# Probability Question

**URL:** <https://boards.straightdope.com/t/probability-question/551958>\
**Category:** Factual Questions\
**Created:** [August 30, 2010, 2:43am UTC](https://boards.straightdope.com/t/probability-question/551958 "2010-08-30T02:43:28Z")\
**Posts on this page:** 2\
**Page:** 3

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**Author:** ![ColdPhoenix](https://avatars.discourse-cdn.com/v4/letter/c/6bbea6/32.png) [@ColdPhoenix](https://boards.straightdope.com/u/ColdPhoenix)\
**Post date:** [September 10, 2010, 12:47pm UTC](https://boards.straightdope.com/t/probability-question/551958/41 "2010-09-10T12:47:03Z")

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> [@Picard\_Kills\_Kirk](#):
>
> Let’s say I am trying out for 1 of 2 open positions on a golf team, and 6 people are trying out for those 2 positions. What are the odds of me making the team?
> 
> I said 1 in 5, but everyone else around said 1 in 3. I understand where they got the 1/3 (2 divided by 6), but my reasoning for 1 in 5 was that **one other kid is definitely going to get the second position, eliminating him from my odds** , because I am only competing for 1 position.
> 
> If this isn’t a GQ question, sorry for misplacing it.

Assuming I’m understanding you correctly then you’re correct that the odds are 1 in 5.

Are you saying that one particular kid is definitely getting chosen (because he’s easily the best player or the son of the person choosing or whatever)?

If that’s the case and the other 5 of you are on a par… Then it’s effectively you 5 competing for 1 spot and so the odds are 1 in 5.

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**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\
**Post date:** [September 10, 2010, 3:10pm UTC](https://boards.straightdope.com/t/probability-question/551958/42 "2010-09-10T15:10:33Z")

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> [@breezman](#):
>
> > [@ultrafilter](#):
> >
> > If you claim that both draws are independent, you’re asserting that there’s a 1/30 chance that you’ll be selected for both slots. Is that really what you mean?
> 
> To be more clear, I’m modeling this the same way you’d model the following situation. One bag contains 6 balls, another contains 5. Each bag contains exactly one black ball. One ball is selected from each bag. What is the probability of selecting exactly one black ball.
> 
> You’re modeling this the same as using one bag containing 6 balls, exactly one of which is black, and asking if two balls are selected without replacement, what is the probability of selecting the black ball in the first or the second draw.
> 
> In the first setup the two draws can happen simultaneously and in the second one they are sequential (which I mentioned in my first post). These are two different models, but the particular question we are asking is equivalent.

Your model assigns positive probability to the event that a player will be picked for both positions, which will never happen in the set up we have here. Therefore, it is not a valid model. The fact that it gives the right answer for one question is not relevant; it needs to give the correct answer for every question.

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