1. ## dice

hello everyone!

If i roll two dice, how many times(on average) will I have to roll it in order to get at least one 6?

here's what i did, the probability is 1/36 for getting at least one 6 from two dice. So one divided by 1/36 gives 36, so I will have to roll them 36 times on average.
Yes?
thanks!

2. Is the probability of getting AT LEAST one 6 1/36?

There are three disjoint possibilities:
EVENT A: Dice-1 gets a 6 and Dice-2 doesn't,
EVENT B: Dice-2 gets a 6 and dice 1 doesn't,
EVENT C: Both dice get a 6.

So the probability of getting AT LEAST one 6 is the sum of the probabilities of these three events. P(A) + P(B) + P(C) = 1/6*5/6 + 5/6*1/6 + 1/6*1/6 = 11/36.

You can get this same number by considering the only other possibility which is:
EVENT D: Neither dice rolls a 6.

This probability is P(D) = 5/6*5/6 = 25/36, and of course we want 1-P(D) = 11/36.

3. Originally Posted by iknowone Is the probability of getting AT LEAST one 6 1/36?

There are three disjoint possibilities:
EVENT A: Dice-1 gets a 6 and Dice-2 doesn't,
EVENT B: Dice-2 gets a 6 and dice 1 doesn't,
EVENT C: Both dice get a 6.

So the probability of getting AT LEAST one 6 is the sum of the probabilities of these three events. P(A) + P(B) + P(C) = 1/6*5/6 + 5/6*1/6 + 1/6*1/6 = 11/36.

You can get this same number by considering the only other possibility which is:
EVENT D: Neither dice rolls a 6.

This probability is P(D) = 5/6*5/6 = 25/36, and of course we want 1-P(D) = 11/36.
Thank you so much, I must say that was really very well explained. I understood it straightaway!!

So, 1/ (11/36) will give me the average number of times I need to throw the dice, which is approx 3.3

4. I like that average better. Whenever you arrive at a result I think it is useful to ask yourself if your asnwer is reasonable. 36 throws of two dice before you see ANY 6's seems hard to believe. #### Search Tags

dice 