Secret Santa Problem Watch

traintracks1995
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#1
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There's 30 employees where I work, every ones name is on a small piece of paper, and person A is invited to take a name at random from the hat. There is a 1/30 chance that they will take their own name, in which case they return it to the hat and pick again. What is the proablity that no one person ever picks their own name?
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TenOfThem
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(Original post by traintracks1995)
There's 30 employees where I work, every ones name is on a small piece of paper, and person A is invited to take a name at random from the hat. There is a 1/30 chance that they will take their own name, in which case they return it to the hat and pick again. What is the proablity that no one person ever picks their own name?
1/30
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traintracks1995
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(Original post by TenOfThem)
1/30
I don't think you quite understand. Yes the first person has a 1/30 chance of drawing their name, but what is the second persons chance of drawing their own name and so on until all 30 people have taken a draw. What is the chance that noone in the 30 drwas their own name ?
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Phichi
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(Original post by traintracks1995)
I don't think you quite understand. Yes the first person has a 1/30 chance of drawing their name, but what is the second persons chance of drawing their own name and so on until all 30 people have taken a draw. What is the chance that noone in the 30 drwas their own name ?
I may be wrong, and believe me, I suck at stats in comparison to any other area of maths, but surely it depends on what name has been pulled out? If the first person drew the second persons name, then the second person has a 100% chance of drawing a name that isn't his own. But if person 1 was to not draw it, person 2's chance of drawing a card that isn't his own would be 28/29. So i'm not really sure if its possible, perhaps look at it from this aspect? I'm thinking a massive tree diagram with every possible combination and outcome.
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traintracks1995
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(Original post by Phichi)
I may be wrong, and believe me, I suck at stats in comparison to any other area of maths, but surely it depends on what name has been pulled out? If the first person drew the second persons name, then the second person has a 100% chance of drawing a name that isn't his own. But if person 1 was to not draw it, person 2's chance of drawing a card that isn't his own would be 28/29. So i'm not really sure if its possible, perhaps look at it from this aspect? I'm thinking a massive tree diagram with every possible combination and outcome.
Exactly. There is a 1/30 chance that the first person will draw their own name. There is a (29/30)*(1/29) chance that the second person will draw their own name. There is a (29/30)*(28/29)*(1/28) chance that the third person will draw their own name etc etc
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Mr M
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(Original post by traintracks1995)
I don't think you quite understand.
Right back at you.
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BabyMaths
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(Original post by traintracks1995)
...
Google derangement.
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Bronco2012
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Let X = number of people who pick their own name


Let A = event of any given student picking their own name

P(A) = 1/30

Using indicator variable, we can trivially prove that E(X) = 1

ImX = [0, 30]
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Racoon
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Never mind about that, whatcha gonna buy?

I suggest one of those fake tattoo sleeves (especially if you pick the ceo's name)
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Holby_fanatic
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Use this website and it ensures no one gets their own name:

http://www.secretsantaelf.co.uk/

That awks moment when I don't realise it's just a maths problem. :lol:
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elohssa
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Yea you have a case of derangements here. The answer is !30 (derangements in 30 slots)/30! (total number of permutations).

97,581,073,836,835,772,079,377,0 81,171,968/30! = 0.36787944117

https://www.mathcelebrity.com/derang...e+Derangements
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