# On splitting up dinner companions and randomness: a probability question **URL:** \ **Category:** Factual Questions\ **Created:** [May 16, 2013, 1:18am UTC](https://boards.straightdope.com/t/on-splitting-up-dinner-companions-and-randomness-a-probability-question/658474 "2013-05-16T01:18:11Z")\ **Posts on this page:** 5\ **Page:** 1
**Author:** ![El\_Famous\_Burrito](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/el_famous_burrito/32/3954_2.png) [@El\_Famous\_Burrito](https://boards.straightdope.com/u/El_Famous_Burrito)\ **Post date:** [May 16, 2013, 1:18am UTC](https://boards.straightdope.com/t/on-splitting-up-dinner-companions-and-randomness-a-probability-question/658474/1 "2013-05-16T01:18:11Z")
I encountered a probability question recently that has given me fits: Six friends – Alex, Beth, Chad, Don, Emily, Fred, and George – are headed to popular restaurant for dinner. When they arrive they learn that there are no tables available to accommodate the whole bunch. They must split into a group of **two** and a group of **four**. They divide themselves into these two groups randomly (via some no doubt clever, but irrelevant mechanism). _What is the probability that Alex and Beth end up at the same table?_ I whipped up a solution that I believed to be correct. My solution, however, was not among the multiple choice answers. The nature of my encounter with this problem was such that I did not get to see the correct answer. Where did I go wrong? (Or am I correct and the test makers wrong?) My solution is below, in the spoiler box. While mathematical solutions are more than welcome, I figured the problem was small enough that I could just use brute force. Below find every combination of diners that could be seated at the two-person table. The remaining diners presumably sat together at the four-person table. A=Alex. B=Beth. AB\* AC AD AE AF AG BC BD BE BF BG CD\*\* CE\*\* CF\*\* CG\*\* DE\*\* DF\*\* DG\*\* EF\*\* EG\*\* FG\*\* One star (\*) denotes the combination that seats Alex and Beth together at the two-person table. Two stars (\*\*) denote that combinations that seat neither Alex nor Beth at the two-person table, thereby seating them together at the four-person table. There are 21 possible seating combinations. 11 of these seat Alex and Beth at the same table. My answer is therefore **11/21**. This was not among the selectable answers. ---
**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:** [May 16, 2013, 1:27am UTC](https://boards.straightdope.com/t/on-splitting-up-dinner-companions-and-randomness-a-probability-question/658474/2 "2013-05-16T01:27:51Z")
I don’t know if you’re wrong, but the test makers certainly are. Alex, Beth, Chad, Don, Emily, Fred, and George are seven friends, not six. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [May 16, 2013, 1:35am UTC](https://boards.straightdope.com/t/on-splitting-up-dinner-companions-and-randomness-a-probability-question/658474/3 "2013-05-16T01:35:06Z")
Are there six or seven friends? You’ve got seven in your list, A–G. If you do your method with six, you’ll get 15 possible combinations at the small table, of which 7 would result in A and B seated together. Thus, 7/15 would be the answer. ---
**Author:** ![Kimble](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/kimble/32/19185_2.png) [@Kimble](https://boards.straightdope.com/u/Kimble)\ **Post date:** [May 16, 2013, 1:55am UTC](https://boards.straightdope.com/t/on-splitting-up-dinner-companions-and-randomness-a-probability-question/658474/4 "2013-05-16T01:55:27Z")
The OP’s method is correct for 7 friends and would give the correct answer of 7/15 if there were 6 friends – just eliminate all the listed possibilities where George is at the 2-top\*, like MikeS said while I was typing this out. Alternately: (1/3 chance Alex is at 2-top) \* (1/5 chance Beth is, too) + (2/3 chance Alex is at 4-top) \* (3/5 chance Beth isn’t there, too) = 1/15 + 6/15 = 7/15 ---
**Author:** ![El\_Famous\_Burrito](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/el_famous_burrito/32/3954_2.png) [@El\_Famous\_Burrito](https://boards.straightdope.com/u/El_Famous_Burrito)\ **Post date:** [May 16, 2013, 2:22am UTC](https://boards.straightdope.com/t/on-splitting-up-dinner-companions-and-randomness-a-probability-question/658474/5 "2013-05-16T02:22:07Z")
Asked and answered. My trouble isn’t with probability but with counting. The question specified **six** friends, and I went ahead and solved for **seven**. Turns out “F” is the 6th letter of the English alphabet, not “G.” Who knew? (The original question did not list everyone’s names. It just said “six friends.” I added the names in my retelling to add a little character and to set up the notation I used in my solution.)