I am running a raffle. There are 10 prizes numbered 1-10 and 100 tickets. The raffle is run in two different ways:

A) Prize 1 is allocated to the first ticket drawn, prize 2 to the second and so on until 10 tickets are drawn. Tickets are not replaced.

B) Prize 10 is allocated to the first ticket drawn, prize 9 to the second and so on until 10 tickets are drawn. Tickets are not replaced.

Question: Is the overall probability of a ticket winning Prize 1 different in either scenario A or B?

Please do not complicate the answer with discussion of the change in probabilities during the draw.

I know the answer - but this is to resolve a very real PTA dispute... please upvote any correct answers to help resolve this pressing issue! There are similar questions, but I can't find any exactly like this, and I want this exact phrasing so the PTA is able to follow without confusion!


No - there is no difference in the probability of a particular ticket winning Prize 1.

In scenario A the probability is obviously $\dfrac{1}{100}$.

In scenario B it is $\dfrac{99}{100}\cdot\dfrac{98}{99}\cdot\dfrac{97}{98}\cdot\dfrac{96}{97}\cdot\dfrac{95}{96}\cdot\dfrac{94}{95}\cdot\dfrac{93}{94}\cdot\dfrac{92}{93}\cdot\dfrac{91}{92}\cdot\dfrac{1}{91}=\dfrac{1}{100}$

From personal experience Scenario A allows the top prize winner to choose their prize from the set of ten. It also allows people with several tickets have who won earlier prizes to waive further prizes if they have already won one, leading to a redraw. But it reduces the tension in the draw compared with Scenario B.

  • $\begingroup$ Yes, we want the tension of the big prize going last so the excitement doesn't peter out! Thanks! $\endgroup$ – James World Dec 4 '13 at 10:42

In case A, of course $p=\frac{1}{100}$. Once the first ticket is drawn, no other ticket can win 1st prize.

In case B, you can calculate the probability of not winning prizes 10 to 2:

$$ \frac{99}{100}\cdot\frac{98}{99}\cdot\cdots\cdot\frac{91}{92} = \frac{91}{100} $$

Now there are 91 tickets left, so the probability of winning the remaining prize is $\frac{1}{91}$.

Of course, multiplying the probability that your ticket was still in the box, by the probability that it is extracted at that step is

$$ \frac{91}{100}\cdot\frac{1}{91} = \frac{1}{100} $$ as desired.

Maybe this very explicit argument will help.

  • $\begingroup$ Very clear that the probabilites are identical, thank you. $\endgroup$ – James World Dec 4 '13 at 10:43

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