Help with probability

Blisterman

I did this in the leaving cert, but I've forgotten how to do it.

If the odds of something happening is X% and you do it Y times, what are the odds that it will happen at least once.

For example, the odds of a die rolling a 3 is 1 in 6.
If you rolled the die 4 times, what are the chances of getting a 3 at least once.

What's the formula for this?

Capt'n Midnight

you do this backwards - on each throw of the dice you have 5 chances of NOT getting a three, so after four throws what is the chance of still not having a three -
5/6 x 5/6 x 5/6 x 5/6 = 0.4822

so the chance of getting a three is ( 1 - 0.4822) ie. slightly better than 50:50

Sev

What is interesting to note, is that..

If the the odds of getting a 4 are 1/6, and you roll the dice 6 times, the odds of getting a 4 at least once is a little over 66%. You can work this out the same way as above.

So what if the odds of getting your result are 1/777 and you attempt it 777 times? (or more generally, the odds are 1/N and you attempt N times)

Well you get roughly the same answer, and in the limit where N goes to infinity, the odds of getting your result at least once approaches the limit of roughly 63%, or (e-1)/e.

I like to use this as a general rule. For example, in poker. The odds of being dealt pocket aces are 1/221. So what if you deal cards to a player 221 times in a row? what are the odds of getting aces at least once? about 63%.

Or if the odds of winning the lotto are 1 in 4 million, and 4 million people each buy a lotto ticket and each independently choose random numbers (much like in reality), whats the odds that at least one person wins? again its roughly 63%.

Blisterman

Ok, thanks, thats all I wanted to know.