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The question:

Alfonso and Colin each bought one raffle ticket at the state fair. If $50$ tickets were randomly sold, what is the probability that Alfonso got ticket $14$ and Colin got ticket $23$?

According to this, the answer should be $ \frac{1}{2450}$ which presumably comes from $\frac{1}{50}\times \frac{1}{49}$. But it seems that the order does not count. I did not assume that Alfonso got ticket $14$ first then Colin got ticket $23$ second.

Update: What is wrong with this reasoning. When I said that I did not assume order, I meant that it's possible

  • Alfonso got ticket $14$ first, then Colin got ticket $23$,
  • Colin got ticket $23$ first, then Alfonso got ticket $14$.

Both of these possibilities are possible before the tickets are given out, so we can make an 'or' statement. Label the event Alfonso got ticket $14$ by $A_{14} $ and Colin got ticket $23$ by $A_{23}$. Then by the addition rule

$ \Pr(\text{ ($A_{14}$ first and $C_{23}$ second) or ($C_{23}$ first and $A_{14}$ second}) ) \\ = \Pr(A_{14}) \times \Pr(C_{23} \mid A_{14}) + \Pr(C_{23}) \times \Pr(A_{14}\mid C_{23}) = \frac 1 {50} \times \frac 1 {49} \times 2.$

I realize that once the tickets are sold, then only one of $ \{ A_{14}C_{23}~ , ~ C_{23}A_{14} \}$ must occur, but before the tickets are sold both possibilities are plausible. Why would the probability change before and after the tickets are sold.

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3 Answers 3

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It's not about the order in which they get the tickets, it's about who gets which ticket. When you write $1\over{}_{50} C_2$, you are saying that there is only one acceptable pair of tickets out of ${}_{50} C_2$ possible pairs. But after you choose the correct pair, you still have to say who gets ticket $14$ and who $23$, and there is only one way to do that correctly out of two possibilities.

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  • $\begingroup$ So you have $ \frac{1} { _{50} C_2 } \times \frac 1 2 $ $\endgroup$
    – john
    Commented May 22, 2018 at 1:31
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    $\begingroup$ @john Multiply it by one half. This is $P(A\cap B)=P(A)P(B|A)$ where $A=$"choosing the correct pair" and $B|A=$"dealing the right card to the right person after you chose the correct pair". (after your edit: yes exactly). $\endgroup$ Commented May 22, 2018 at 1:33
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"I did not assume that Alfonso got ticket 14 then Colin got ticket 23" - order in this sense has nothing to do with time. It is a question of whether or not

Alfonso got ticket 14 and Colin got ticket 23

is different from

Alfonso got ticket 23 and Colin got ticket 14.

And clearly it is different, and only the first is asked for in your problem. There are $50\times49$ ways in which tickets could be allocated to A and C, with order (in the sense I have just explained) being important; and only one successful case; so the probability is $$\frac1{50\times49}\ .$$

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  • $\begingroup$ Is it possible to frame the question so that time does matter. In other words, it does matter that first Alfonso gets ticket 14 then Colin gets ticket 23 second, versus Colin first getting ticket 23 then Alfonso getting ticket 14 second. $\endgroup$
    – john
    Commented May 22, 2018 at 1:09
  • $\begingroup$ @John I don't think this is possible without making it a totally different question. For example "one ticket is sold every minute from 2:10 to 2:59 and tickets are sold in random order, Alfonso arrives at a random time from 2:00 to 3:00, Colin at a random time from 2:30 to 3:30, what is the probability that Colin buys his ticket first and gets no.23 and Alfonso gets no.14". But this is getting a bit ridiculous :) $\endgroup$
    – David
    Commented May 22, 2018 at 1:13
  • $\begingroup$ So you don't agree with my update to the question. Thanks for your patience, by the way ;D. $\endgroup$
    – john
    Commented May 22, 2018 at 1:22
  • $\begingroup$ No, I don't agree. You can't even talk about which event happened first (in time) unless you have some assumptions about when tickets are sold, when people arrive etc. $\endgroup$
    – David
    Commented May 22, 2018 at 2:39
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    $\begingroup$ Yes I agree with the answer. I don't agree with your saying it's analogous, it is clearly different from your original question. $\endgroup$
    – David
    Commented May 22, 2018 at 3:43
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To say that order does not count means there is no difference between the following:

  • Alfonso got ticket $14$ and Colin got ticket $23$,
  • Alfonso got ticket $23$ and Colin got ticket $14$.

You found the probability that between them they got a certain set of two tickets. After they get those two tickets, there still the choice of which gets which, and there are two choices, each with (conditional) probability $1/2.$

You can also look at it like this: $$ \Pr(\text{Alfonso got }\#14) \times \Pr(\text{Colin got }\#23\mid \text{Alfonso got }\#14) = \frac 1 {50} \times \frac 1 {49}. $$

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  • $\begingroup$ But how do you address the fact that Colin could get ticket 23 first, then Alfonso gets ticket 14 second. I am a little confused because time seems to be implicit in the problem. $\endgroup$
    – john
    Commented May 22, 2018 at 1:12
  • $\begingroup$ $\Pr(\text{Alfonso got }\#14) \times \Pr(\text{Colin got }\#23\mid \text{Alfonso got }\#14) + \Pr(\text{Colin got }\#23) \times \Pr(\text{Alfonso got }\#14\mid \text{Alfonso got }\#14) = \frac 1 {50} \times \frac 1 {49}.$ $\endgroup$
    – john
    Commented May 22, 2018 at 1:14
  • $\begingroup$ Here's your comment formatted a bit differently: $$\begin{align} & \Pr(\text{Alfonso got }\#14) \times \Pr(\text{Colin got }\#23\mid \text{Alfonso got }\#14) \\ {}+{} & \Pr(\text{Colin got }\#23) \times \Pr(\text{Alfonso got }\#14\mid \text{Alfonso got }\#14) = \frac 1 {50} \times \frac 1 {49}. \end{align}$$ $\endgroup$ Commented May 22, 2018 at 1:15
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    $\begingroup$ Chronology is not taken into account in the answer as I wrote it. Note that $\Pr(\text{Colin got }\#23 \mid \text{Alfonso got }\#14)$ does not depend on which happened first. And $\Pr(\text{Alfonso got }\#14)$ also does not depend on which happened first. $\endgroup$ Commented May 22, 2018 at 1:18
  • $\begingroup$ Whoops, I meant to write $\Pr(\text{Alfonso got }\#14) \times \Pr(\text{Colin got }\#23\mid \text{Alfonso got }\#14) \\ + \Pr(\text{Colin got }\#23) \times \Pr(\text{Alfonso got }\#14\mid \text{Colin got }\#14) = \frac 1 {50} \times \frac 1 {49} \times 2. \\ \text{and this is clearly false because the answer should be } \frac 1 {50} \times \frac 1 {49} $ $\endgroup$
    – john
    Commented May 22, 2018 at 1:56

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