# help with this math question **URL:** \ **Category:** Factual Questions\ **Created:** [February 20, 2014, 3:11pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791 "2014-02-20T15:11:44Z")\ **Posts on this page:** 17\ **Page:** 1
**Author:** ![cornopean](https://avatars.discourse-cdn.com/v4/letter/c/8c91f0/32.png) [@cornopean](https://boards.straightdope.com/u/cornopean)\ **Post date:** [February 20, 2014, 3:11pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/1 "2014-02-20T15:11:44Z")
if i have 2 red candies 1 green and 1 purple and i’m picking two at a time, what’s the possibility of getting two reds? Explain please. ---
**Author:** ![Bullitt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bullitt/32/5725_2.png) [@Bullitt](https://boards.straightdope.com/u/Bullitt)\ **Post date:** [February 20, 2014, 3:21pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/2 "2014-02-20T15:21:59Z")
There are four equally-likely possible outcomes: (a) 2 red are picked (b) a red and a green are picked © a red and a purple are picked (d) a green and a purple are picked Therefore, 25%. ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [February 20, 2014, 3:28pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/3 "2014-02-20T15:28:17Z")
> [@cornopean](#): > > if i have 2 red candies 1 green and 1 purple and i’m picking two at a time, what’s the possibility of getting two reds? It’s a roll of the die - 1 in 6. You can always just count the possibilities (especially easy when there are so few options). Even though there are 2 red candies, you should count each separately when enumerating. There’s a problem with saying 1 in 4. There are 4 outcomes but they are NOT equally likely: (A) Two red - exactly one way to do this (B) 1 red, 1 green - two ways to do this (each red) - it’s twice as likely as (A). (C) 1 red, 1 purple - two ways to do this (each red) - it’s twice as likely as (A). (D) 1 green, 1 purple - exactly one way to do this. So, of the 4 outcomes, only 1 (A) matches what you want, but it is half as likely as (B) and (C). For purposes of computing a probability, it’s easier to list each red separately. So, we have Red A, Red B, Green, and Purple. The ways to pick 2 of these are: Red A, Red B (what we want) Red A, Green Red B, Green Red A, Purple Red B, Purple Green, Purple So, there are 6 ways of choosing 2 candies and only 1 of them match our criteria. For a more formal way of expressing exactly the same information, there are C(4,2) combinations of 4 things taken 2 at a time, or: C(4,2) = 4! / (2! \* (4-2)!) = 4 ! / (2!_2!) = 24 / (2_2) = 24/4 = 6 So, there are 6 total ways of picking 2 of the candies. Of these, there are C(2,2) ways of picking 2 out of 2 red candies. C(2,2) = 1, so that’s simple. There are also C(2,0) ways of picking 0 out of the 2 non-red candies C(2,0) = 1, so that’s also simple. So, there are C(2,2)\*C(2,0) ways of picking 2 of the 2 red candies and 0 of the 2 non-red candies. In this case, that’s 1 way, as C(2,2)_C(2,0) = 1_1 = 1. So, the overall probability is 1 / 6, as we noted earlier. The same concept can be used to compute many similar sorts of probabilities, including lotto probabilities. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [February 20, 2014, 3:28pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/4 "2014-02-20T15:28:52Z")
Six equally-likely picks (r1 r2, r1 g, r1 p, r2 g, r2 p, g p), so 1/6 chance. [ninjaed] ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [February 20, 2014, 3:32pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/5 "2014-02-20T15:32:31Z")
In short, to list the 4 possible outcomes with probabilities, we have: (A) 2 red - 1/6 (B) 1 red, 1 green - 2/6 = 1/3 © 1 red, 1 purple - 2/6 = 1/3 (D) 1 green, 1 purple - 1/6 ---
**Author:** ![Great\_Antibob](https://avatars.discourse-cdn.com/v4/letter/g/e47c2d/32.png) [@Great\_Antibob](https://boards.straightdope.com/u/Great_Antibob)\ **Post date:** [February 20, 2014, 3:40pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/6 "2014-02-20T15:40:34Z")
Sorry to keep posting, but there’s another way of expressing the idea these are not equally likely outcomes. Say we have 4 people from 3 different states: Tom and Anne from Texas Bill from Montana Susan from Kentucky If randomly chosen, what is the probability we pick the two from Texas? Tom and Bill is a different from Anne and Bill, even though both are “Texas + Montana”, so each should be counted separately for probability purposes. The Red candies are no different. ---
**Author:** ![cornopean](https://avatars.discourse-cdn.com/v4/letter/c/8c91f0/32.png) [@cornopean](https://boards.straightdope.com/u/cornopean)\ **Post date:** [February 20, 2014, 3:56pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/7 "2014-02-20T15:56:45Z")
Very helpful. thanks, ---
**Author:** ![SpyOne](https://avatars.discourse-cdn.com/v4/letter/s/82dd89/32.png) [@SpyOne](https://boards.straightdope.com/u/SpyOne)\ **Post date:** [February 20, 2014, 4:48pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/8 "2014-02-20T16:48:55Z")
Just want to point out that it doesn’t matter if you are choosing them one at a time or together: the odds of the final outcome would be the same. ---
**Author:** ![Thudlow\_Boink](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/thudlow_boink/32/320_2.png) [@Thudlow\_Boink](https://boards.straightdope.com/u/Thudlow_Boink)\ **Post date:** [February 20, 2014, 5:02pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/9 "2014-02-20T17:02:15Z")
> [@SpyOne](#): > > Just want to point out that it doesn’t matter if you are choosing them one at a time or together: the odds of the final outcome would be the same. Right, and this gives us an alternative way of approaching the problem: The probability that the first candy selected is a red one is 2/4 (= 1/2). The probability that the second candy is red if the first one was is 1/3 (since we’re selecting from among the three remaining). So, by the [multiplication rule for dependent events](http://www.mathgoodies.com/lessons/vol6/dependent_events.html), the probability that the first and second are both red is (1/2)x(1/3) = 1/6. ---
**Author:** ![Learjeff](https://avatars.discourse-cdn.com/v4/letter/l/94ad74/32.png) [@Learjeff](https://boards.straightdope.com/u/Learjeff)\ **Post date:** [February 20, 2014, 10:37pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/10 "2014-02-20T22:37:12Z")
> [@Bullitt](#): > > There are four equally-likely possible outcomes: > > (a) 2 red are picked > (b) a red and a green are picked > (c) a red and a purple are picked > (d) a green and a purple are picked > > Therefore, 25%. The above illustrates how easy it is to have what looks like a really good argument about probability and be totally wrong. > [@Thudlow\_Boink](#): > > Right, and this gives us an alternative way of approaching the problem: > > The probability that the first candy selected is a red one is 2/4 (= 1/2). > The probability that the second candy is red if the first one was is 1/3 (since we’re selecting from among the three remaining). > So, by the [multiplication rule for dependent events](http://www.mathgoodies.com/lessons/vol6/dependent_events.html), the probability that the first and second are both red is > (1/2)x(1/3) = 1/6. That’s how I attacked it, getting 1/6 in just a moment’s thought. But I wasn’t sure I was right until reading the replies, even for such a simple problem. Probability stuff is SO easy to get wrong! ---
**Author:** ![Tim\_R.Mortiss](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/tim_r.mortiss/32/142_2.png) [@Tim\_R.Mortiss](https://boards.straightdope.com/u/Tim_R.Mortiss)\ **Post date:** [February 20, 2014, 10:56pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/11 "2014-02-20T22:56:57Z")
> [@Thudlow\_Boink](#): > > Right, and this gives us an alternative way of approaching the problem: > > The probability that the first candy selected is a red one is 2/4 (= 1/2). > The probability that the second candy is red if the first one was is 1/3 (since we’re selecting from among the three remaining). > So, by the [multiplication rule for dependent events](http://www.mathgoodies.com/lessons/vol6/dependent_events.html), the probability that the first and second are both red is > (1/2)x(1/3) = 1/6. This is a good example of a [hypergeometric](http://en.wikipedia.org/wiki/Hypergeometric_distribution) distribution, the one you use to compute odds of winning the Lotto. ---
**Author:** ![Lumpy](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/lumpy/32/446_2.png) [@Lumpy](https://boards.straightdope.com/u/Lumpy)\ **Post date:** [February 21, 2014, 1:17am UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/12 "2014-02-21T01:17:51Z")
Ignore the colors for a moment and just label the four candies A, B, C, and D. There are six ways you can pick two of them: AB, AC, AD, BC, BD, CD. If two of them are red, say A and B, then only 1/6 of the possibilities give you both reds. ---
**Author:** ![Bullitt](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bullitt/32/5725_2.png) [@Bullitt](https://boards.straightdope.com/u/Bullitt)\ **Post date:** [February 21, 2014, 9:07am UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/13 "2014-02-21T09:07:24Z")
Ahh, it is 1 in 6, not 1 in 4. I stand corrected, thanks guys. ---
**Author:** ![Jragon](https://avatars.discourse-cdn.com/v4/letter/j/e19b73/32.png) [@Jragon](https://boards.straightdope.com/u/Jragon)\ **Post date:** [February 21, 2014, 9:10am UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/14 "2014-02-21T09:10:09Z")
100%. Or 0% if you don’t feel like picking two reds I guess. I don’t know, maybe we should condition the problem based on some prior about the chooser’s color preferences? … what? The question never specified I was picking them at random with no idea of which had which color. 😛 ---
**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\ **Post date:** [February 21, 2014, 9:35pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/15 "2014-02-21T21:35:30Z")
We beat that one to death. ---
**Author:** ![ThisUsernameIsForbidden](https://avatars.discourse-cdn.com/v4/letter/t/b4bc9f/32.png) [@ThisUsernameIsForbidden](https://boards.straightdope.com/u/ThisUsernameIsForbidden)\ **Post date:** [February 21, 2014, 9:51pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/16 "2014-02-21T21:51:58Z")
> [@cornopean](#): > > if i have 2 red candies 1 green and 1 purple and i’m picking two at a time, what’s the possibility of getting two reds? > > Explain please. 1/2 x 1/3 is 1/6 ---
**Author:** ![watchwolf49](https://avatars.discourse-cdn.com/v4/letter/w/e9c0ed/32.png) [@watchwolf49](https://boards.straightdope.com/u/watchwolf49)\ **Post date:** [February 21, 2014, 10:01pm UTC](https://boards.straightdope.com/t/help-with-this-math-question/681791/17 "2014-02-21T22:01:57Z")
Picking two reds is a certainty if you’re allowed to look. Which level of Candy Crush Saga did you find this puzzle?