(just a real life story, slightly to do with calculators; you may like it)



Once a year the "nationale wetenschapsquiz" (science quiz) is being held in the Netherlands. It consists of 20 questions that can be answered by knowing, reasoning or guessing. Every year in December the questions and answers are discussed in a television broadcast.

Back in the year 2000 question nr.11 was about 26 people (in a café, but that part is not relevant to know): "what is the chance that 2 of these people have their anniversary on the same date?" You could choose from 30, 60 or 80 percent.



After some heated discussions about this question between several colleagues of mine, I simply reasoned the answer out. After this it was a matter of calculating the formula, were it not that most calcs cannot do this in one go.



Now my questions are: What is the answer and how did you compute it on which HP calc?





Bram

This is known as the "birthday problem" or "birthday paradox", though in my opinion there's no paradox involved.

The probability that two people out of n share the same date is 1 minus the probability that they each have a unique date.
Disregarding leap days, for n people, the probability of all having unique dates is 365Cn/365^n.

Here is an untested program for the HP-41, run with n in X:

LBL BP

1

LBL 01

366

RCL ST Z

-

*

365

/

DSE ST Y

GTO 01

1

X<>Y

-

END

A similar program (also untested) should work on most other RPN calculators by storing the iteration variable in a numbered register rather than on the stack:

LBL B

STO 0

1

LBL 1

366

RCL 0

-

*

365

/

DSZ 0

GTO 1

1

X<>Y

-

R/S

Quote:

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the probability of all having unique dates is 365Cn/365^n.

---

Shouldn't this be 365Pn/365^n?

1 - 365Pn/365^n = 0.598240820 or 59.8% (for n=26), in accordance with your (tested) HP-41 program.

Thanks Eric, for your contribution.

My comments were already formulated (the last part may be the paradox you mentioned):



To start with, the heated discussions were about the precise meaning of the question. Was or wasn't it allowed to have a date with more than 2 anniversaries? Was or wasn't it allowed to have (more than) 2 couples of anniversaries? To my opinion the question means in normal every day language that we are interested in knowing the chances of an occurrence that at least two people celebrate on the same day. I came to that conclusion (and to a computable answer) by reversing the question: how about each and every of the lot celebrating on different days?

That's very easy to answer by the definition of 'chance':

# wanted situations / # possible situations.



# wanted situations = 365x364x363...x340 (365!/(365-26)!)

# possible situations = 365x365x365...x365 (365**26)

(of course you'll have to change 365 by 366 for a leap year, but that doesn't make much difference)



With for example Free42 (a real HP-42S is slightly too "limited") you can easily calculate the quotient directly, but otherwise you'll have to program (almost any HP calc will do).

Chance=1 ; FOR i=0 TO 25 ; Chance=Chance*(365-i)/365 ; NEXT

which may translate for the HP-41C(V) into this:

*LBL CHANCE

1

-

1000

/

STO 00

1

STO 01

*LBL 01

365

RCL 00

INT

-

ST* 01

365

ST/ 01

ISG 00

GTO 01

RCL 01

Quote:

---

Shouldn't this be 365Pn/365^n?

---

Yes, the "C" was a misteak.