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bigmac366
08-31-2005, 06:57 PM
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.

BruceZ
08-31-2005, 07:02 PM
[ 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)!.

BruceZ
08-31-2005, 07:53 PM
[ 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%.

Lexander
08-31-2005, 10:51 PM
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.

BruceZ
08-31-2005, 11:21 PM
[ 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.

UATrewqaz
08-31-2005, 11:24 PM
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.

BruceZ
08-31-2005, 11:26 PM
[ 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.

SumZero
08-31-2005, 11:29 PM
[ 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.

UATrewqaz
09-01-2005, 12:41 AM
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

Lexander
09-01-2005, 01:05 AM
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).

BruceZ
09-01-2005, 01:15 AM
[ QUOTE ]
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.

[/ QUOTE ]

I could care less about an exact calculation. I took into account Feb. 29th for the same reason I took into account the other 365 days.

Lexander
09-01-2005, 01:18 AM
*smiles*

Fair enough BruceZ, wasn't trying to rain on your parade. Just figured it might be interesting to consider that the actual chance might be higher and favor the bet even more.

09-01-2005, 03:24 PM
I was under the impression that Amarillo Slim performed this bet with 30 people, not 25. I think this increases the odds to about %72.

TwoNiner
09-01-2005, 04:04 PM
He wrote about thirty in his book.

BruceZ
09-01-2005, 04:36 PM
[ QUOTE ]
I was under the impression that Amarillo Slim performed this bet with 30 people, not 25. I think this increases the odds to about %72.

[/ QUOTE ]

70.6%, assuming 365 equally likely b-days.

Here are some other stats:

mean: 24.6
median: 23
mode: 20

The mean is the expected value, or the arithmetic average of the number of people it takes to get a match if we repeated this experiment over and over with different people. The median is the number of people it takes to get a match 50% of the time (actually 50.7% here). The mode is the most likely number of people it takes to get a match, or the most common number that results in a match.