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SuperRookie
Male Standard Member
116 Posts


Brooklyn, New York
USA

Ducati

'04 M620ie

Posted - 12/14/2010 :  12:42 PM                       Like
I have a probability question and am trying to figure out "how to figure it out".

I know this is TOTALLY random, but look at it like a brain teaser...so here goes:

We're doing Secret Santa at my workplace and we have 41 people signed up. If you choose your own name, obviously you must put it back in the bag and choose again. What are the odds that the last person to choose will pick their own name?

James R. Davis
Male Administrator
17375 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 12/14/2010 :  1:02 PM Follow poster on Twitter  Join poster on Facebook as Friend  
Without rigorous thought, I'll take a stab at it.

There are only two possible outcomes: either your odds of drawing your own name are one in 41 if you are last to draw and the second to last drew a name and kept it, or zero if the second to last did return the name he drew.
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aidanspa
Male Advanced Member
1740 Posts
[Mentor]


Omaha, NE
USA

Harley-Davidson

Road King

Posted - 12/14/2010 :  1:46 PM
I would say zero. If every person returns their own name and draws again until they select another person, then by the last draw the remaining name must belong to somebody else.

Or not.
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SuperRookie
Male Standard Member
116 Posts


Brooklyn, New York
USA

Ducati

'04 M620ie

Posted - 12/15/2010 :  9:18 AM
Hmm...why did I think it would not be that simple? There's no way to truly know when your name was picked, but the odds increase every time a name is picked and yours remains in the hat. I thought there would be some multiplier of the odds because of that. I clearly don't understand probability theory
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James R. Davis
Male Administrator
17375 Posts
[Mentor]


Houston, TX
USA

Honda

GoldWing 1500

Posted - 12/15/2010 :  10:25 AM Follow poster on Twitter  Join poster on Facebook as Friend  
LOL

The idea that the odds are zero can only be true if the second to last to draw returns his name.

Assume only three players (A, B, and C) and you are 'C'. A draws B, B draws A. Then it's your turn and you WILL draw C.
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Moses
Male Senior Member
377 Posts
[Mentor]


Grand Rapids, Michigan
USA

Harley-Davidson

FX Softail

Posted - 12/15/2010 :  11:17 AM
I'd say 50-50. Either someone else has picked your name by now, or they haven't, regardless of when it happened. When it gets to the last name, THEN the odds are 50-50 that it's already happened.

That's my guess.
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aidanspa
Male Advanced Member
1740 Posts
[Mentor]


Omaha, NE
USA

Harley-Davidson

Road King

Posted - 12/15/2010 :  12:07 PM
Probability (Odds) That Player Index N Draws Ticket Number N
By Ion Saliu, Probability Scientist At-Large

Problem:
There are n people, each of whom writes their name on a piece of paper and puts that piece of paper into a hat. Each person takes it in turn to remove a piece of paper from the hat, if the piece of paper they remove has their own name written on it they put it back in the hat and choose again. What is the probability that the last person chooses their own name?

The final probability becomes:

p= 1 / {N2 - N + 1}

p= 1 / {412 - 41 + 1}

p= 1 / {1,681 - 42}

p= 1 / 1,639

p= .0006
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SuperRookie
Male Standard Member
116 Posts


Brooklyn, New York
USA

Ducati

'04 M620ie

Posted - 12/15/2010 :  12:59 PM
Well, no, it can't be 50/50 until there are only two names left and one of them is yours. Incidentally, as it so happened, only one person drew their own name in our drawing, and that was pretty early on. But, using James' ABC example...suppose we have seven participants A thru G where ("-->" = chooses)
A-->B
B-->C
C-->D
D-->E
E-->F and
F-->A,
the odds are 100% at that point that G chooses G, their own name because that's all that's left. As long as there's an odd number of participants the possibility exists for this to happen, but only to the person choosing last, of course, right? I just don't understand how to represent it and it seems there should be some ridiculously small number to articulate the odds of this happening, especially if a high number of people are involved, no? I know that number gets bigger as long as your name remains in the hat.

Yeesh, this is such low-level logic...what am I missing??



EDITED...again lol

Aidanspa, Thanks...not such low-level logic after all eh? So, the odds are 3/5000.

Edited by - SuperRookie on 12/15/2010 1:13 PM
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aidanspa
Male Advanced Member
1740 Posts
[Mentor]


Omaha, NE
USA

Harley-Davidson

Road King

Posted - 12/15/2010 :  1:21 PM
Yup, or 1/1639 according to that guy...but that was using a hat and not a bag.

Edited by - aidanspa on 12/15/2010 2:05 PM
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SuperRookie
Male Standard Member
116 Posts


Brooklyn, New York
USA

Ducati

'04 M620ie

Posted - 12/15/2010 :  1:41 PM
Ok, now I'm losing it... Feels like Monday for me. I haven't been able to think straight since I put my bike in storage two weeks ago
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aidanspa
Male Advanced Member
1740 Posts
[Mentor]


Omaha, NE
USA

Harley-Davidson

Road King

Posted - 12/15/2010 :  2:06 PM
I feel your pain, believe me I do.
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