Sorry for a stupid question, but it is bugging me a lot.

Let's say there are $30$ classmates in my class and one of us has to clean the classroom. No one wants to do that. So we decided to draw a lot - thirty pieces of paper in a hat, one of which is with "X" on it. The one who draws "X" has to do the cleaning. Each one starts to draw...

Is this kind of lot drawing fair or not fair? It looks to me like the first one's chances to get an "X" are equal to $1/29$, while the second one's chances would be equal either to $1/28$ (in case the first one didn't draw an "X") or zero $0/29 = 0$ (in case the first one drew an "X"). However, neither $1/28$, nor $0/29$ is equal to $1/29$.

  • $\begingroup$ Why not draw names instead of an X? Seems like that would be significantly easier, and gives everyone a 1/30 chance. $\endgroup$ – Carcigenicate Jul 29 '16 at 21:35
  • $\begingroup$ Say we flip a coin and write down on a piece of paper whether it came up heads or tails. Then we look at the piece of paper. The odds it will say "heads" are either 100% (if the coin came up heads) or 0% (if the coin came up tails). But neither of those is 50%. Does that make the error in the reasoning more obvious? $\endgroup$ – David Schwartz Jul 29 '16 at 21:49
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    $\begingroup$ The odds for the first person are 1:29, but that's a probability of 1/30. $\endgroup$ – Hurkyl Jul 29 '16 at 23:40
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    $\begingroup$ This is essentially the Gambler's fallacy. $\endgroup$ – argentpepper Jul 30 '16 at 3:28

P(first one draws $X) = \dfrac1{30}$
P(first one doesn't draw $X) = \dfrac{29}{30}$

The possibility of the second one drawing $X$ arises if and only if the first one doesn't, so
P(second one draws $X) = \dfrac{29}{30}\cdot\dfrac1{29} = \dfrac1{30}$ again !

You will find, if you continue in the same way, that each one will have the same probability, $\dfrac1{30}$, and you needn't compute at all !

Imagine all papers arranged randomly in a line.
The $X$ is as likely to be in the first place as the last (or any other)

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    $\begingroup$ I don't mean to be pedantic because I really love this answer, but for the OP: remember that $P(\text{Student 2 draws X})$ is implicitly stating that Student 2 can only draw $X$ if and only if Student 1 didn't draw $X$. This is why we have to multiply the probabilities together, which seemed to be your confusion in the comment to another answer. $\endgroup$ – anonymouse Jul 29 '16 at 17:01
  • $\begingroup$ @anonymouse: Thanks, I have added a line to my answer. I thought the confusion OP had was not about the multiplication , but why Pr for the second person was 1/29, why not 1/28, (given that first hadn't drawn X). 1/29 vs 1/28 became the dominant discussion a bit later. $\endgroup$ – true blue anil Jul 29 '16 at 17:43
  • $\begingroup$ Thanks for your answer. But I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? $\endgroup$ – brilliant Aug 1 '16 at 2:07
  • $\begingroup$ If you can get the insight contained in the last two lines of the answer, there is no need to calculate, and no need to compare, because (using the technically correct terms), the probability is the same for all of them at $\frac1{30}$, or the odds at $1:29$ $\endgroup$ – true blue anil Aug 1 '16 at 10:03

It's fair. The easiest way to see that is to imagine the people drawing their papers but not looking until all have got theirs.

Each person's chance is $1/30$.

In your way of thinking about the problem the first person's chance of getting the X (and having to clean) is 1/30 (not 1/29 as you state in your question - those are the odds, not the probability).

If the first person doesn't get the X then the second person's chance is 1/29, but it wasn't 1/29 from the start - it only became 1/29 once you knew the first person didn't get the X.

To find the true probability for the second person you have to reason backwards from the 1/29 chance he sees, to take into account the fact that he wouldn't have to draw at all when the first person gets the X. That happens 1/30 of the time, so the second person has to draw just 29 times out of 30. So the probability (from the start) that the second person draws the X is $$ \frac{29}{30} \times \frac{1}{29} = \frac{1}{30} $$

This is the essential algebra in many of the other correct answers.

If by chance none of the first 29 people gets the X then the chance that the last person does is 1/1, which is no surprise. But it wasn't 1/1 from the start, of course. You can make the algebra show you that too.

Edit: To address the OP's question about odds.

Odds and probabilities are two different ways to express the same mathematics. In your problem the first person to draw has a probability of 1/30 to draw the X. That means he (or she) cleans 1 time out of 30. His odds are 1:29, which you read aloud as "1 to 29". The odds mean he cleans 1 time for each 29 times he doesn't. Odds aren't usually written as fractions, because that often leads to confusion.

You can figure out problems like yours working with probabilities or with odds, but it's easier with probabilities as all the answers show.

Your logical error about the odds is this.

If the first person doesn't get the X then the second person's odds are indeed just 1:28, as you say, but they weren't 1:28 from the start - they only became 1:28 once you knew the first person didn't get the X.

To find the true odds at the start for the second person you have to reason backwards from the 1:28 odds in his draw to take into account the fact that he wouldn't have to draw at all when the first person gets the X. That's what using probability instead of odds lets you do, as above.

You might be able to see the error in your logic if you imagine there are three people instead of 30. The first person has odds of 1:2. If he doesn't get the X then the second person has odds of 1:1 (even chance, which makes sense) but that isn't the case from the start since he actually has to draw only 2/3 of the time.

  • $\begingroup$ Was going to say the same thing about not looking at the paper. $\endgroup$ – shawnt00 Jul 31 '16 at 2:15
  • $\begingroup$ Thank you for your answer, but it looks like your calculating the probability here and not the chances (odds). I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? $\endgroup$ – brilliant Aug 1 '16 at 2:08
  • $\begingroup$ @brilliant You're welcome. Please see my edit explaining your logical error in terms of odds rather than probabilities. It's the same error. $\endgroup$ – Ethan Bolker Aug 1 '16 at 13:17
  • $\begingroup$ For some reasons, I read this answer in Kevin Spacey's voice. +1 though! $\endgroup$ – MonK Aug 1 '16 at 13:36

The draw is fair. It just seems counter-intuiutive because of the way you are thinking of it.

Suppose that all thirty pieces of paper were handed out simultaneously and all at once the papers were flipped and the person with the $X$ would clean. This would seem fair, right?

Well, drawing a paper one at a time is virtually the same thing, the only difference is that the draw stops once the $X$ is revealed. Everyone still has a $\frac{1}{30}$ chance of getting the $X$.



Give all pieces of paper a distinct letter (among them $A$ and $X$).

What is the probability that - let's say the one drawing as $23$rd - get's e.g. letter $A$?

Would this be different for letter $X$?

No. Every letter has the same chance of falling in hands of number $23$.


The first one has a chance of $1/30$, not $1/29$, as there are 30 pieces of paper in the box (if there were only 29 pieces in the box, it would obviously not be fair, as the last one would never get the cross: the other 29 classmates would already have removed all the pieces).

Obviously, if one classmate is chosen at random, he should have a chance of $1/30$ to be the chosen one, therefore it's fair to him.

Now after he has drawn the paper and found no cross (and assuming he didn't put it back into the box— another way to make it unfair), there are only 29 candidates left. Therefore it is obvious that if everyone of them has to have a fair chance, now for each of them the probability to be the chosen one is to be $1/29$. And indeed, that's the chance that the next one taking a piece from the box has, as there are only 29 pieces of paper left.

Note that this is a conditional probability (on the condition that the first one didn't get the cross). To calculate the total probability, you have to multiply the conditional probability with the probability of the condition. That condition is that the first one did not get a cross, and the probability of that condition is $29/30$ (since with probability $1/30$ he did get the cross). So the total probability that the second one gets selected is $$\frac{29}{30}\cdot\frac{1}{29} = \frac{1}{30}$$ as it should be. So everything is still fair.

Now it isn't hard to check that it remains fair also for the remaining classmates, as at any time there are exactly as many pieces of paper left in the box as there are classmates that didn't yet draw one.


From your comments on other answers I gather that you don't understand why the probability for the first one is $1/30$ and not $1/29$. Well, I guess you are simply confusing the probability with the odds.

The probability tells you how often the event will happen if you repeat the experiment many times. That is, if the event (in this case, the first classmate getting the "X") has probability $p$, and you repeat the whole procedure $N$ times, then the expected number of times the even happens is $pN$. In this case, there are 30 different pieces of paper, and on average he will draw each one equally often. Therefore at one of 30 draws he will get the cross, and thus the probability is $1/30$

The odds on the other hand give the relation of the event happening versus the event not happening. I this case, there's one piece with a cross, and 29 without, each of which he will draw with the same probability, and therefore his odds to draw the cross are $1:29$.

Note the different way I wrote this, as this is not really a division (although it is close to it). In particular, if the box only contained the piece with the cross, his odds would be $1:0$; this is not a division by zero error, but a completely valid odds. On the other hand, the probability to draw the cross is $1/1 = 1$. This is a division, and a well-defined one. The value $1$ means he will draw the cross every time ($pN = N$), and that's indeed what will happen if the only piece of paper in the box is the one with the cross.

  • $\begingroup$ This reminds me of homogenous coordinates, which is a way of representing points in projective geometry that does not suffer from the problem of infinite gradient (division by zero). $\endgroup$ – user21820 Jul 30 '16 at 4:55
  • $\begingroup$ Thank you for you answer and kind explanations.I see now that there is a difference between probability and chances (odds). I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? $\endgroup$ – brilliant Aug 1 '16 at 2:06
  • $\begingroup$ @brilliant: Most of things in probability theory have been done with probabilities (well, the name is already a dead giveaway!). If you would want to do everything in odds, you'd have to first translate all the rules from probabilities to odds. I'm also not sure if odds can be rigorously defined when going beyond finite number of possible outcomes. $\endgroup$ – celtschk Aug 1 '16 at 7:08
  • $\begingroup$ (+1)I found your answer more helpful than the other answers...(I don't know why).... $\endgroup$ – tatan Aug 12 '16 at 5:03

The second person's chance is still $1/30$. Indeed, Person 2 draws an X exactly when Person 1 does NOT draw an X, and so: \begin{align*} \Pr[P_2 = X] &= \Pr[P_1 \neq X] \cdot \Pr[P_2 = X \mid P_1 \neq X] \\ &= \frac{29}{30} \cdot \frac{1}{29} \\ &= \frac{1}{30} \end{align*} Likewise, Person 3 draws an X exactly when the previous two people do NOT draw an X: \begin{align*} \Pr[P_3 = X] &= \Pr[P_1 \neq X] \cdot \Pr[P_2 \neq X \mid P_1 \neq X] \cdot \Pr[P_3 = X \mid P_2 \neq X] \\ &= \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{1}{28} \\ &= \frac{1}{30} \end{align*}

This continues all the way up to Person 29 and Person 30: $$ \Pr[P_{29} = X] = \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{27}{28} \cdot \ldots \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \\ \Pr[P_{30} = X] = \frac{29}{30} \cdot \frac{28}{29} \cdot \frac{27}{28} \cdot \ldots \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{1} = \frac{1}{30} $$

  • $\begingroup$ Can you, please, explain this to me in a more simple way? It looks to me that if I have a box with 30 apples and one of them is rotten, while my friend has a box of 29 apples with one of them rotten, he seems to have more chances to draw a rotten apple than me. No? $\endgroup$ – brilliant Jul 29 '16 at 15:37
  • $\begingroup$ This is, of course, correct - but probably not useful to the OP - if he knew how to reason with conditional probabilities he wouldn't have had to ask the question. $\endgroup$ – Ethan Bolker Jul 29 '16 at 15:38
  • $\begingroup$ @EthanBolker - Yes. You are right. I am like an idiot here. I see that all people here see right away that the draw is fair, while to me it looks unfair. And I just can't understand where my logic is failing me. $\endgroup$ – brilliant Jul 29 '16 at 15:48
  • $\begingroup$ @brilliant I've edited my answer to try to show where your logic fails and how to correct it. $\endgroup$ – Ethan Bolker Jul 29 '16 at 15:58
  • $\begingroup$ I wonder why do we need to calculate probability at all? Why not just compare the chances - the way I did in my question? $\endgroup$ – brilliant Aug 1 '16 at 2:09

Actually, it looks like its fair to me.

The probability that person 1 chooses an "X" is $\frac{1}{30}$, since only one of the 30 lots has an "X" on it.

The probability that person 2 chooses an "X" is $\frac{29}{30}*\frac{1}{29} =\frac{1}{30} $, since Person 1 must have not chosen a lot with an "X" on it.

The probability that person 3 chooses an "X" is $\frac{29}{30}*\frac{28}{29}*\frac{1}{28} =\frac{1}{30}$

You can show that this is true for every successive person, and to further prove this, we can look at the 30th person.

The probability that person 30 chooses an "X" is the probability that everybody else did not choose an "X", which is $\frac{29!}{30!} = \frac{1}{30}$

So it is indeed fair for every person.

  • $\begingroup$ "The probability that person 2 chooses an "X" is (29/30)*(1/29) = 1/30 since Person must have not chosen a lot with an "X" on it" - Why is it (29/30)*(1/29) and not 1/28? After all, the second person is now drawing from the pool of 29 lots (that is, not 30 anymore)? $\endgroup$ – brilliant Jul 29 '16 at 15:31
  • $\begingroup$ The first person is drawing from a pool of $30$, and the second person is drawing from a pool of $29$, and so on... $\endgroup$ – true blue anil Jul 29 '16 at 15:34
  • $\begingroup$ @trueblueanil - Exactly! $\endgroup$ – brilliant Jul 29 '16 at 15:34
  • $\begingroup$ When the first person draws, 1 out of the 30 lots are going to have an "X" on it. Thus his odds of drawing the "X" are 1/30 . The second person draws from a pool of 29 lots. Thus the probability that he chooses an "X" is 1/29 , iff the first person did not draw an "X". The probability that the first person does not draw an "X" is 29/30 since there are 29 lots without an "X" on it when he draws. Thus Person 2's probability of drawing an X is (29/30)*(1/29) = 1/30 . Does that help? $\endgroup$ – Ninja_Coder Jul 29 '16 at 15:37
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    $\begingroup$ @brilliant Which becomes obvious if you think about the opposite side. What is the probability of drawing anything other than the ace of spades? Well, there are 51 cards that aren't the ace of spades, and only 1 that is. By your logic the probability would be 51/1, but it should be 51/52. $\endgroup$ – Paulpro Jul 29 '16 at 15:48

You've identified three outcomes for the second person:

  • Person 2 doesn't draw
  • Person 2 draws and doesn't get the X
  • Person 2 draws and gets the X

and you're interested in the probability that the last of the three outcomes happens.

You then divided the outcomes into two categories:

  • Person 2 doesn't draw
  • Person 2 draws

You've even (nearly) correctly computed the probabilities

  • Assuming person 2 doesn't draw, he has probability 0 of getting the X
  • Assuming person 2 does draw, he has probability 1/29 of getting the X

But where you went off the rails is that you forgot your original goal was "What is the probability he gets the X?", and neither of the calculations you did were to compute that. They're similar-sounding questions, but similar does not mean the same. This is especially true for questions about probability, counting, and statistics, where sloppy reasoning often leads to wildly incorrect results.

If you are to use the work you've done, the problem now is how to correctly combine the two probabilities. This can be done with an additional piece of information:

  • The probability that person 2 doesn't draw is 1/30
  • The probability that person 2 does draw is 29/30

And thus we can compute

  • The proportion of samples that person 2 doesn't draw is 1/30
    • Of those, the proportion where person 2 gets the X is 0
      • The total proportion of this case is $\frac{1}{30} \cdot 0 = 0$
  • The proportion of samples that person 2 does draw is 29/30
    • Of those, the proportion where person 2 gets the X is 1/29
      • The total proportion of this case is $\frac{29}{30} \cdot \frac{1}{29} = \frac{1}{30}$

and then adding up all of the different cases gives $0 + \frac{1}{30} = \frac{1}{30}$.


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