# What is the probability of rolling a one in X throws of a die?

Geoff Millikan used Ask the Experts™
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The probability of not throwing a 1 in two die throws is 5^2 / 6^2  = 25/36 = 69.4%.   So the odds of throwing a 1 are just the reverse, 36/36 - 25/36 = 11/36 or 30.6%.  The formula scales nicely out to the Xth toss for example the probability of not throwing a 1 in ten die throws is 5^10 / 6^10  = 9765625/60466176 = 16.1%.

But instead of figuring out the probability of something NOT happening and doing subtraction, can the formula be written purely from the positive perspective?  Such that, "What's the probability of rolling a one in X throws of a die?"
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Commented:
The forward formula for probablilty of a 1 in two throws is the probability for each event minus the probability of BOTH events (which would otherwise be counted twice)...

i.e.
1/6  + 1/6   - 1/36  =  11/36

More genericly:

Probability of A or B
P(A or B) = P(A) + P(B).

In other words, the probability of A or B occurring is the sum of the probability of A and the probability of B (this assumes A + B cannot both occur). If there is a probabiilty of A and/or B occuring, then you must subtract the overlap (the "snake eyes" case)...

This scales as expected to 3 (etc.)...
For 3:
1/6 + 1/6 + 1/6 - (3 x 1/64)  - 1/216 = 91/216

i.e the probablity of each roll having a 1 MINUS the overlaps of TWO ones PLUS the overlap of THREE ones... etc.

This might be easier to visualize with Venn diagrams...

Commented:
This case has binomial distribution. So function you would use is:

As you can see, this works quite nice if you know x. In the specific case you're talking of, it's much easier to find x if you're looking at negative event (NOT 1) instead of positive (1).

But you can of course write another formula to find x in the positive.

HTH.

Commented:
I like the Venn diagram and I'm not understanding the overlap yet.

Maybe I'm not getting the overlap because I'm not sure there's actually 36 outcomes.  It seems like the net unique outcomes on two separate rolls of the die are 21, not 36.  Because throwing a 5 the first roll and a 6 on the second roll is the same net outcome as throwing a 6 the first roll and a 5 on the second roll.

http://www.edcollins.com/backgammon/diceprob.htm

By the way, it seems like throwing one die repeatedly is different math than throwing multiple die once....

Ugh, this is killing me.
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Commented:
throwing one die repeatedly is the same math as throwing multiple die once
Commented:

If you roll one die once, the odds of a "1" appearing are 1/6...

On the second roll it is also a one in six chance...

But the question was "what are the odds of getting a 1?"  !!

If you enumerate all 36 choices of 2 rolls, how many rolls have a 1?
How would you count them?

1-1, 1-2, 1-3, 1-4, 1-5, 1-6 and
1-1, 2-1, 3-1, 4-1, 5-1, 6-1

There are TWELVE listed (i.e. 1/6 + 1/6) but the FIRST in each group is actually a DUPLICATE of a SINGLE ROLL, so it must NOT be counted TWICE!  Thus it is SUBTRACTED (1/36)...

If you draw two overlapping circles (Venn diagram) the OVERLAP would be the 1-1...

Similarly if you drew THREE overlapping circles, you could see the areas where 2 at a time are added and the intersection of the 3 must be re-added back... and so on...
Commented:
PS: There ARE 36 outcomes!

If you don't understand THAT basic concept, do NOT go to a casino & try to play craps!

Although the "doubles" may confuse you, as they ARE each unique, the combo of (for example) 5-6 or 6-5 are actually 2 distinct rolls...

Hence the odds of rolling a TWO (1-1) is 1/36
but the odds of rolling an ELEVEN (5-6 OR 6-5) is 2/36!!

Commented:
Wow, thank you all!

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