Interview Questions And Answers 
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You are at a party with a friend and 10 people are present including you and the friend. your friend makes you a wager that for every person you find that has the same birthday as you, you get $1; for every person he finds that does not have the same birthday as you, he gets $2. would you accept the wager?
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Unless it’s a birthday party where everyone was born on the same date, no.
Assuming the general population has birthdays evenly distributed across the 365 days of the year, and the party population is sampled randomly from the general population…
Sounds vaguely like the statistics class example where the teacher asks how many students he will have to sample before finding any two with the same birthday. As I recall the answer comes out to an average of something like 15-20. But this is because it doesn’t matter which day - any two the same is fine.
In the question, the people must have the same *specific* birthday, lowering the odds very much (don’t remember enough statistics to calculate it). The odds are way against you on finding a match - and on top of that, you’re being paid half as much for each match. If you were being paid like $200 for each match you might have statistics on your side but in this situation it is obvious that everything is going against you. Pass on the bet.
Yes, if,
1. Let me find out first.
2. He cannot credit for anyone who I already find out, whether having or not having the same birthday of me.
my expected income:
15{(1/365)*1+(364/365)*(-2)}
which is negative!
i dont expect a profit and so won take the chance!
For my friend, with a probability of having no same B’Days, as the chances are more, gets $2 for every Bdays that doesnt match. For him, even I can be taken into consideration as its not mentioned anywhere that we two need to be excluded. This is the only case that he doesnt get his bet of $2. But in neither case will I be benefitted. Hence will not take the bet.
No. No need to even use a calculator or a scratch pad to come out with the answer.
The reason is that the probability of each person having the same birthday as mine is small compared to the complement event, say epsilon. Now,
Espilon * 1 + (1-Epsilon)*(-2)
if all of them have the same b’day he will get $9.but the
friend will get $18, so its loss for him.
Here its not mentioned who is going to pay for a success or a failure if there is a 3rd person who will pay two both of us then I will accept the wager since in this case I will always be in profit however, my frnd has probability of winning more than me.
My expected Profit = 1*(1/365)+2*(1/365)^2+3*(1/365)^3+…+10*(1/365)^10 = 1 (approx)
The above expression is AGP. Refer the formula for
sum of n terms
Friend’s expected profit = 2*(364/365) + 4*(364/365)^2 + .. + 20*(364/365)^10 = -18*365 (approx)
My profit (+ve) > My friend profit (-ve)
Hence I accept the bet.
http://en.wikipedia.org/wiki/Birthday_paradox
So, “statistically”, I’m almost guaranteed that every 57th person I meet has the same birthday as me.. But that still leaves the other 56 people who don’t..
hmm.. so, if I take the bet, my friend makes at least $112 more than me (or $56 more even if it’s an equal wager)
The question does not say who asks the people. Thus, if I ask all of the party attendees, I may luckily find someone with the same birthday as mine. But my friend has no chance to find by himself.
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