Quote from: Azlan on June 06, 2014, 10:00:33 PM

Quote from: dooglus on June 05, 2014, 08:38:41 AM

Consider rolling a 6 sided die until you roll a six. The expected number of rolls to get a six is 6.

True, but incomplete

Only because you deleted the next line that I wrote:

Quote from: dooglus on June 05, 2014, 08:38:41 AM

But if you only roll 4 times, you have about a 51% of seeing a six, even though you've rolled less than the expected number.

The whole point I am making is that there is a difference between "the expected number of rolls to see a 6" and "the number of rolls you need to have a 50% chance of seeing a 6".

Quote from: Azlan on June 06, 2014, 10:00:33 PM

Take your die: You want to roll a six. You roll 6 times. Whats the probability of not seeing the six?
5/6 ^ 6 = 33.48% So a third of the time (with six observations) you will not see the 6 at all.

Read about what "expected value" means.

It sounds like you think it means "most likely value", but it doesn't.

Take your 1000 sided die. Let's play a game. If you roll a 6, you win a million dollars, but if you roll anything else, you lose a dollar.

What's the expected value of rolling the die? Almost every time you roll (99.9% of the time), you lose a dollar. So you 'expect' to lose almost every time.

But that's not what expected value means. Your expected value for each roll is +$999.001:

(1 * 1e6 - 999 * 1) / 1000 = 999.001