probability poser

[merlinmynx]

100 people had left their sweat shirts in a pile. They returned later and each person picked a sweat shirt from the pile, entirely at random.
Which is more likely:
a) everyone took a wrong shirt; or
b) at least one person took their own shirt ?

I believe it is b) but I can't explain why.
Any brief explanation would be appreciated.

[AngloScot]

I know a way to figure this out but my batteries in my calculator would run out or id still be pressing buttons next Julember...

 

So if anyone can tell me a formula to find the sum of (0.99)x(0.99+0.98)x(0.99+0.98+0.97)x...x(0.99+0.98+...+0.01+0.00) then I can definitely tell you but I don't know how to find the sum of a non-aritmetic series.

[jenstar]

The probability that everyone got the wrong shirt is (99/100)^100. I haven't worked it out, but I'm pretty sure that works out at a very small probability indeed, in which case it is, as you say, (b).

[Marsh]

Hi AngloScot.  I started to calculate (0.99) x (0.99+0.98 x (0.99+0.98+0.97) x�, using a spreadsheet, but after two brackets it came to 1.98 and 2,604,918.952 after ten. 

Hi Jenstar.  I would have thought so too but the probability for two people isn't (1/2)^2 (= �), it's � ; for three people it's not (2/3)^3 (= 8/27), it's  2/6; and for 4 people its not. (3/4)^4 (= 0.316�), its 3/8 (= 0.375)  (
http://www.theanswerbank.co.uk/Quizzes-and-Puzzles/Quest ion87877.html)  So I don't think the formula willl work for 100 people either.

Hi  merlinminx.  I too was following magicdice's advice  and found you had got here before me.  If you have to produce an answer on what you have at the moment you could still suggest that (99/100)^100 (=.0366..) is a reasonable-sounding approximation even though you know it's not the exact answer.  Let us know if you find out what that is, please.

[NetSquirrel]

The first thing to note about this question is that the two possible answers are mutually exclusive - that is, answers a and b cannot both be true at the same time.  Furthermore, a and b are opposites of each other in terms of probability, i.e. the probability that a is true must be 1- (one minus) the probability that b is true.  If probability (a=true) is 0.3 then probability (b=true) must be 0.7.

 

The probability that everyone takes a wrong shirt can be logically deduced quite easily, and I believe the answer is 1/100, i.e. 0.01.

 

To prove this to yourself, consider the first person to take a shirt.  Of the 100 shirts, 99 are not his (or hers).  The probability of taking a wrong shirt is therefore 99/100.  The second person sees 99 shirts, 98 of which are not his, so his probability is 98/99.  This continues until just 2 shirts remain.  The probability that the 99th person picks the wrong shirt is 1/2.  (The 100th person is ignored because there is only 1 shirt left at this stage.)  The calculation goes:

 

(99/100) x (98/99) x (97/98) ... (3/4) x (2/3) x (1/2)

 

[continued]

[NetSquirrel]

(Writing the above down on paper will make understanding the next part easier.)
 
If this series was written as a single fraction, it would go:
 
(99 x 98 x 97 ... x 3 x 2 x 1) / (100 x 99 x 98 ... x 4 x 3 x 2)
 
Notice that the bottom line of the fraction is almost exactly the same as the top, but just one number out of step.  This allows all the 99s to be cancelled, all the 98s to be cancelled, all the 97s...all the 4s, all the 3s and all the 2s.  In fact, everything cancels out except the 1 and the 100, leaving the final probability as 1/100.  Since a and b are opposite, the probability that at least one person took their own shirt is 99/100.

[Peter Pedant]

The chance of b is 1 - (.99)^100

This is almost unity

sorry this is the same of Jenstar answer

actually it should be1 -  .99x.98x ,97

ll the way back to one but that is still very close to one

[merlinmynx]

Hi NetSquirrell,

Many thanks for your lucid answer and to all the other contributors who also helped with their suggestions.

Happy new year,

Merlinmynx

[jenstar]

I think NetSquirrel's answer falls into the same trap as the one I gave. For four people, there are 24 ways of distributing the shirts, of which 9 result in one or more people getting their own (as Marsh says, that's 3/8). Unfortunately NetSquirrel's formula would give 3/4 x 2/3 x 1/2 ... cancel out gives 1/4. I think the problem may lie in the fact that the probabilities are not actually separate but depend on each other. By the time that, say, person no.32 comes to pick a shirt, their shirt might not even be in the pile anymore, which would mean their chance was not 1/68, but zero.

[mr. piper]

my small foray into probability, leads me to think that the working out is (1+2+3......+99+100)/*. when * = no. of selections remaining which gives the possible no of permutations divided by the amount of selections left-as the selections decrease the probability gets higher of picking your own, lest it is still remaining unselected, which thought nudges me into thinking that "*" in my example would still need to be 100, that also prompts me now to think that i have a high probability of being wrong for once .

[cuan2002]

Net squirrel has alittle fault in his answer.

The first person to pick has 100 sweaters to choose from, 99 of which are not his. Therefore has a 99/100 chance of not picking his own.

The second person has 99 sweaters to choose from. All 99 may not be his (i.e. the first person may have picked it already). He may have a 99/99 chance of not picking his own or a 98/99 of not picking his own.

The rest of the logic seems sound

[AngloScot]

I agree with Jenstar 'By the time that, say, person no.32 comes to pick a shirt, their shirt might not even be in the pile anymore, which would mean their chance was not 1/68, but zero.'

And I took this into consideration before my first answer, which is why i couldn't find an answer because the working would take far too long and formulae (there must be some) would be needed to prove a correct answer without sitting pressing buttons all day. 

Also with the amount of buttons that need to be pressed on the calculator means that the probability of finding the correct answer would be very high! 

[AngloScot]

Sorry I meant the wrong answer whoops i pressed the wrong buttons on my keyboard!

[NetSquirrel]

I see where I went wrong, now (damn!) This question hurts my head.

[cuan2002]

The answer by repetition is:

for 3 people,

[cuan2002]

for 3 people there are 4/6 ways at least one will have a right sweater.

for 4 people : 15 / 24

for 5 people 72 / 120

the series is:

for n people : ((n/2)+1/2)/n => for n = 100

((100/2)+(1/2))/100

= 50.5/100 = .505

It's slighly more likely that at least one person will pick up their own sweater.

[NetSquirrel]

Brilliant, cuan2002!  I attempted the same method but I counted 70 ways for b to be true instead of the correct 72, which kind of messed things up.  Your answer appears to be the first correct one!

 

However, the solution still remains to be worked out through logic and thought, rather than trial-and-error.  It's a real bug*er to do.

[merlinmynx]

to all who have contributed to this poser of the shirts:

I have stumbled upon the correct method of solving this riddle.

I will submit my answer on or after Saturday 15th January.

I can give an Excel spreadsheet solution, as well as A ONE LINE SOLUTION.

Be patient !!

Merlinmynx

[Gnu]

The problem that you have posed is a version of the misaddressed letter problem that was posed by Niclaus Bernoulli.

With two people, the probability of getting the wrong sweatshirt is 1/2, for three people it is 2/6 and for four people it is 9/24.  The general formula for the probability of n people all getting the wrong shirt is:

1/2! - 1/3! + 1/4! - 1/5! + 1/6! ... + 1/100!

The probability methods discussed fail because the probabilities do depend on each other, as jenstar says.  A listing method is tedious but does generate the first few cases easily enough e.g.

 

person A  B

s/shirt  a   b        i.e. 1/2 where all are wrong

           b   a

 

person A  B  C

s/shirt  a   b  c           i.e. 2/6 where all are wrong

           a   c  b

           b   a  c

           b   c  a

           c   a   b

           c   b   a

The problem is in 'The Penguin Book of Curious and Interesting Puzzles' and probably other places as well.