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Secret Santa Math problem
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Originally posted by texansfan View PostLet's assume Sue picks 1st and John picks 2nd (because we're gentlemen).
Let's also assume they are not allowed to pick their own name.
If we have 29 names in the hat
then Sue has 1/28 chance of picking John
given that Sue actually picked John,
John will have 1/28 chance of picking Sue
so the probability is 0.0012755102
But if John and Sue don't get #1 and #2 picks then all bets are off and you have more splaining to do
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The answer I got using the different number of combinations formula with a given set of people is .246%
The formula is n!/(r! (n-r)!) where n=number of people and r= number of people in each group.
This is simplified to (n*(n-1)*(n-2)...*(n-r+1))/r!
So (29*28)/(2*1), because n-r+1=29-2+1=28
So that gives 812/2=406
There are 406 different combos of people in that office
Now, only one of the combos includes you and your selected secret Santa, so 1 out of 406 combos is what you're looking for
1/406=.00246
.246% is my answer
This is the simple formula. The problem with figuring the chances is that you have to consider the fact that it isn't a total arrangement of secret santas, it's a drawing (I'm guessing)
But, for the sake of argument, I think that this is a fair representation of the chance you have of being paired with your secret Santa
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Originally posted by stinkbelly View PostOr increase if they are the last two to draw and the only names left.
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Originally posted by Rounder View PostIn my head this should be easy to figure out but wanted to get with the math gurus on here to check my work.
29 people in an office and each draws one name out of a hat for secret santa. What are the odds that two people draw each others names (John draws Sue and Sue draws John).
My thought is each has 1/29 chance to draw the other so .0345% and you multiply that number together to give you a .00119% chance to draw each other. Is that correct?
Thanks for the help and please show your work
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Originally posted by popup_menace View PostThe answer I got using the different number of combinations formula with a given set of people is .246%
The formula is n!/(r! (n-r)!) where n=number of people and r= number of people in each group.
This is simplified to (n*(n-1)*(n-2)...*(n-r+1))/r!
So (29*28)/(2*1), because n-r+1=29-2+1=28
So that gives 812/2=406
There are 406 different combos of people in that office
Now, only one of the combos includes you and your selected secret Santa, so 1 out of 406 combos is what you're looking for
1/406=.00246
.246% is my answer
This is the simple formula. The problem with figuring the chances is that you have to consider the fact that it isn't a total arrangement of secret santas, it's a drawing (I'm guessing)
But, for the sake of argument, I think that this is a fair representation of the chance you have of being paired with your secret Santa
There are 14 combos of that 406 that will allow for a pair of people to have the others name. So 14/406 = 3.4%
I stick by my number, if you are asking the odds overall and not for just you.
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