# 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:** 2
**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 31, 2010, 1:08am UTC](https://boards.straightdope.com/t/probability-question/551958/21 "2010-08-31T01:08:39Z")
> [@Sinisterniik](#): > > > [@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. It makes no difference whether the spots are chosen one at a time or simultaneously. With the same number of candidates for the same number of spots the odds (calculated before the process starts, of course) of you winning a spot are the same. Odds calculated along the way change with each pick of course - either to 1/1 if you are chosen, otherwise they increase to 1/2 as you outlined. But that’s not relevant to the OP. Your two methods of calculation end up with the same odds, 100/101, when done before any results come in. ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [September 1, 2010, 2:40am UTC](https://boards.straightdope.com/t/probability-question/551958/22 "2010-09-01T02:40:06Z")
> [@Chessic\_Sense](#): > > … 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 that second line, you’re leaving out one term. 1/6 + 1/5 = 11/30, not 1/3. There is a 1/6 chance of getting the first slot and a 1/5 chance of getting the second, but you must subtract 1/5\*1/6=1/30, because you can’t get both slots. Another way the OP can think of this is that there are 15 (unordered) pairs of people that can be chosen, 5 of which contain the OP, so there’s a 5/15=1/3 chance of being selected. ---
**Author:** ![Sinisterniik](https://avatars.discourse-cdn.com/v4/letter/s/51bf81/32.png) [@Sinisterniik](https://boards.straightdope.com/u/Sinisterniik)\ **Post date:** [September 1, 2010, 2:53am UTC](https://boards.straightdope.com/t/probability-question/551958/23 "2010-09-01T02:53:32Z")
> [@Askance](#): > > It makes no difference whether the spots are chosen one at a time or simultaneously. > [@](#): > > Odds calculated along the way change with each pick of course You realize that those two statements are contradictory, right? ---
**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:** [September 1, 2010, 2:54am UTC](https://boards.straightdope.com/t/probability-question/551958/24 "2010-09-01T02:54:32Z")
> [@breezman](#): > > > [@Chessic\_Sense](#): > > > > … 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 that second line, you’re leaving out one term. 1/6 + 1/5 = 11/30, not 1/3. There is a 1/6 chance of getting the first slot and a 1/5 chance of getting the second, but you must subtract 1/5\*1/6=1/30, because you can’t get both slots. No, that’s not quite the right way of looking at it. The 1/5 probability is a conditional probability: you have a 1 in 5 chance of being chosen on the second round **provided you have no been chosen on the first.** So the right equation is: 1/6 + (5/6 \* 1/5) = 1/3 The 5/6 term is the probability that you will **not** be chosen on the first round. ---
**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:** [September 1, 2010, 3:13am UTC](https://boards.straightdope.com/t/probability-question/551958/25 "2010-09-01T03:13:54Z")
> [@Sinisterniik](#): > > > [@Askance](#): > > > > It makes no difference whether the spots are chosen one at a time or simultaneously. > > > [@](#): > > > > Odds calculated along the way change with each pick of course > > You realize that those two statements are contradictory, right? The odds as calculated _before anything happens_ are the same regardless of whether the spots are _going to be_ filled all at once or one at a time. If it does end up being done one at a time, the odds change (and so must be re-calculated) after each pick, as with each pick one possibility has collapsed to 1/1 and others to 0/1. ---
**Author:** ![Sinisterniik](https://avatars.discourse-cdn.com/v4/letter/s/51bf81/32.png) [@Sinisterniik](https://boards.straightdope.com/u/Sinisterniik)\ **Post date:** [September 1, 2010, 4:06am UTC](https://boards.straightdope.com/t/probability-question/551958/26 "2010-09-01T04:06:11Z")
> [@Askance](#): > > The odds as calculated _before anything happens_ are the same regardless of whether the spots are _going to be_ filled all at once or one at a time. If it does end up being done one at a time, the odds change (and so must be re-calculated) after each pick, as with each pick one possibility has collapsed to 1/1 and others to 0/1. Ok. I think we’ve been in agreement the whole time. ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [September 1, 2010, 9:17pm UTC](https://boards.straightdope.com/t/probability-question/551958/27 "2010-09-01T21:17:16Z")
> [@Giles](#): > > No, that’s not quite the right way of looking at it. It’s a perfectly fine way of looking at it. You’re looking at filling the slots sequentially and I’m looking at filling them simultaneously. Both are valid ways of determining the probability. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [September 1, 2010, 10:39pm UTC](https://boards.straightdope.com/t/probability-question/551958/28 "2010-09-01T22:39:37Z")
> [@breezman](#): > > > [@Giles](#): > > > > No, that’s not quite the right way of looking at it. > > It’s a perfectly fine way of looking at it. You’re looking at filling the slots sequentially and I’m looking at filling them simultaneously. Both are valid ways of determining the probability. Your approach actually does work, but it’s convoluted enough that I needed a couple minutes with scratch paper to convince myself that it’s correct, and based on your explanation, I’m not sure that you understand it. It has nothing to do with filling the slots sequentially or simultaneously; instead, **Giles** is using a standard approach that will work for any two events, and you’re using a formula I’ve never seen before that only works for mutually exclusive events. In more formal probabilistic terms, if A is the event that a player is picked for the first slot, and B is the event that a player is picked for the second slot, what **Giles** is computing is P(A) + P(A[sup]c[/sup])P(B|A[sup]c[/sup]), and what you’re computing is P(A) + P(B|A[sup]c[/sup]) - P(B|A[sup]c[/sup])P(A). Those are both equal if A and B are mutually exclusive (as they are here), but as I mentioned above, I’ve never seen the second approach before. If I ever teach an undergrad probability class, I’ll definitely use this as the basis for a fairly difficult test question, so I owe you one. ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [September 2, 2010, 2:36am UTC](https://boards.straightdope.com/t/probability-question/551958/29 "2010-09-02T02:36:14Z")
> [@ultrafilter](#): > > …what you’re computing is P(A) + P(B|A[sup]c[/sup]) - P(B|A[sup]c[/sup])P(A). Those are both equal if A and B are mutually exclusive (as they are here), but as I mentioned above, I’ve never seen the second approach before. I’m using the standard method of calculating probabilities using the principle of inclusion exclusion. In this case there are two events, and the P(A or B) = P(A) + P(B) - P(A and B). It doesn’t matter if the events are mutually exclusive. I’m thinking of this as two separate events, selecting one of six and selecting one of five. It’s possible for the person to be selected either time, but not both (since the OP is only one person), so the events are A=selected from the six and B=selected from the five. What we want is P(A) + P(B) - P(A and B). You can think of this as two selections happening simultaneously. (It’s basically picking apart the idea of counting the total number of pairs of people and the number of pairs that include the OP). Using conditional probabilities is sequential in that you calculate the probability of getting the first slot, and then the probability of getting the second provided that you didn’t get the first. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [September 2, 2010, 6:19pm UTC](https://boards.straightdope.com/t/probability-question/551958/30 "2010-09-02T18:19:14Z")
> [@breezman](#): > > > [@ultrafilter](#): > > > > …what you’re computing is P(A) + P(B|A[sup]c[/sup]) - P(B|A[sup]c[/sup])P(A). Those are both equal if A and B are mutually exclusive (as they are here), but as I mentioned above, I’ve never seen the second approach before. > > I’m using the standard method of calculating probabilities using the principle of inclusion exclusion. In this case there are two events, and the P(A or B) = P(A) + P(B) - P(A and B). It doesn’t matter if the events are mutually exclusive. > > I’m thinking of this as two separate events, selecting one of six and selecting one of five. It’s possible for the person to be selected either time, but not both (since the OP is only one person), so the events are A=selected from the six and B=selected from the five. What we want is P(A) + P(B) - P(A and B). You can think of this as two selections happening simultaneously. (It’s basically picking apart the idea of counting the total number of pairs of people and the number of pairs that include the OP). > > Using conditional probabilities is sequential in that you calculate the probability of getting the first slot, and then the probability of getting the second provided that you didn’t get the first. Two comments: [ol] [li]What exactly are your events A and B here? If you define them the way you want to, the formula I wrote down is the correct representation of what you’re calculating.[/li][li]P(A and B) = P(A)P(B) only if A and B are [independent](http://en.wikipedia.org/wiki/Independence_%28probability_theory%29). Any two events that are mutually exclusive are not independent, so you can’t use that equality in this case.[/li][/ol] ---
**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:** [September 2, 2010, 7:15pm UTC](https://boards.straightdope.com/t/probability-question/551958/31 "2010-09-02T19:15:17Z")
> [@](#): > > P(A and B) = P(A)P(B) only if A and B are independent. Any two events that are mutually exclusive are not independent, so you can’t use that equality in this case. I don’t think he is using that equality. ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [September 3, 2010, 3:52am UTC](https://boards.straightdope.com/t/probability-question/551958/32 "2010-09-03T03:52:04Z")
> [@ultrafilter](#): > > Two comments: > What exactly are your events A and B here? If you define them the way you want to, the formula I wrote down is the correct representation of what you’re calculating. Here’s how I’m looking at the situation. A is the event that the person is selected from a set of six. B is that the person is selected from a set of five. I’m considering these as two independent selections (which I’ll call drawings) and determining the probability of being selected exactly once. This is equivalent to considering the two drawings as you did, with the second being dependent on the first. The event (A and B) is the event that the person is selected both times, so we must subtract the probability that it occurs. So what we want is P(A) + P(B) - P(A and B). ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [September 4, 2010, 5:09pm UTC](https://boards.straightdope.com/t/probability-question/551958/33 "2010-09-04T17:09:00Z")
> [@breezman](#): > > > [@ultrafilter](#): > > > > Two comments: > > What exactly are your events A and B here? If you define them the way you want to, the formula I wrote down is the correct representation of what you’re calculating. > > Here’s how I’m looking at the situation. > > A is the event that the person is selected from a set of six. > > B is that the person is selected from a set of five. > > I’m considering these as two independent selections (which I’ll call drawings) and determining the probability of being selected exactly once. This is equivalent to considering the two drawings as you did, with the second being dependent on the first. The event (A and B) is the event that the person is selected both times, so we must subtract the probability that it occurs. So what we want is P(A) + P(B) - P(A and B). Saying that two events are independent means that the outcome of one has no influence on the other. Since the probability that you’re selected in the draw of five depends on whether you were selected in the draw of six, they are not independent, and so you can’t write P(A and B) = P(A)P(B). ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [September 6, 2010, 2:09am UTC](https://boards.straightdope.com/t/probability-question/551958/34 "2010-09-06T02:09:34Z")
> [@ultrafilter](#): > > Saying that two events are independent means that the outcome of one has no influence on the other. Since the probability that you’re selected in the draw of five depends on whether you were selected in the draw of six, they are not independent, and so you can’t write P(A and B) = P(A)P(B). I think that you are missing the fact that my A and B aren’t the same as your A and B. I’m modeling this situation as two independent (and I’m well aware of what that means) drawings–one of five and one of six–and looking at the probability of being selected in exactly one of the drawings. To do that, P(A xor B) = P(A) + P(B) - P(A AND B), which in this situation is 1/6 + 1/5 - (1/6)(1/5). Using conditional probabilities is equivalent. If you factor out the 1/5 above, you get 1/6 + 1/5 (1 - 1/6) = 1/6 + (1/5)(5/6). In both we have to eliminate the possibility of being selected for both slots. One method adds 1/5 then subtracts (1/6)(1/5), the other multiplies (1/5) by (5/6) before adding it. Both are valid ways of modeling the situation. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [September 6, 2010, 3:43pm UTC](https://boards.straightdope.com/t/probability-question/551958/35 "2010-09-06T15:43:46Z")
> [@breezman](#): > > I think that you are missing the fact that my A and B aren’t the same as your A and B. I’m modeling this situation as two independent (and I’m well aware of what that means) drawings–one of five and one of six–and looking at the probability of being selected in exactly one of the drawings. To do that, P(A xor B) = P(A) + P(B) - P(A AND B), which in this situation is 1/6 + 1/5 - (1/6)(1/5). 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? ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [September 7, 2010, 3:27am UTC](https://boards.straightdope.com/t/probability-question/551958/36 "2010-09-07T03:27:28Z")
> [@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? Yes, and it is the reason why I subtracted 1/30. ---
**Author:** ![breezman](https://avatars.discourse-cdn.com/v4/letter/b/4491bb/32.png) [@breezman](https://boards.straightdope.com/u/breezman)\ **Post date:** [September 9, 2010, 11:40am UTC](https://boards.straightdope.com/t/probability-question/551958/37 "2010-09-09T11:40:23Z")
> [@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. ---
**Author:** ![RealityChuck](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/realitychuck/32/195_2.png) [@RealityChuck](https://boards.straightdope.com/u/RealityChuck)\ **Post date:** [September 9, 2010, 12:17pm UTC](https://boards.straightdope.com/t/probability-question/551958/38 "2010-09-09T12:17:45Z")
> [@TriPolar](#): > > How well do you play golf? This is the right track. It’s not a probability question. Straight probability assumes that all those competing have an equal chance, but obviously, they would not. A pro golfer is going to have a better chance than a weekend duffer. While the latter may play the round of his life and make the team, it’s impossible to determine the probability of that. ---
**Author:** ![Irishman](https://avatars.discourse-cdn.com/v4/letter/i/b487fb/32.png) [@Irishman](https://boards.straightdope.com/u/Irishman)\ **Post date:** [September 9, 2010, 2:04pm UTC](https://boards.straightdope.com/t/probability-question/551958/39 "2010-09-09T14:04:07Z")
**CurtC** said: > [@](#): > > 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. That might be true of a sport like baseball, or maybe basketball, where there are defined positions with different skill sets. Golf doesn’t work that way. Golf “teams” are just individual players that pool their individual scores for a total. It’s not like there’s Position 1 that handles all the tee-offs and Position 2 that handles the putting. ---
**Author:** ![alindallas](https://avatars.discourse-cdn.com/v4/letter/a/f17d59/32.png) [@alindallas](https://boards.straightdope.com/u/alindallas)\ **Post date:** [September 10, 2010, 11:57am UTC](https://boards.straightdope.com/t/probability-question/551958/40 "2010-09-10T11:57:20Z")
It’s amazing how much probability is influenced by who is your friend. Do you know these guys? Do any of the other “possibilities” hang out with these guys? If your competition is a drinking buddy of two or more choosers, I would estimate your odds of being chosen as about .000000025%. [Previous page](https://boards.straightdope.com/t/probability-question/551958.md?page=1) [Next page](https://boards.straightdope.com/t/probability-question/551958.md?page=3)