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[TEXASGUY]

If there are 30 people in a room, what are the odds that two of the 30 has the same birthday? And if you get that one, this one will probably be easy. What is the probability that if you name three cards, preflop, say 10,2 and 4, that one of these three will come on the flop? I cant figure either out, so if you can help it will be appreciated, also the math would be great.

[Tom1975]

The birthday problem has been addressed many times on here. If you do a search you'll probably find something. If not, just google 'birthday problem'.

As for the second question, the odds are 1-C(40,3)/C(52,3)=55.2%

[BruceZ]

[ QUOTE ]
If there are 30 people in a room, what are the odds that two of the 30 has the same birthday? And if you get that one, this one will probably be easy. What is the probability that if you name three cards, preflop, say 10,2 and 4, that one of these three will come on the flop? I cant figure either out, so if you can help it will be appreciated, also the math would be great.

[/ QUOTE ]

Note that these problems are different in that cards are drawn without replacement, while b-days are "drawn" with replacement, meaning that multiple people can have the same b-day, but once a certain card is drawn, there is one less of that denomination.

The probability that 2 or more people have the same b-day out of 30 people is 1 minus the probability that all 30 have different b-days, or

1 - (365*364*363*...*336)/365^30

= 1 - P(365,30)/365^30 =~ 70.6%.

Where P(365,30) = C(365,30)*30! is the number of permutations of 30 days chosen out of 365 days. The difference between P and C is that C ignores order, while P does not. P can be computed with the Excel function =PERMUT.

This solution ignores leap year (which makes very little difference as this calculation for 25 people showed) and assumes that b-days are uniformly distributed among 365 days. If in fact some days are more likely than others, then the probability that 2 or more people have the same b-day will be higher.

[AaronBrown]

Another difference between the two problems is in the second one you're fixing the numbers of cards that must be matched, while in the birthday problem each person who doesn't match a previous person's birthday gives a new opportunity to match. That's why it takes surprisingly few people to have a 50% chance of a match. If you pick a single birthday, you need 183 random people to have a 50% chance of someone matching it.

If you deal cards until you get a pair, it takes 5 cards to have a 50% chance of a pair, after 9 cards you have a 95% chance of a pair.