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noncom Regular user Birmingham, UK 125 Posts |
One of those strangely counterintuitive statistics I have heard relates to the following question - what number of randomly selected people does it require before the odds of two of them share a birthday exceed 50%? What number before the odds exceed 90%?
Contrary to what anyone I've ever asked would guess, the answers are (I think) 24 and 36 respectively. My question is: can anyone think of a neat way to use this in a performing situation? From a mentalism perspective, I could imagine using it in a group of 30 or so people and "divining" that two people in the group share a birthday, before going on to do something else relating to starsigns etc. Any other ideas? Thanks Andy
It ain't what you do, it's the way that you do it - Bananarama.
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rgranville Elite user Boston area 463 Posts |
By my calculations, the odds of at least two people having the same birthday hit 50% at 23, 90% at 41, and 99% at 57. With 30 people, your chances are about 70%. While you can divine that two people from a group of 30 share the same birthday, I would think determining which two would rapidly get tedious in a performance.
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Nir Dahan Inner circle Munich, Germany 1390 Posts |
It is also quite a known fact these days.
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landmark Inner circle within a triangle 5194 Posts |
A much harder, though similar looking problem is this: The spectator pulls out a dollar bill. What are the odds that two of the eight serial number digits match? What are the odds that three of the digits match? I've driven math PhD’s crazy with this question, Karl Fulves has, I believe, an incorrect answer in Self Working Number Magic (I think that's the name of it). My brother, quite the math whiz (MIT graduate) claims to have solved this problem, and his proof while somewhat complicated, looks correct to me (I'm just okay in math, but I can follow a probability argument, since I've taught high school math for many years).
I'll give you a chance to come up with your answer, so I won't post my brother's answer for a while. Jack Shalom Okay, I'm editing this because I'm mis-stating the problem. See my next post. Actually this problem is easy: Probability of no matches is (9/10) x (8/10) x (7/10) . . . x (3/10). The probability of at least one match is simply the above subtracted from 1, which gives us a more than 98% chance that there will be two numbers the same. The probability that three numbers match is more difficult to figure out, but do-able if you have a strong math background. I'll leave that one to you. However the really tough problem, the one that has stumped PhD’s is the one described in my next post.
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
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Nir Dahan Inner circle Munich, Germany 1390 Posts |
Jack,
Are there any restrictions on the digits because of some laws of bills serial numbers, e.g. the last digit is a parity digit and so on. Nir |
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landmark Inner circle within a triangle 5194 Posts |
I totally screwed up here. I misstated the problem. The problem as it appears in Fulves (with the incorrect answer), and which my brother solved is this:
You name two digits at random. The spectator pulls out a dollar bill with an eight-digit serial number. (Assume serial numbers are totally random). Now, what is the probability that both named digits appear on the bill? Now, suppose you name three digits at random. What is the probability that all three show up in the serial number? Jack
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
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TomasB Inner circle Sweden 1144 Posts |
I assume the two chosen random digits are different and called A and B.
P(at least one A AND at least one B) = P(at least one A) + P(at least one B) - P(at least one A OR at least one B) = (1-(9/10)^8) + (1-(9/10)^8) - (1-(8/10)^8) = 0.30683774 And if we assume that we instead select three different digits A, B and C the probability is P(at least one A AND at least one B AND at least one C) = P(at least one A) + P(at least one B) + P(at least one C) - P(at least one A OR at least one B) - P(at least one A OR at least one C) - P(at least one B OR at least one C) + P(at least one A OR at least one B OR at least one C) = 3*(1-(9/10)^8) - 3*(1-(8/10)^8) + (1-(7/10)^8) = 0.15426684 /Tomas |
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landmark Inner circle within a triangle 5194 Posts |
Very nice indeed, Tomas. You are absolutely correct. Just curious, what's your math background?
Jack Shalom
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
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TomasB Inner circle Sweden 1144 Posts |
Jack, just to be sure of the answers I cheated and made 10 million simulations each of the tasks. I've got a M.Sc in Applied Physics and Electrical Engineering, profile of specialization Signal and Image Processing.
/Tomas |
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Ben Blau Inner circle 1475 Posts |
Bill Simon published some ideas along these lines (same birthday odds). Some of that material can be found in his "Mathematical Magic" book.
Ben Blau
Ben Blau
http://www.benblaumentalism.com |
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Zeiros New user 59 Posts |
Quote:
On 2003-09-06 00:24, rgranville wrote: This is quite right, and although I know that this isn't strictly related to the first post, there is a nice site that demonstrates how this is calculated here: http://www.people.virginia.edu/~rjh9u/birthday.html Anyway, I found it interesting and hopefully you will too |
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TomasB Inner circle Sweden 1144 Posts |
A related problem: You stand in line with a bunch of other people. The first one to share a birthday with any of the people in front gets a prize. Where in the line should you stand to get the biggest probability of being the first to share your birthday with someone in front of you?
/Tomas |
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rgranville Elite user Boston area 463 Posts |
Let's see if I can do this. As with all of the above calculations, we'll ignore leap years.
If there are n people in front of you in the line, the probability of winning the prize is the probability of n people having different birthdays times the probability of you having the same birthday as one of those people. Note this is NOT the same as 1 minus the probability of n+1 people having different birthdays, that's the probability of at least two people having the same birthday, and that won't (necessarily) get you the prize. With n people in front of you, the different possible combinations of birthdays is 365**n. The number of different possible combinations of n unique birthdays is 365Pn or 365! / (365 - n)! So the probability of n people having different birthdays is 365Pn / 365**n The number of combinations of n different birthdays where one of them matches yours is n * 364P(n-1) (If one of them matches you, the other n-1 people have to have birthdays from the remaining 364, and any one of the n could be the match.) We already know there are 365Pn combinations of possible unique birthdays for the n people, so the probability of one of them matching you is (n * 364P(n-1)) / 365Pn Hey! The numerator of the first probability cancels out the denominator of the second, so for our final probability of winning the prize, we get (n * 364P(n-1)) / 365**n Let the computer do the number crunching, and we find that being in position 20 (19 people in front of you) has the highest chance of winning, with a probability of 0.0323. |
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TomasB Inner circle Sweden 1144 Posts |
Very nice, and I'm not just saying that to get my 50:th post.
/Tomas |
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Scott Cram Inner circle 2678 Posts |
Quote:
On 2003-09-25 14:34, TomasB wrote: Welcome to the Banquet room! Enjoy Secret Sessions and more! |
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landmark Inner circle within a triangle 5194 Posts |
Awesome, rgranville.
Jack
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
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Nathan Kranzo V.I.P. 2955 Posts |
The big question is this...
How do you go about finding the matches? The elimination process seems long and full of dead time obviously. I'm wondering if there is some simple way to do this for a large group and quickly figure out a few matches.... Anyone? Kranzo
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TomasB Inner circle Sweden 1144 Posts |
Nathan,
You can start by asking everyone born in January to stand up and simultaneously call out their birthdates in that month. You (and them) will hear if one or more says the same number. Continue with February etc. At some months no one might stand up and the match will probably not happen in December so the procedure would often be pretty quick I think. /Tomas |
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landmark Inner circle within a triangle 5194 Posts |
Tomas has been running a simulation for the last four years to test this method out.
Jack Shalom
Click here to get Gerald Deutsch's Perverse Magic: The First Sixteen Years
All proceeds to Open Heart Magic charity. |
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TomasB Inner circle Sweden 1144 Posts |
...and I found that in a group of 1 person they all have the same b-day.
/Tomas |
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