Make money with probability

smiles

Well, it's not so much a trick as an interesting bit of probability that our lecturer showed us.

He asked would anyone bet against him that in the class of 30 people that as least 2 people would share the same birthday... in fact he offered to put a fiver up if anyone would bet 1 euro.

He won (and he still gave the guy the god damn fiver) -- there's a 70.6% chance that in any random group of 30 people that they will have the same birthday.

cute or what?

I can provide the proof if anyone is so interested, but try it with a big group of mates, it's fun

<< Fio >>

GuanYin

Mneh... with only 365 days in the year the chance of anyones birthday falling on a particular day is 1/365.

In a large enough group of people its not really surprising that there are shared brithdays...which is a testament to how many bloody birthday presents I had to buy this month.

People have a skewed version of probability. I remember a few years back in F1 they marvelled over how two drivers set identical lap times in qualifying.... "what are the odds" they cried.

Strangely when nearly all 20 drivers were lapping within 100ths of a second of each other, the answer is "quite high".

dod

If you have the time and the inclination to post the proof Smiles, I'd like to read it. Sort of thing you could have some fun with in the right company after a few bottles of wine some night.

ecksor

You should be able to find a proof by searching for the 'Birthday theorem'. There's an associated attack on cryptographic hashes for finding collisions. Chapter 5 of 'Security Engineering' by Ross Anderson has a great explanation.

rainbow kirby

My stats lecturer did that last week, couldnt find 2 in a class of 60 who had the same birthday. probability of that is teeny!

Syth

Also I think that the odds of say 4 coming up at least once in the roll of 4 dice is bigger than the odds of one not coming up

dudara

it's similar to the probability of clustering in the lotto numbers, which is actually higher than people think.

GuanYin

would the probability of sequential lotto numbers not be the same as non sequestial? seeing as the sample spaces would both be the same?

dudara

No the odds are actually greater.

For instance draw the first number (1/42). Assuming now for simplicity that it isn't 1 or 42, to get a sequential number, either before or after, the odds are now 2/41 (greater than 1/41) and so on...

ecksor

The odds of picking an adjacent number are better than picking any other arbitrary number, but if you don't care precisely what the other number is, then the chances are 2/41 vs 39/41 for the next number, no?

GuanYin

but hold on, a sample space is all possible events and in the sample space for a lotto draw we are interested in the set of all sequences of 6 different numbers between 1 and 42. Then the probability of any sequence of numbers occuring is equal to any other.

Now, this is totally different to the individual probability of to the drawing of sequential numbers.

Capt'n Midnight

30 people in a room

so for first person chances of sharing birthday with another =29/364
so for first person chances of sharing birthday with another =28/363 (363 cos it can't be any of the other days)
etc.

How to put it together
do backwards - the chances of first person NOT sharing birthday are (365-30)/365 =0.9178 - multiply this by the odds of the next person not having a birthday and when the result gives you the odds... - ie less than 0.5 means better than evens...

Capt'n Midnight

Lotto Draw - 42 numbers 6 balls
1/42 * 1/41 * 1/40 * 1/39 * 1/38 * 1/37 =odds of sequence
- order of balls does not matter

so it's 42*41*40*39*38*36 / (1.2.3.4.5.6) = big number

GuanYin

Originally posted by Capt'n Midnight
Lotto Draw - 42 numbers 6 balls
1/42 * 1/41 * 1/40 * 1/39 * 1/38 * 1/37 =odds of sequence
- order of balls does not matter

so it's 42*41*40*39*38*36 / (1.2.3.4.5.6) = big number

hrmm so sample spaces don't apply to lotto draws?

Ok put it like this.

If my lotto numbers are: 5, 7, 9, 23, 29 and 40

and my friends are 1, 2, 3, 4, 5, 6

which has the greater probability of occuring. (I know the answer, I'm just making a point).

ecksor

I can't see any contradiction between the two of ye.

GuanYin

Oops, I didn't take in cpt's post properly:)

I've gotten very paranoid and and defensive since I started posting in humanities