The Loop

Tuesday, July 29, 2008

Statistics

Dr Belen and I were learning a g'mera this morning which resulted in a battle over statistics. Question; If 5 people have 5 objects in front of them and only one object is appropiate to each person. Each person has a 20% chance of picking the correct object. Dr Belen said that before anyone picks -there is a greater than 20% chance that at least one of the five will pick the appropiate object. I said NO. After the first person picks, the second person has 4 objects left to choose and the appropiate one may already have been taken so there is a less than 20% chance of at least one picking the appropiate object. Who is right?

The g'mera was talking about 5 men who didnt know who there parents were but did know for sure that each had one brother. All 5 men married and died childless and their 5 wives fell to yibum but no one brother knew which woman feel to whom. Common senario in Detroit.

posted by dad at 8:18 AM

9 Comments:

  • I'm pretty sure the first one has a 20% chance, the second one has a 16%, the third has about a 13% chance, etc. so you are right.

    By Blogger stillruleall, at 11:29 AM  

  • Dr. Belen is correct.
    The 1st person has a 20% probability to get his object. This includes all future outcomes.
    If the 1st picked incorrectly, the 2nd has a certain chance of picking correctly , so the 20% can only go up.
    To be exact:
    The probability that the 1st picks incorrectly but the 2nd picks correctly is 60%*25%=15%. 60% is the probability that the 1st one picked an object that belongs to one of the last 3 people (so he's incorrect but he also didn't pick the 2nd), 25% is the probability that the 2nd picks correctly (1 of the 4 remaining).
    So we already have a 35% probability that one of the first two picks correctly, before we even add the probability that one of the next 3 gets correctly...

    By Blogger Shlomo, at 2:55 PM  

  • I say they all have a 50 percent chance, either they pick the right one or they don't

    By Blogger Just Shu, at 3:23 PM  

  • Hi folks

    Think about five men standing before five doors. The five women line up behind the doors . At same time all doors open

    The chance that at AT LEAST one man is facing his wife ( which includes man1 facing his wife or man2 facing wife or man3 facing wife or both man1 and man2 or both man2 and man3 ,etc ,etc ) is 1 minus the chance that no man is
    facing his wife

    For each person the chance that
    he is not facing his wife is .8

    The chance that none of the 5 is
    facing his wife is
    .8 x.8 x.8 x.8 x.8 = .33

    So the chance that at least one
    man is facing wife is
    1 - .33 = .67

    I never new I could understand Gamarra , or even spell it

    Thanks for telling me about this blog

    By Blogger Len_Chicago, at 10:58 PM  

  • hi again

    Sorry , the solution I gave to g'mara problem was wrong . It
    assumed that when each door was
    opened the chance of no match was
    4/5. This is not correct because
    the chance of a match on the third door depends on who specifically was matched on the first and second doors.

    You can see this if you think about the problem as a random
    group of 5 men , 5 women bumping into each other in a room . When the first couple is formed and does not match you are left with 4 men and 4 women and there are then 3 possible matches out of the 16 possible pairings of the 8 people remaining in the room because the first pairing left an odd man and an odd woman in the room . If the next pairing is between the odd man and the odd woman the remaining 3 men and 3 women left in the room will contain 3 matches out of the 9 remaiing pairings,and the chance of a match on the next pairing will be 3/9 however if the second pairing did not pair the odd man and the odd woman , the 9 remaining pairings in the room will only contain 2 possible matches so the chance of the next pairing being a match is 2/9

    This adds a lot of complexity to the problem and I do not see a
    quick way to solve this for the general case of N men and N women

    In the case of 5 men and 5 women if you go through all of the possibilities you will find that
    the chance that at least one match exists in the 5 pairings is 61.6 % which is a little less than the
    67 % I gave in my prior post

    If I find a general solution to the case of N men and N women I will let you know

    Good Night

    PS - If you have read this far you are probably interested that Mitchel's wife Laura is pregnant with their third child and is due in Feb 09

    This reminds me of an interesting g'mara problem During invitro fetilization a woman with 5 ova is impregnated with 5 sperm , and only certain sperm can fertilize certain ova what are the chances of delivering quintuplets ....

    By Blogger Len_Chicago, at 3:04 AM  

  • hi, i'm back with the answer

    The probabliity that there is a last one match with N men is

    1 - 1/2!+1/3!-1/4!+1/5! ,etc

    continue series to N on bottom
    in there are N men
    The ! is called factorial .
    5 ! = 5x4x3x2x1

    The probability that at least one man is matched with his wife when there are 5 men is almost the same as when there are 10 men or 1000 men . This is .63

    The problem is called the "Hat Check" problem or the "wife swapping problem "

    It was first solved by the famous mathematican Euler in 1740, a few years after probability was invented

    For more info look at the
    article "Derangements "
    by Conrey at www.geometer.org/mathcircles

    i guess you do not sign blogs?

    By Blogger Len_Chicago, at 7:09 PM  

  • You don't have to sign teh blog, because your screen name gives you away

    By Blogger Just Shu, at 1:22 PM  

  • wow,
    I should have posted my statistics homework on the blog. Maybe I would have gotten better grades...

    B'sha'a Tova (It should be in a good time(?)) to Mitch and Laura!

    By Blogger stillruleall, at 2:09 PM  

  • Hi Uncle Leonard - Welcome to the Zacks Loop

    By Blogger Air Time, at 4:01 AM  

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