f course, you recognized at once that the wager is an elementary exercise in probability theory, and you remembered the solution: That, given as few as 23 persons selected at random, there is a better than an even chance that at least two of them have the same birthday {HyperNote}.
Both you and the statistician were able to count only 20 persons, so it was not an even bet. Based on numbers, you would likely lose -- a situation made decidedly worse by the evidence of a birthday being celebrated at the nearby table, unless..., unless...
You would expect that, even while losing her own bet, your friend might have wished you a happy birthday.

HyperNote

he solution is traditionally found by first reversing the statement of the problem:

Selecting at random, how many people are needed to assure an even bet that at least two of them share the same birthday?
Selecting at random, how many people are needed to assure an even bet that no two of them share the same birthday?
For convenience, let's let...
n = the unknown number of persons,
p = the probability that two or more persons share the same birthday, and
q = the probability that no two persons share the same birthday.
You seek the minimim value for n such that p > 1/2. You know that p = 1 - q, and you can easily find the value of q for two extreme values of n...
If n = 1, then q = 1 and p = 0 -- impossible: no other person with whom to share a birthday.
If n = 365, then q = 0 and p = 1 -- certainty: no more birthdays to spread around without duplicates.
For every value of n, there are 365n possible combinations of birthdays, which is absolutely all there can be. Accordingly, that is what goes into the denominator of the fraction used to calculate your probability q. Your numerator requires only slightly more sophistication in numbers.
Any person you might select at random will have one of 365 birthdays. That leaves only 364 birthdays for the second person, in order to satisfy the part of the conditions mandated by probability q which says "no two of them share the same birthday." Having given one of those birthdays to the second person, you will have only 363 birthdays for the third person. And so forth.
You can make a table and a graph as follows:

 n q p 1 (365) (365) 0 2 (365)(364) (365)(365) 0.00274 3 (365)(364)(363) (365)(365)(365) 0.00820 4 (365)(364)(363)(362) (365)(365)(365)(365) 0.01636 ... and so forth ... 23 Hoo-ha! 0.507297