why those two effects cancel each other out exactly, giving a fair game in the end.

To see that the game is indeed fair, for the moment forget about the colours, and just assume that the balls are numbered $1$ to $4$. Now consider the following questions:

  • What ball is Andy most likely to draw? Well, obviously he'll draw any ball with equal probability, that is, with probability $1/4$.

  • What ball is Bruce most likely to draw? Well, he'll draw one of the balls Andy didn't, but since Andy draws any ball with equal probability, each ball will be missing with equal probability, and therefore this doesn't cause any bias; Bruce will again draw any ball with equal probability, $1/4$.

  • Finally, what ball is Charles most likely to draw? Well, again, no ball is more likely to be missing than any other, so the probabilities are again equal, and every ball has probability $1/4$ to be drawn by Charles.

OK, so now we paint ball $4$ red. But how we paint the balls doesn't affect the probability of the ball being drawn, therefore still Andy, Bruce and Charles each have a probability of $1/4$ to draw the red ball (the remaining $1/4$ is where neither draws the red ball).

Thus they all have equal chances of winning.

$\endgroup$

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.