Would you accept any of these bets? Which one(s)? Why? Which would you be least inclined to take? Why?

## January 31, 2008

### You bet

You're at a party, one of twenty-five guests. You're the betting sort. One of the other guests walks up to you and offers you three possible bets:

That you share your birthday with someone else at the party.

That you share your birthday with her.

That at least two people at the party share a birthday.

Would you accept any of these bets? Which one(s)? Why? Which would you be least inclined to take? Why?

Would you accept any of these bets? Which one(s)? Why? Which would you be least inclined to take? Why?

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## 9 comments:

Long time reader, first time commentator.

I will go for the third bet, because the other two cases are special cases of this one, so it has the highest probability of happening.

I wouldn't go for the second one, since I wouldn't expect her to want to bet with me and lose.(she has definitely come up to me, knowing her birthday as well as mine, and hence knows that she will win).

But if the payouts were different, then it's a different matter altogether.

See,this is why I do not party.

No party, no bets. A blogger I am. (Or am I?)- I have no life.:)

Differently Challenged is right on money.

Accept (3) as probability says it takes 23 people to have any two people sharing birthday.

(1) is 24 times more attractive than (2), but not as good as (3).

Obviously I too will choose the third option, since it has the highest probability to success out of all the three. But where's the party man? I'm all set to go...

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The third option, for the following reason:

The second option has prob 1/365 (her birthday can be any 1 out of 365)

The first option has prob 24/365 (i.e. I have 1/365 prob to match with any of 24 people).

The third option has prob Sum(i from 1 to 24) i/365

(the first person has 24/365 prob of match (1/365 among 24 people), the second 23/365 (1/365 among 23 people, b/c we cant count the first person), and so on.

For a group of 25 this sums to about 82%. So its a decent bet!

Sorry, the above analysis is wrong.

The prob that all are different is:

1-(1-1/365)(1-2/365)(1-3/365)...(1-24/365)

which is about 57%.

The idea is, the second guy can have 364 birthdays, the third 363 birthdays, etc. to be different. One minus the prob that all are different is that at least one shares birthdays. I had seen this problem in a probability class a long time ago but forgot the answer!!

The probability of 1 (mine is same as one of the others) is

1-prod i .. 24 (364/365) as if all are different, each other person just needs a different birthday than me. This is not too different from the naive sum though.

Which bet I take depends on the Odds.

In the first case, the probability of the event occuring is 1 - (364/365)^24 = 6.37%

In the second case, the probability of the event occuring is 1/365 = 0.274%

In the third case, the probability of the event occuring is [(365-1)*(365-2)*(365-3)*......*(365-25)] / 365^25 = 43.1%

Give me right odds and I am willing to bet on any of these events!

Note: This is a simplistic calculation and I have not included the consideration of 29th Feb of the leap years.

Sorry, the third case in the above case should be

In the third case, the probability of the event occuring is 1 - [(365-1)*(365-2)*(365-3)*......*(365-25)] / 365^25 = 1 - 43.1% = 56.9%

wow i cant even toss~~~no coins hav 3 sides

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