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  #1  
Old 08-31-2005, 06:57 PM
bigmac366 bigmac366 is offline
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Default amarillo slim\'s favorite bet

In the august 2005 article of cardplayer magazine, there is an interview with evel knievel. In this interview he states that " one of slims favorite tricks was to bet that two of any 25 people chosen at random would have the same birthday. he always won that bet because the math was rediculously in his favor". can anyone here figure the exact odds on this bet? thanks.
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  #2  
Old 08-31-2005, 07:02 PM
BruceZ BruceZ is offline
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Default Re: amarillo slim\'s favorite bet

[ QUOTE ]
In the august 2005 article of cardplayer magazine, there is an interview with evel knievel. In this interview he states that " one of slims favorite tricks was to bet that two of any 25 people chosen at random would have the same birthday. he always won that bet because the math was rediculously in his favor". can anyone here figure the exact odds on this bet? thanks.

[/ QUOTE ]

Ignoring Feb. 29th, the probability of a match is:

1 - P(365,25)/365^25 =~ 56.9%

where P(365,25) = 365*364*363*...*341 (25 terms) = 365!/(365-25)!.
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  #3  
Old 08-31-2005, 07:53 PM
BruceZ BruceZ is offline
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Default Re: amarillo slim\'s favorite bet

[ QUOTE ]
[ QUOTE ]
In the august 2005 article of cardplayer magazine, there is an interview with evel knievel. In this interview he states that " one of slims favorite tricks was to bet that two of any 25 people chosen at random would have the same birthday. he always won that bet because the math was rediculously in his favor". can anyone here figure the exact odds on this bet? thanks.

[/ QUOTE ]

Ignoring Feb. 29th, the probability of a match is:

1 - P(365,25)/365^25 =~ 56.9%

where P(365,25) = 365*364*363*...*341 (25 terms) = 365!/(365-25)!.

[/ QUOTE ]

Here is the adjustment for Feb. 29th, which makes the answer closer to 56.8%:

P(match) = 1 - P(no match) =

1 - [ P(no Feb 29th)*P(no match given no Feb 29th) +
P(1 Feb 29th)*P(no match given 1 Feb 29th) ]

1 - { [1 - 1/(4*365.25)]^25 * P(365,25)/365^25 +
25*1/(4*365.25)*[1 - 1/(4*365.25)]^24 * P(365,24)/365^24 }

=~ 56.8%.
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  #4  
Old 08-31-2005, 10:51 PM
Lexander Lexander is offline
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Default Re: amarillo slim\'s favorite bet

I won't agree completely that this is the correct number. The calculation is fine, but the assumption of equally distributed birthdays is invalid (amusingly, this very topic was recently discussed in a recent issue of Amstat they print for students such as myself). Without thinking at all about the math, my first guess would be the chances are higher.
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  #5  
Old 08-31-2005, 11:21 PM
BruceZ BruceZ is offline
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Default Re: amarillo slim\'s favorite bet

[ QUOTE ]
I won't agree completely that this is the correct number. The calculation is fine, but the assumption of equally distributed birthdays is invalid (amusingly, this very topic was recently discussed in a recent issue of Amstat they print for students such as myself). Without thinking at all about the math, my first guess would be the chances are higher.

[/ QUOTE ]

The birthday problem is an old and famous problem, and it is generally assumed in the problem statement that there are 365 equally distributed birthdays. This case is all you need to realize the essential point that 25 people give you a huge advantage on an even-money bet.
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  #6  
Old 08-31-2005, 11:24 PM
UATrewqaz UATrewqaz is offline
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Default Re: amarillo slim\'s favorite bet

We did this problem in statistics class, apparently once you get over 20 something people, I forget, you have over a 50% chance and the teacher said you could win alot of bar bets with this.

We did this in a class of about 30 people and there were indeed 2 people with the same birthday.
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  #7  
Old 08-31-2005, 11:26 PM
BruceZ BruceZ is offline
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Default Re: amarillo slim\'s favorite bet

[ QUOTE ]
We did this problem in statistics class, apparently once you get over 20 something people, I forget, you have over a 50% chance and the teacher said you could win alot of bar bets with this.

[/ QUOTE ]

23.
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  #8  
Old 08-31-2005, 11:29 PM
SumZero SumZero is offline
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Default Re: amarillo slim\'s favorite bet

[ QUOTE ]
I won't agree completely that this is the correct number. The calculation is fine, but the assumption of equally distributed birthdays is invalid (amusingly, this very topic was recently discussed in a recent issue of Amstat they print for students such as myself). Without thinking at all about the math, my first guess would be the chances are higher.

[/ QUOTE ]

Yes, the worst case for the person betting there will be a match is uniformly distributed birthdays. If the birthdays are not uniformly distributed then the odds of a match become better.
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  #9  
Old 09-01-2005, 12:41 AM
UATrewqaz UATrewqaz is offline
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Default Re: amarillo slim\'s favorite bet

Birthdays are nto uniformly distributed but pretty damn close to it, this is because people have alot more sex during some periods.

I remember in 1996 whent hey had the huge blizzard int he northeast that lasted like a whole month, 9 months later there was a huge boom in newborns. Apparently during the blizzard people were forced to stay inside and decided to pass the time screwing
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  #10  
Old 09-01-2005, 01:05 AM
Lexander Lexander is offline
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Default Re: amarillo slim\'s favorite bet

Admittedly, I was just being picky, but when somebody takes into account leap year, they are trying for an exact calculation. The problem is that the exact calculation makes an assumption that is convenient, but not true, so exact calculations are a bit meaningless anyhow.

[BruceZ has pointed out that in fact he was not going for an exact calculation, which is pretty obvious once I see that nice little symbol for approximation. My bad).
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