Diceand the Laws of Probability

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by

EdwardD. Collins

For more questions and problems regarding dice (and coin)probabilities, please seethis page.

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Let"simagine you are playing a game which uses dice. You are about to rollthree of them. You NEED to roll at least one 6. A 6 appearing onany one (or more) of the three dice will win the game for you! What areyour chances?

33.3%

42.1%

50%

66.6%

Quitesome time ago, I was over at a friend"s house watching him and anotherfriend play a board game called Axis & Allies.At one point this exact scenario came up - Kent was planning on rollingthree dice and reallywanted at least one 6 to appear. He made a comment that with threedice, his chances were 3/6 or 50%.

Kent"sreasoning was, with one die, the chances of rolling a 6 were 1/6 which is correct. Healso believed if he were to roll two dice, his chances were double thisor 2/6. This is INCORRECT andthis is where his faulty reasoning begins.

Knowinga little bit about the laws of probability, I quickly knew the fraction"2/6" for two dice and "3/6" for three dice was incorrect and spent a brief moment computing and then explaining the true percentages. Unfortunately, I do notbelieve I was successful in explaining to Kent why my figures werecorrect. Maybe I can do so here. The knowledge gained could certainly be very useful if you wish toplay free craps games.

Obviously,with Kent"s logic above, if the chances of rolling a 6 with two dice is2/6 and the chances ofrolling a 6 with three dice is 3/6, then the chances of rolling a 6with six dice would be 6/6 !! 100%?? Of course, this is obviouslyincorrect. I don"t care how many dice you roll, the chances of rollinga 6 will never be 100%.

Whenyou roll just one die, there are six different ways the die can land,as shown by the following graphic:


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Whentwo dice are rolled, there are now 36 different and unique ways thedice can come up. This figure is arrived at by multiplying the numberof ways the first die can come up (six) by the number of ways thesecond die can come up (six). 6 x 6 = 36.

Thisgraphic shows this very nicely. I"ve used two different colored dies tohelp show a roll of 2-1 is different from a roll of 1-2.


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Ifyou use the above graphic and count the number of times is 6 appearswhen two dice are rolled, you will see the answer is eleven. Eleventimes out of 36 or 30.5 %, slightly less than the 33.3% (2/6) Kent thought. Whenyou roll two dice, you have a 30.5 % chance at least one 6 will appear.

Thisfigure can also be figured out mathematically, without the use of thegraphic. One way to do so is to take the number of ways a single diewill NOT show a 6 when rolled (five) and multiply this by the number ofways the second die will NOT show a 6 when rolled. (Also five.) 5 x 5 =25. Subtract this from the total number of ways two dice can appear(36) and we have our answer...eleven.

So,let"s use this same method to answer our question and determine thechances of at least one 6 appearing when three diceare rolled.

Takethe chances of a 6 NOT appearing on the first die...

5 / 6

andmultiply this by the chances of a 6 NOT appearing on the second die...

5 / 6 x 5 / 6 = 25 / 36

andmultiply this by the chances of a 6 NOT appearing on the third die...

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25 / 36 x 5 / 6 = 125 / 216

So,there are 125 out of 216 chances of a 6 NOT appearing when three diceare rolled. Simply subtract 125 from 216 which will give us the chancesa 6 WILL appear when three dice are rolled, which is 91. 91 out of 216or 42.1 %, not quite the 50% Kent originally thought.

Hereis a table showing the fractions and percentages of a 6 appearing (orany other single digit for that matter) and notappearing with several different numbers of dice: