Probability of Rolling a 6 on One Die

Date: 30 Apr 1995 02:51:31 -0400
From: Anonymous
Subject: Dice

The following questions came up during a game of Risk...

First, we know that the odds of rolling a six with one die are 1 in six. What
are the odds of rolling a six in two, three, four (etc.) rolls? It can't be
1/6 + 1/6 + ... or else you would have a 100% chance of a six in 6 rolls. How
do the probabilities add up?

Secondly, my friend says that if you roll several low numbers first, your
chances are then better of getting a high roll. While he admits that this
makes no logical sense probability-wise, he swears that there has been
research done to back it up. Ever hear of anything like this?

Thanks in advance,
cantalupo@aol.com

Date: 1 May 1995 15:00:35 GMT
From: Daniel Eisenbud
Subject: Re: Dice

The odds of rolling a six with each of two consecutive rolls is
1/6 * 1/6 = 1/36.  This makes sense, if you realize that when you roll
the second die, you will only have rolled a 6 on the first one out of
six times.  So there's only a 1/6 chance that if you do roll a six
on the second die, it will be your second six.
The probability of rolling n sixes in a row is in fact (1/6)^n, for
similar reasons.

>Secondly, my friend says that if you roll several low numbers first, your
>chances are then better of getting a high roll. While he admits that this
>makes no logical sense probability-wise, he swears that there has been
>research done to back it up. Ever hear of anything like this?

No.  The chance of rolling a six is always 1/6, no matter what you roll
before it, no matter what.

-Dan "Dr. Math" Eisenbud

Date: 11 Feb 2000 13:26:22 GMT
From: Dr. TWE
Subject: Re: dice

Hi! I think Dan misunderstood your first question. Are you asking what is the
probability of rolling _at least_ one six on two, three, four (etc.) rolls?
Let's examine this roll by roll.

In one roll, the probability of rolling a 6 is 1/6.

For two rolls, there is a 1/6 probability of rolling a six on the first roll.
If this occurs, we've satisfied our condition. There is a 5/6 probability
that the first roll is not a 6. In that case, we need to see if the second
roll is a 6. The probability of the second roll being a 6 is 1/6, so our
overall probability is 1/6 + (5/6)*(1/6) = 11/36. Why did I multiply the
second 1/6 by 5/6? Because I only need to consider the 5/6 of the time that
the first roll wasn't a 6. As you can see the probability is slightly less
than 2/6.

For three rolls, there is a 1/6 probability of rolling a six on the first
roll. There is a 5/6 probability that the first roll is not a 6. In that
case, we need to see if the second roll is a 6. The probability of the second
roll being a 6 is 1/6, giving us a probability of 11/36. There is a 25/36
probability that neither of the first two rolls was a 6. In that case, we
need to see if the third roll is a 6. The probability of the third roll being
a 6 is 1/6, giving us a probability of 1/6 + (5/6)*(1/6) + (25/36)*(1/6) =
91/216. Again, this is less than 3/6.

The general formula for rolling at least one 6 in n rolls is 1 - (5/6)^n.

One final note: the probability of rolling _exactly_ one 6 in n rolls is
different, and a little more difficult to compute.

- Doctor TWE, The Math Forum
  http://mathforum.org/dr.math/