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(The following problem is from MAML, Meet 3, Round 1, December 2012, Problem 3.)

In the game of Yahtzee one has a chance to get Yahtzee (5 of the same kind, such as 5 sixes) in the throw of 5 dice. One has three chances to produce the 5 of a kind by throwing the dice three times, picking up all or some for the next throw. Denton threw two fives in his first throw and left those. He picks up the other three dice and proceeds. What is the probability that he gets a Yahtzee of fives?

My thinking is that there are four cases:

  1. On the second throw, we get no fives. In this case, we need to throw all of the dice again and get three fives on the third throw.
  2. On the second throw, we get one five. In this case, we need to throw the other two dice without a five again and get two fives on the third throw.
  3. On the second throw, we get two fives. In this case, we need to throw the last die without a five again and get one five on the third throw.
  4. On the second throw, we get three fives. In this case, we have a Yahtzee after the second throw, so we do not need to do a third throw.

The probability that we get a five on the second throw is obviously $\frac 1 6$. Therefore, the probability that we fail on the second throw and get to the third throw is $\frac 5 6$. Thus, the probability that we get a five on the third throw is $\frac 5 6*\frac 1 6=\frac{5}{36}$. Thus, for each case, we get the following probabilities:

  1. We need all three dice to succeed on the third throw, so the probability is $\left(\frac{5}{36}\right)^3$.
  2. We need one on the second throw and two on the third throw. There are three ways to arrange the dice in this way, so we get $3\left(\frac 1 6\right)\left(\frac{5}{36}\right)^2$.
  3. We need two on the second throw and one on the third throw. There are three ways to arrange the dice in this way, so we get $3\left(\frac 1 6\right)^2\left(\frac{5}{36}\right)$.
  4. We need all three dice to succeed on the second throw, so the probability is $\left(\frac 1 6\right)^3$.

Thus, the answer is: $$\left(\frac{5}{36}\right)^3+3\left(\frac 1 6\right)\left(\frac{5}{36}\right)^2+3\left(\frac 1 6\right)^2\left(\frac{5}{36}\right)+\left(\frac 1 6\right)^3$$

Now, this is a problem that is supposed to be simplified without a calculator in under twelve minutes. Can someone give me some tips on how to make this a solution feasible under twelve minutes? Is there a trick to simplifying the above?

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Notice that this expression is in the form of: $$a^3+3ba^2+3b^2a+b^3$$ This is a form that you might see a lot in math problems like this, so you'll simply have to learn to recognize it, just like people recognize $a^2-b^2$ and $a^2+2ab+b^2$. Pascal's Triangle can help with this.

Now, we can simplify this to: $$\left(a+b\right)^3=\left(\frac 1 6+\frac{5}{36}\right)^3=\left(\frac{11}{36}\right)^3=\frac{11^3}{36^3}=\frac{11^3}{6^6}$$ At this point, you have to know your powers of small numbers well enough in order to recognize that $11^3=1331$ and $6^6=46656$ so the answer is $\frac{1331}{46656}$. (For $36^3$, you can also use $36^3=216^2$ to save some time if you don't have it memorized.)


However, there is an easier way to do this problem than the four cases you set up, as I explain in this YouTube video.

Let's consider each die individually. There are two cases:

  • The die succeeds on the second throw, so the probability is $\frac 1 6$.
  • The die succeeds on the third throw, so the probability is $\frac{5}{36}$.

Therefore, the probability that the die has a five by the end is $\frac 1 6+\frac{5}{36}$. We have three dice, so our answer is: $$\left(\frac 1 6+\frac{5}{36}\right)^3$$ We still need to do the whole $11^3$ and $6^6$ ordeal, but now, we don't need to use the binomial theorem and we don't need to consider four different cases, which makes this problem a lot eaiser.

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    $\begingroup$ Knowing some cubes helps. Seeing that $36^3 = 216^2$ might save some multiplying. $\endgroup$ – John Jun 15 '16 at 0:12
  • $\begingroup$ @John Good point! I've added that to the answer above. $\endgroup$ – Noble Mushtak Jun 15 '16 at 0:45
  • $\begingroup$ Are you adressing yourself in the second person ("the four cases you set up")? Didn't you set those cases up yourself in the question? Or am I missing something? $\endgroup$ – joriki Jun 15 '16 at 15:24
  • $\begingroup$ @joriki Yes, I'm addressing myself in the second person. $\endgroup$ – Noble Mushtak Jun 15 '16 at 15:26

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