Problems
The Birthday GameThis rather popular party game seems counterintuitive. So your intuition was wrong. How could you have anticipated the actual result?.
Problem as Given
The people at a party agree to tell their birthdays (day and month only), one at a time. They do this until a birthday coincides with one previously given, or there are no more people to play. There are of course 365 days in the year (ignoring Leap Year Day for our purposes), and thus 365 different possible birthdays. Yet in any decently lively party, say with two dozen people present, it happens more often than not that two people actually do have the same birthday. What is going on?
AnalysisWhat appears to happen is that A and B have the same birthday, despite the seemingly tiny chance (1 in 365) that this should be the case. The actual problem has a different shape; it is a matter of each successive birthday not being the same as all previously mentioned birthdays. The model of successive events which interfere with each other's probabilities is relevant here.
We can visualize it better this way. Suppose we have a game table divided into 365 pockets. A ball thrown on the table must land in 1 of the pockets. If we define "win" as having a ball land in an unoccupied pocket, and if one person after another tosses a ball onto the table, how long before the chances are at least even that a ball will land in an already occupied pocket? There is some player in the series of winners, the nth, for whom the chance of also winning is for the first time less than even, meaning less than 0·500. What number is n?
The first toss of course is free; there are no occupied pockets. Thus the chance of landing in an unoccupied pocket is 365/365 (or 1·000, complete certainty). For the second person, 1 pocket is taken, and there are only 364 pockets available if a duplication is to be avoided. Thus the chance of landing in an unoccupied pocket is now 364/365 (or 0·997, still very good). For the third person, the chance of avoiding both the previous two is 363/365 (or 0·994). And so on down, the numerator being reduced by 1 at each step. This is thus a nonreplacive game, but here that affects only the numerator of the fraction; the denominator (the number of days in the year; the number of pockets in our model of the game, the number of places where the ball could land) stays the same throughout.
Since these are series events (each player in turn must be successful in evading all previous birthdays), we must multiply the probabilities together. It is this factor of multiplication of probabilities that makes the chance of avoidance decrease much faster than you think it is going to. Thus, the 20th player finds 19 pockets on the game table occupied by the dates given by the previous 19 players, and thus seems to have a 346/365 = 0·948 chance of winning (of avoiding all of them). This is indeed the chance for that one player. But when we consider the need for all previous players to have won (otherwise the game would have stopped before that point), the picture changes. Here are the first few chances, and the cumulative probability of success that they imply:
Player
1 2 3 4 5 6 7 8 9 10 Days Open
365 364 363 362 361 360 359 358 357 356 Total
365 365 365 365 365 365 365 365 365 365 Chance
1·000 0·997 0·995 0·992 0·989 0·986 0·983 0·981 0·978 0·975 Cum Chance
1·000 0·997 0·992 0·984 0·973 0·960
0·944 0·926 0·905 0·883
The line to watch is not the highlighted line showing the chance of that player to win, but the cumulative chance for that player and all previous players to have won. The chance of winning (that is, of continuing not to duplicate) for the 10th player is 356/365 = 0·975. But the joint chance that all 10 players in sequence will win is the product of the first ten probability fractions, which is the somewhat lower number 0·883. But that is still a winning situation (any odds better than 0·500 is a winning situation). We conclude that if 10 people tell their birthdays, the chance is that none of the birthdays will be the same. But let's keep on calculating:
Player
11 12 13 14 15 16 17 18 19 20 Days Open
355 354 353 352 351 350 349 348 347 346 Total
365 365 365 365 365 365 365 365 365 365 Chance
0·973 0·970 0·967 0·964 0·962 0·959 0·956 0·953 0·951 0·948 Cum Chance
0·859 0·832 0·806 0·777 0·747 0·716 0·685 0·653 0·621 0·589
The bottom line, again, is the one to watch. We are obviously getting close to the critical point, a probability of 0·500, where by definition it will be an even chance whether the next player wins or loses. Here are the figures for the next ten players:
Player
21 22 23 24 25 26 27 28 29 30 Days Open
345 344 343 342 341 340 339 338 337 336 Total
365 365 365 365 365 365 365 365 365 365 Chance
0·945 0·942 0·940 0·937 0·934 0·932 0·929 0·926 0·923 0·921 Cum Chance
0·556 0·524 0·492 0·462 0·431 0·402 0·373 0·346 0·319 0·294
So we see that though every player here calculated continues to have a better than 9 in 10 chance of winning individually, the chance that all players in turn will win declines much more rapidly. If there are 22 people at the party, the chances are slightly better than even that no two of them will have the same birthday. But if there are 23 or more, the chances are in favor of at least one duplication of birthdays.
The answer to the Birthday Problem is thus 23.
TipTo multiply strings of fractions on an ordinary 8-place desk calculator, which is the kind we assume you are using, the trick is not to exceed the calculator's capacity at any point in the proceedings. So rather than write out all the fractions and then start by multiplying all the numerators together:
365 · 364 · 363 · 362 . . . which will get you in trouble in no time at all, instead use successive multiplication and division
365 / 365 · 364 / 365 · 363 / 365 · 362 / 365 . . . and you will be OK. At least within the limits of this problem.
MoralMultiplying nearly identical fractions together for half an hour may seem like an unprofitable exercise, but in fact you can learn a lot from it. It is far superior to counting sheep. You get a feel for the way the product keeps declining, slowly at first, then increasingly faster. At 23 guests, the odds are just better than even (5 in 10) that two of them will have the same birthday. Toss in another six, and the odds increase to 7 in 10. It can all be read off the chart, and the chart can be compiled with a little time and a desk calculator. There is no mystery here, once you have the sense of how a series of multiplied fractions behaves. It is worth the half an hour.
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