# Probability Question **URL:** \ **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:** 20\ **Page:** 1
**Author:** ![Picard\_Kills\_Kirk](https://avatars.discourse-cdn.com/v4/letter/p/dec6dc/32.png) [@Picard\_Kills\_Kirk](https://boards.straightdope.com/u/Picard_Kills_Kirk)\ **Post date:** [August 30, 2010, 2:43am UTC](https://boards.straightdope.com/t/probability-question/551958/1 "2010-08-30T02:43:28Z")
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. ---
**Author:** ![TriPolar](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tripolar/32/3008_2.png) [@TriPolar](https://boards.straightdope.com/u/TriPolar)\ **Post date:** [August 30, 2010, 2:49am UTC](https://boards.straightdope.com/t/probability-question/551958/2 "2010-08-30T02:49:00Z")
How well do you play golf? ---
**Author:** ![friedo](https://avatars.discourse-cdn.com/v4/letter/f/8edcca/32.png) [@friedo](https://boards.straightdope.com/u/friedo)\ **Post date:** [August 30, 2010, 2:49am UTC](https://boards.straightdope.com/t/probability-question/551958/3 "2010-08-30T02:49:48Z")
It depends on how the spots are filled. If they are filled serially, your chances are 1/6 for the first spot and 1/5 for the second spot. But if they are filled simultaneously, your chances are 2/6 or 1/3. ---
**Author:** ![Peremensoe](https://avatars.discourse-cdn.com/v4/letter/p/ce7236/32.png) [@Peremensoe](https://boards.straightdope.com/u/Peremensoe)\ **Post date:** [August 30, 2010, 3:07am UTC](https://boards.straightdope.com/t/probability-question/551958/4 "2010-08-30T03:07:45Z")
> [@Picard\_Kills\_Kirk](#): > > I am only competing for 1 position. You wouldn’t accept the other one? ---
**Author:** ![Picard\_Kills\_Kirk](https://avatars.discourse-cdn.com/v4/letter/p/dec6dc/32.png) [@Picard\_Kills\_Kirk](https://boards.straightdope.com/u/Picard_Kills_Kirk)\ **Post date:** [August 30, 2010, 3:20am UTC](https://boards.straightdope.com/t/probability-question/551958/5 "2010-08-30T03:20:53Z")
> [@spark240](#): > > > [@Picard\_Kills\_Kirk](#): > > > > I am only competing for 1 position. > > You wouldn’t accept the other one? Well, I would, but I was assuming someone else already won it. ---
**Author:** ![Asympotically\_fat](https://avatars.discourse-cdn.com/v4/letter/a/e47c2d/32.png) [@Asympotically\_fat](https://boards.straightdope.com/u/Asympotically_fat)\ **Post date:** [August 30, 2010, 3:25am UTC](https://boards.straightdope.com/t/probability-question/551958/6 "2010-08-30T03:25:31Z")
> [@friedo](#): > > It depends on how the spots are filled. If they are filled serially, your chances are 1/6 for the first spot and 1/5 for the second spot. But if they are filled simultaneously, your chances are 2/6 or 1/3. Please explain, unless I’m misunderstanding what you mean by serially the chance is still 1/3, assuming everything else is equal and he is competing for two spots. How could it be different? chance of getting 1st spot = 1/6 chance of not getting 1st spot, but getting 2nd spot = 5/6\*1/5 = 1/6 1/6+1/6 = 1/3 ---
**Author:** ![DanBlather](https://avatars.discourse-cdn.com/v4/letter/d/f4b2a3/32.png) [@DanBlather](https://boards.straightdope.com/u/DanBlather)\ **Post date:** [August 30, 2010, 3:38am UTC](https://boards.straightdope.com/t/probability-question/551958/7 "2010-08-30T03:38:02Z")
> [@Picard\_Kills\_Kirk](#): > > > [@spark240](#): > > > > You wouldn’t accept the other one? > > Well, I would, but I was assuming someone else already won it. No, he won your spot. Now you can get his. ---
**Author:** ![statsman1982](https://avatars.discourse-cdn.com/v4/letter/s/e79b87/32.png) [@statsman1982](https://boards.straightdope.com/u/statsman1982)\ **Post date:** [August 30, 2010, 3:38am UTC](https://boards.straightdope.com/t/probability-question/551958/8 "2010-08-30T03:38:34Z")
Here’s another way to look at it. There are 6!/(4!\*2!) = 15 ways to choose the two members of the team. Let’s say the individuals are represented by letters as ABCDEF, and assuming that you’re “A” (it doesn’t matter what letter we assume you are). Then there are 5 pairings favorable to you: AB, AC, AD, AE, AF. That’s 5/15 = 1/3. ---
**Author:** ![Askance](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/askance/32/8281_2.png) [@Askance](https://boards.straightdope.com/u/Askance)\ **Post date:** [August 30, 2010, 3:42am UTC](https://boards.straightdope.com/t/probability-question/551958/9 "2010-08-30T03:42:55Z")
> [@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. By that thinking your odds are the same whether there are 1 or 100 open positions, and regardless of how many competitors there are (as long as there are at least as many as there are positions), which obviously cannot possibly be the case. If there were 100 positions and 101 people going for them, are you saying you’d only have a 1/100 chance ? If there were one open position only your odds would obviously be 1/6. Doubling the number of openings must affect your odds; if your 1/6 chance of getting position #1 doesn’t come off you now still have a chance at position #2, which you didn’t before. Another way of thinking of it is to add up the odds for all 6 people - if each only has a 1/6 chance then their total odds only adds up to 1, whereas you have 2 positions to fill so the odds must add up to 2 - so 1/3 for each person. ---
**Author:** ![friedo](https://avatars.discourse-cdn.com/v4/letter/f/8edcca/32.png) [@friedo](https://boards.straightdope.com/u/friedo)\ **Post date:** [August 30, 2010, 3:43am UTC](https://boards.straightdope.com/t/probability-question/551958/10 "2010-08-30T03:43:39Z")
> [@These are my own pants](#): > > > [@friedo](#): > > > > It depends on how the spots are filled. If they are filled serially, your chances are 1/6 for the first spot and 1/5 for the second spot. But if they are filled simultaneously, your chances are 2/6 or 1/3. > > Please explain, unless I’m misunderstanding what you mean by serially the chance is still 1/3, assuming everything else is equal and he is competing for two spots. How could it be different? > > chance of getting 1st spot = 1/6 > > chance of not getting 1st spot, but getting 2nd spot = 5/6\*1/5 = 1/6 > > 1/6+1/6 = 1/3 That’s true. I forgot to consider that you would have to lose the first spot to be considered for the second spot. So it’s 1/3 no matter how it’s chosen. Unless it’s chosen by skill. 😛 ---
**Author:** ![Picard\_Kills\_Kirk](https://avatars.discourse-cdn.com/v4/letter/p/dec6dc/32.png) [@Picard\_Kills\_Kirk](https://boards.straightdope.com/u/Picard_Kills_Kirk)\ **Post date:** [August 30, 2010, 3:52am UTC](https://boards.straightdope.com/t/probability-question/551958/11 "2010-08-30T03:52:15Z")
> [@Askance](#): > > > [@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. > > By that thinking your odds are the same whether there are 1 or 100 open positions, and regardless of how many competitors there are (as long as there are at least as many as there are positions), which obviously cannot possibly be the case. If there were 100 positions and 101 people going for them, are you saying you’d only have a 1/100 chance ? > > If there were one open position only your odds would obviously be 1/6. Doubling the number of openings must affect your odds; if your 1/6 chance of getting position #1 doesn’t come off you now still have a chance at position #2, which you didn’t before. > > Another way of thinking of it is to add up the odds for all 6 people - if each only has a 1/6 chance then their total odds only adds up to 1, whereas you have 2 positions to fill so the odds must add up to 2 - so 1/3 for each person. Well, out of the 100 positions and 101 people, originally I would have said I have a 1/2 change of getting a position. I understand now why that reasoning is wrong. Thanks everyone for helping me clear that up. ---
**Author:** ![CurtC](https://avatars.discourse-cdn.com/v4/letter/c/ce73a5/32.png) [@CurtC](https://boards.straightdope.com/u/CurtC)\ **Post date:** [August 30, 2010, 1:47pm UTC](https://boards.straightdope.com/t/probability-question/551958/12 "2010-08-30T13:47:02Z")
From my understanding of the OP, there are two separate positions on the team, call them position A and position B. You want one of them, and of the six total people, you don’t know which of those (other than yourself) are going out for A or B. If you’re the only one who wants A, then the probability of your getting it is 1. If all six people want A, your probability is 1/6. If it’s unknown, and there’s no reason to think there is a preference, then the middle assumption is that three are going for A, three are going for B, so that gives you a 1/3 chance of getting it. ---
**Author:** ![Giles](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/giles/32/60_2.png) [@Giles](https://boards.straightdope.com/u/Giles)\ **Post date:** [August 30, 2010, 2:16pm UTC](https://boards.straightdope.com/t/probability-question/551958/13 "2010-08-30T14:16:44Z")
> [@TriPolar](#): > > How well do you play golf? This. The answer is going to vary depending on the relative abilities of the people competing, and your knowledge of that. (1) If you know nothing about the relative abilities, then the odds are 2/6 = 1/3. (2) If one competitor is Tiger Woods, and the other five have an ability equal to the average amateur player, then the odds at 1/5, because Tiger Woods will get a position even on a bad day. (3) If five have average ability, and the sixth is me, with my ability at golf, then I have zero chance of making the team. (I’ve played golf once, on a 9-hole course, nearly 50 years ago – I know which end of the club to hold, but that’s about it.) ---
**Author:** ![Chessic\_Sense](https://avatars.discourse-cdn.com/v4/letter/c/7c8e57/32.png) [@Chessic\_Sense](https://boards.straightdope.com/u/Chessic_Sense)\ **Post date:** [August 30, 2010, 2:27pm UTC](https://boards.straightdope.com/t/probability-question/551958/14 "2010-08-30T14:27:16Z")
Basically, OP, your error was in assuming you can’t get the other spot. You didn’t count in your chances of getting that spot, so you got a result that resembles what you’d get if there was one spot and five people going for it. In other words, you assumed that Tiger was playing for one of the spots. I realize why you thought that. You know someone else is going to go into the other spot no matter what. But you shouldn’t say “probability of getting that spot”, but instead “probability of making the team”. In the case where you’re first pick, you labeled the second spot “0 probabililty” instead of “1 probability” because you forgot what you were counting. So you did: 0_1/6 + 1_ 1/5 = 1/5 Instead of: 1\* 1/6 + 1\*1/5 = 1/3 In one more wording: If it turns out to be the 1 in 6 times you get picked for the first position, then the second five options are all successes, not failures as you counted them. ---
**Author:** ![pancakes3](https://avatars.discourse-cdn.com/v4/letter/p/ba9def/32.png) [@pancakes3](https://boards.straightdope.com/u/pancakes3)\ **Post date:** [August 30, 2010, 2:36pm UTC](https://boards.straightdope.com/t/probability-question/551958/15 "2010-08-30T14:36:38Z")
good thing you’re not trying out for the mathletes. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 30, 2010, 5:14pm UTC](https://boards.straightdope.com/t/probability-question/551958/16 "2010-08-30T17:14:57Z")
Which spot is the “other” spot? If you get Spot A, then obviously the “other spot” was B. But if you get Spot B, then obviously the “other spot” was A. And if you get neither, which one was it then? ---
**Author:** ![kopek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kopek/32/337_2.png) [@kopek](https://boards.straightdope.com/u/kopek)\ **Post date:** [August 30, 2010, 5:20pm UTC](https://boards.straightdope.com/t/probability-question/551958/17 "2010-08-30T17:20:20Z")
If either spot is in Florida, all bets are off. 😉 ---
**Author:** ![Sinisterniik](https://avatars.discourse-cdn.com/v4/letter/s/51bf81/32.png) [@Sinisterniik](https://boards.straightdope.com/u/Sinisterniik)\ **Post date:** [August 30, 2010, 5:45pm UTC](https://boards.straightdope.com/t/probability-question/551958/18 "2010-08-30T17:45:22Z")
> [@Askance](#): > > By that thinking your odds are the same whether there are 1 or 100 open positions, and regardless of how many competitors there are (as long as there are at least as many as there are positions), which obviously cannot possibly be the case. If there were 100 positions and 101 people going for them, are you saying you’d only have a 1/100 chance ? Well you have odds of 1/101 for the first position, then 1/100 for the second, 1/99 for the third, 1/98 for the fourth… Assuming you don’t get the first spot. Eventually the odds would be 1/2, assuming you never got chosen. Each time a position closes, you have a better chance of getting the next position. If they choose the spots all at once, the odds would be 100/101 That is why it is either 1/6 for the first spot and 1/5 for the second (assuming you don’t get the first) or 2/6 (or 1/3) if the spots are chosen (announced) simultaneously. Right? I mean, that’s the way that poker odds work. If you’re on the turn (in Hold 'Em) you calculate odds out of 47 remaining cards, but on the river it’s 46. But you never calculate out of 52 when you know that 5 cards have been revealed. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [August 30, 2010, 5:54pm UTC](https://boards.straightdope.com/t/probability-question/551958/19 "2010-08-30T17:54:52Z")
Right, _if five cards have been revealed_. But at the start, they haven’t been yet, and we might not get to the situation where all but one of the spots have been filled and you still don’t have one. ---
**Author:** ![Little\_Nemo](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/little_nemo/32/3120_2.png) [@Little\_Nemo](https://boards.straightdope.com/u/Little_Nemo)\ **Post date:** [August 30, 2010, 8:32pm UTC](https://boards.straightdope.com/t/probability-question/551958/20 "2010-08-30T20:32:20Z")
> [@Chessic\_Sense](#): > > Basically, OP, your error was in assuming you can’t get the other spot. You didn’t count in your chances of getting that spot, so you got a result that resembles what you’d get if there was one spot and five people going for it. In other words, you assumed that Tiger was playing for one of the spots. > > I realize why you thought that. You know someone else is going to go into the other spot no matter what. But you shouldn’t say “probability of getting that spot”, but instead “probability of making the team”. In the case where you’re first pick, you labeled the second spot “0 probabililty” instead of “1 probability” because you forgot what you were counting. > > So you did: 0_1/6 + 1_ 1/5 = 1/5 > Instead of: 1\* 1/6 + 1\*1/5 = 1/3 > > In one more wording: If it turns out to be the 1 in 6 times you get picked for the first position, then the second five options are all successes, not failures as you counted them. A way to demonstrate the fallacy of this reasoning is to assume every person has an equal opportunity to make the team. So the probabilities should be the same for every player. But by applying the OP’s logic, none of them will get the first spot because it will be filled by “the other kid”. Obviously this is not accurate. One of the six must be that other kid. [Next page](https://boards.straightdope.com/t/probability-question/551958.md?page=2)