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I've recently got into playing Yahtzee with my brother. During a game we had on my first roll for my last round I got (5 4 4 2 1). I only had my large straight left, so naturally I was hopeful with what I got. I quickly thought about it, and my instinct was to roll the 4 and the 1, at a good chance for a 3. I figured better than just one chance at a 3, and if I got it, that would leave me having a good chance at a 6 or a 1 for my last roll.

My question is this, how do you calculate the over all chance of either strategies in order to choose which is better?

I used to be great with calculus, even in electrical calculations, however probabilities are really giving me a headache.

I think I got the first strategy of rolling just one die with the calculation of 1 - 1/6 = 5/6 (not happening) for each single die roll, for a sum of 25/36 (not happening) or 11/36 (happening) or 30.56% chance. yeah?

The problem I have is with the second strategy since the first roll is with two dice, but then second roll is either one or two di(c)e depending on the outcome. How do you factor that in mathematically?

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2 Answers 2

up vote 3 down vote accepted

An interesting question!

Strategy 1: Keep 1, 2, 4, 5

You are right: Here, you have a success rate of $$1/6 + 5/6\cdot 1/6 = 11/36 \approx 30.6\%.$$

Strategy 2: Keep 2, 4, 5

Now we have to do a case-by-case analysis, depending on the outcome of the first roll. I give an overview of the possible cases in the following table (the rows keeping track of the result of the first dice, the columns of the result of the second dice): $$\begin{array}{c|cccccc} & 1 & 2 & 3 & 4 & 5 & 6 \\\hline 1 & B & B & A & B & B & B \\ 2 & B & D & C & D & D & B \\ 3 & A & C & C & C & C & A \\ 4 & B & D & C & D & D & B \\ 5 & B & D & C & D & D & B \\ 6 & B & B & A & B & B & B\end{array}$$

Case A) You roll 3 and either 1 or 6. There are 4 possible rolls (13, 31, 36, 63), so the chance is $4/36 = 1/9$.

Case B) You roll either 1 or 6, but no 3. There are 16 possible rolls (11,12,14,15,16,21,26,41,46,51,56,61,62,64,65,66), so the chance is $16/36$. Now you have a $1/6$ chance to roll a $3$ on the second roll. Total success rate in this case: $16/36 \cdot 1/6 = 2/27$

Case C) You roll 3, but neither 1 nor 6. There are 7 possible rolls (23,32,33,34,35,43,53) which is a chance of $7/36$. On the second roll you need either 1 or 6, for which you have a chance of $1/3$. In total: $7/36 \cdot 1/3 = 7/108$.

Case D) You roll none of the numbers 1, 3 or 6. There are 9 possible rolls (22,24,25,42,44,45,52,54,55) which is a chance of $9/36$. On the second roll, you need to hit 3 as well as 1 or 6. The chance for this is 1/9 (see case A). So the total success rate is here $9/36 \cdot 1/9 = 1/36$.

Summing up the 4 single success rates, we get a total of $$1/9 + 2/27 + 7/108 + 1/36 = 5/18 \approx 27.8\%.$$

So your first strategy is slightly better than the second.

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wow, that was thorough, I was able to follow it first time through. Thanks! Now I have to tell me brother that he's 2.8% more right :\ –  user65015 Mar 5 '13 at 0:49
    
You're welcome. Just to warn you: I've hacked this together right now. There is no guarantee that all the calculations are correct. Better double-check them before you tell your brother! –  azimut Mar 5 '13 at 0:53
    
(+1): My calculations agree with yours--though apparently you finished your answer while I was still working on mine. –  Cameron Buie Mar 5 '13 at 1:03
    
@CameronBuie: Yeah, I immediately upvoted your answer when I saw that you got the same result! –  azimut Mar 5 '13 at 1:04
    
P.S.: I love the way you used the table. It made things so much more self-explanatory than mine did! I'll have to bear that in mind in the future. –  Cameron Buie Mar 5 '13 at 1:10
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Well, if you just rerolled the one of the dice that landed on $4$, then there's a $\frac16$ chance that you get a $3$ right away, and a $\frac56$ chance that you don't. In the latter case, you've got a $\frac16$ chance of getting a $3$ on your last roll, so your chance of a large straight if you just reroll one of the dice that landed on $4$ is $$\frac16+\frac56\cdot\frac16=\frac{11}{36}.$$

If we're rolling two dice, then the following table will be useful:

$$\begin{array}{c|cccccc} & 1 & 2 & 3 & 4 & 5 & 6\\\hline1 & 1,1 & 1,2 & 1,3 & 1,4 & 1,5 & 1,6\\ 2 & 2,1 & 2,2 & 2,3 & 2,4 & 2,5 & 2,6\\3 & 3,1 & 3,2 & 3,3 & 3,4 & 3,5 & 3,6\\4 & 4,1 & 4,2 & 4,3 & 4,4 & 4,5 & 4,6\\5 & 5,1 & 5,2 & 5,3 & 5,4 & 5,5 & 5,6\\6 & 6,1 & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\end{array}$$

If you reroll your $1$ and one of your $4$s, then you're hoping for one of your dice to be a $3$, and one of your dice being a $1$ or a $6$. Checking the table, there is a $\frac4{36}$ chance of doing this in one roll. There's a $\frac7{36}$ chance that you get a $3$, but not your $1$ or $6$, and then a $\frac26$ chance that you get your $1$ or $6$ on the last roll--better than the $\frac4{36}=\frac19$ chance of getting a large straight on the last roll if you reroll both those dice again. There's a $\frac{16}{36}$ that you get your $1$ or $6$ but don't get your $3$, and then a $\frac16$ chance of getting your $3$ on the last roll if you keep the $1$ or $6$--better than the chance of getting a large straight if you reroll both those dice again. Now, there's a $\frac9{36}$ chance that you get neither $3$ nor $1$ nor $6$ on your second roll, in which case you have a $\frac4{36}$ chance of getting the large straight on the last roll. Hence, if you decided to reroll the $1$ and one of your $4$s, and chose your strategy optimally after your second roll, then your chance of getting a large straight is $$\frac4{36}+\frac7{36}\cdot\frac26+\frac{16}{36}\cdot\frac16+\frac9{36}\cdot\frac4{36}=\frac{10}{36}.$$ You'd be a bit better off just rerolling one of the $4$s, and hoping for a $3$ to come up.

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