# Simple (not to me!) probability problem **URL:** \ **Category:** Factual Questions\ **Created:** [November 11, 2011, 1:51pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594 "2011-11-11T13:51:03Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Turek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/turek/32/18782_2.png) [@Turek](https://boards.straightdope.com/u/Turek)\ **Post date:** [November 11, 2011, 1:51pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/1 "2011-11-11T13:51:03Z")
It’s been a LONG time since probability class, and I’m struggling with this this morning. You have a bag with 1000 balls in it. Of those, 250 are red, 750 are black. If you pick one ball, the probability is .25 that it will be red. If you pick two balls, the probability is roughly .25 \*.25 = .0625 that they BOTH will be red. But, if you choose two balls, what is the probability that at least ONE of them will be red? Part of me says it’s .25 +.25 = .5, but that doesn’t seem right because it implies that if I choose 4 balls, the odds of one being red is approaching 1. ---
**Author:** ![ColdPhoenix](https://avatars.discourse-cdn.com/v4/letter/c/6bbea6/32.png) [@ColdPhoenix](https://boards.straightdope.com/u/ColdPhoenix)\ **Post date:** [November 11, 2011, 1:55pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/2 "2011-11-11T13:55:48Z")
To find the probability that at least one of them is red just calculate the posibility that both are black and subtract from 1. 1 - (0.75\*0.75) = 0.4375 To get the probablithy that exactly 1 is red subtract the chances of both being red from the above. 0.4375 - (0.25\*0.25) = 0.375 The above is rough as the odds will change slightly after you’ve removed a ball. ---
**Author:** ![Turek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/turek/32/18782_2.png) [@Turek](https://boards.straightdope.com/u/Turek)\ **Post date:** [November 11, 2011, 2:05pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/3 "2011-11-11T14:05:29Z")
Thanks, that sounds better! ---
**Author:** ![don\_t\_ask](https://avatars.discourse-cdn.com/v4/letter/d/e68b1a/32.png) [@don\_t\_ask](https://boards.straightdope.com/u/don_t_ask)\ **Post date:** [November 11, 2011, 2:31pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/4 "2011-11-11T14:31:57Z")
And why is this very important to understand? Say we are talking about two sports teams, team RED and team BLACK. And BLACK wins 3 out of four games between them. But still in a two game series RED will win at least one game nearly half the time. ---
**Author:** ![Turek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/turek/32/18782_2.png) [@Turek](https://boards.straightdope.com/u/Turek)\ **Post date:** [November 11, 2011, 2:44pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/5 "2011-11-11T14:44:05Z")
Assuming the two teams are evenly matched, yes. But that’s not a valid assumption when it comes to sports. While I understand the point your making, a better example is if you flip 3 heads in a row, the the probability of flipping a head on the 4th toss is still .5. ---
**Author:** ![don\_t\_ask](https://avatars.discourse-cdn.com/v4/letter/d/e68b1a/32.png) [@don\_t\_ask](https://boards.straightdope.com/u/don_t_ask)\ **Post date:** [November 11, 2011, 2:48pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/6 "2011-11-11T14:48:21Z")
Not really. The point I am making is that, even if the teams are far from evenly matched there is no guarantee that the better team will finish on top over a short series of competitions. Most people fail to see this and invoke nonsense theories. ---
**Author:** ![bup](https://avatars.discourse-cdn.com/v4/letter/b/6bbea6/32.png) [@bup](https://boards.straightdope.com/u/bup)\ **Post date:** [November 11, 2011, 2:51pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/7 "2011-11-11T14:51:27Z")
**Turek** , **ColdPhoenix** ’ analysis is of course correct. What’s wrong with your .25 + .25 = .5 analysis is that you double-counted some outcomes : .25 that ball \*one \*is red already counts outcomes where ball \*two \*is red _as well as_ black, and counting .25 that ball two is red counts \*all \*outcomes where ball two is red. In other words, you double-counted the outcomes where both balls are red. So your analysis requires subtracting that probability - which is 1/4 \*1/4 = .0625. .25 + .25 - .0625 = .4375. 🆒 ---
**Author:** ![Pai325](https://avatars.discourse-cdn.com/v4/letter/p/c68b51/32.png) [@Pai325](https://boards.straightdope.com/u/Pai325)\ **Post date:** [November 11, 2011, 3:00pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/8 "2011-11-11T15:00:31Z")
It seems like there are 4 possibilities if you pick 4 balls. Both red, both black, first red second black, first black second red. In three of those cases you have at least one red ball. So 75%? But why do I have the feeling I am missing something? ---
**Author:** ![jonesj2205](https://avatars.discourse-cdn.com/v4/letter/j/ecc23a/32.png) [@jonesj2205](https://boards.straightdope.com/u/jonesj2205)\ **Post date:** [November 11, 2011, 3:10pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/9 "2011-11-11T15:10:38Z")
> [@Pai325](#): > > It seems like there are 4 possibilities if you pick 4 balls. Both red, both black, first red second black, first black second red. In three of those cases you have at least one red ball. So 75%? But why do I have the feeling I am missing something? Your calculation assumes the probabilities of picking red or black ball on any pull are the same, which they aren’t in the OP (three times as many black as red balls). ---
**Author:** ![Pai325](https://avatars.discourse-cdn.com/v4/letter/p/c68b51/32.png) [@Pai325](https://boards.straightdope.com/u/Pai325)\ **Post date:** [November 11, 2011, 3:27pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/10 "2011-11-11T15:27:27Z")
> [@jonesj2205](#): > > Your calculation assumes the probabilities of picking red or black ball on any pull are the same, which they aren’t in the OP (three times as many black as red balls). Thank you! I guess I was confusing it with the four kids, what’s the probability one is a girl question. ---
**Author:** ![don\_t\_ask](https://avatars.discourse-cdn.com/v4/letter/d/e68b1a/32.png) [@don\_t\_ask](https://boards.straightdope.com/u/don_t_ask)\ **Post date:** [November 11, 2011, 3:55pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/11 "2011-11-11T15:55:44Z")
> [@Pai325](#): > > It seems like there are 4 possibilities if you pick 4 balls. Both red, both black, first red second black, first black second red. In three of those cases you have at least one red ball. So 75%? But why do I have the feeling I am missing something? That is basically right though. There are four possibilties: RR RB BR BB and the chances are .25 \* .25 = .0625 .25 \* .75 = .1875 .75 \* .25 = .1875 .75 \* .75 = .5625 ---
**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:** [November 11, 2011, 5:10pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/12 "2011-11-11T17:10:59Z")
Though of course, it isn’t actually that difficult to arrive at the exact probabilty, as opposed to a good approximattion: (250/1000 \* 249/999) + (250/1000 \* 750/999) + (750/1000 \* 250/999) = 1- (750/1000 \* 749/999) = 437250/999000 = 4.37̅6̅8̅ ---
**Author:** ![Derleth](https://avatars.discourse-cdn.com/v4/letter/d/b9e5f3/32.png) [@Derleth](https://boards.straightdope.com/u/Derleth)\ **Post date:** [November 11, 2011, 5:34pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/13 "2011-11-11T17:34:07Z")
> [@Turek](#): > > If you pick two balls, the probability is roughly .25 \*.25 = .0625 that they BOTH will be red. Are you putting the first ball back before you pick the second one? If you are, you’re right; otherwise, you have to account for the fact you have on fewer red ball and one fewer ball total. Anyway, you’re looking at a [hypergeometric distribution](http://stattrek.com/lesson2/hypergeometric.aspx) there and we know [lots and lots](http://en.wikipedia.org/wiki/Hypergeometric_distribution) about them. ---
**Author:** ![Turek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/turek/32/18782_2.png) [@Turek](https://boards.straightdope.com/u/Turek)\ **Post date:** [November 11, 2011, 5:44pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/14 "2011-11-11T17:44:35Z")
> [@Derleth](#): > > Are you putting the first ball back before you pick the second one? If you are, you’re right; otherwise, you have to account for the fact you have on fewer red ball and one fewer ball total. No, I’m not putting the first ball back, which is why I said “roughly” .25 \* .25. ---
**Author:** ![C\_K\_Dexter\_Haven](https://avatars.discourse-cdn.com/v4/letter/c/b2d939/32.png) [@C\_K\_Dexter\_Haven](https://boards.straightdope.com/u/C_K_Dexter_Haven)\ **Post date:** [November 11, 2011, 5:51pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/15 "2011-11-11T17:51:05Z")
> [@Pai325](#): > > It seems like there are 4 possibilities if you pick 4 balls. Both red, both black, first red second black, first black second red. In three of those cases you have at least one red ball. … This is my favorite probability error. Similarly, if I buy a lottery ticket, either it will win or it won’t win: 50/50 chance. ---
**Author:** ![Pai325](https://avatars.discourse-cdn.com/v4/letter/p/c68b51/32.png) [@Pai325](https://boards.straightdope.com/u/Pai325)\ **Post date:** [November 11, 2011, 6:27pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/16 "2011-11-11T18:27:37Z")
> [@C\_K\_Dexter\_Haven](#): > > This is my favorite probability error. Similarly, if I buy a lottery ticket, either it will win or it won’t win: 50/50 chance. If only! ---
**Author:** ![GoodOmens](https://avatars.discourse-cdn.com/v4/letter/g/d26b3c/32.png) [@GoodOmens](https://boards.straightdope.com/u/GoodOmens)\ **Post date:** [November 11, 2011, 7:10pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/17 "2011-11-11T19:10:46Z")
> [@C\_K\_Dexter\_Haven](#): > > This is my favorite probability error. Similarly, if I buy a lottery ticket, either it will win or it won’t win: 50/50 chance. I often would represent it to my students this way: say I’m in a boxing match with Mike Tyson. One of us will win, right? Therefore, I have a 50% chance of winning, right? Most of them already know what when I say something, and then say, “Right?” I purposefully was wrong and I’m hoping they’ll figure out the mistake. ---
**Author:** ![grimpixie](https://avatars.discourse-cdn.com/v4/letter/g/ecb155/32.png) [@grimpixie](https://boards.straightdope.com/u/grimpixie)\ **Post date:** [November 11, 2011, 7:32pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/18 "2011-11-11T19:32:44Z")
> [@C\_K\_Dexter\_Haven](#): > > This is my favorite probability error. Similarly, if I buy a lottery ticket, either it will win or it won’t win: 50/50 chance. Which is why I always buy two, to cover my bases. ---
**Author:** ![Punoqllads](https://avatars.discourse-cdn.com/v4/letter/p/d2c977/32.png) [@Punoqllads](https://boards.straightdope.com/u/Punoqllads)\ **Post date:** [November 11, 2011, 8:56pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/19 "2011-11-11T20:56:10Z")
One thing not specified in the OP is whether the picks are with replacement or without replacement. That is, after you pick the first ball, do you put it back in and randomize the bag before picking out the second ball? Or do you leave the first ball out when you reach in to pick a second ball? Also, one of the ways you can analyze the problem is to calculate the probability that both balls will be black, and then subtract that from 1 to calculate the probability that at least won’t be black; i.e., that one ball will be red. E.g., in the first scenario you’ve got a 750/1000 chance each time to pick a black ball, and 1 - 0.75 \* 0.75 = 0.4375, just as **ColdPhoenix** said. In the second scenario, the probability is 1 - 750/1000 \* 749/999 = 0.437̅6̅8̅, like **These are my own pants** said. ---
**Author:** ![Turek](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/turek/32/18782_2.png) [@Turek](https://boards.straightdope.com/u/Turek)\ **Post date:** [November 11, 2011, 9:03pm UTC](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594/20 "2011-11-11T21:03:45Z")
I didn’t specify the replacement scheme in the OP, but I did specify a few posts later. 🙂 The actual, real life scenario is this: My wife and I are going to a reading by a famous author. For buying a ticket, you get to attend the reading and you get a copy of the book. 1000 tickets were sold. Randomly mixed into the 1000 books being distributed are 250 signed copies. I was trying to figure out the probability of us getting at least one signed copy. [Next page](https://boards.straightdope.com/t/simple-not-to-me-probability-problem/602594.md?page=2)