Mathematically speaking, what's my best choice in yahtzee? I have a 1,3,4,5 on the board on the first roll.

Question

I have two rolls left ... should I roll the 1 and the extra die for an open ended straight possibility or should I just roll the extra die twice for a gutshot straight possibility?

Proving your answer with math will earn you the best answer.

Answer 1

Okay, I'm gonna assume that you're playing with all six-sided dice.

If you roll the one die, you want a 2, and nothing else helps. The probability of getting a two in two rolls is

P(2) = (1/6) + (5/6)(1/6) = 11/36,

as if you get a two on the first roll you won't roll again, and if you don't get it on the first roll you have to roll again, to try to get the two.

Now let's consider if you roll the two dice, keeping the 3,4, and 5. You want, basically, some sort of straight. Since you are keeping the 3, 4, and 5, it's obvious that you need a 2 for any straight, whether it ends in a 5 or a 6. Then you need either a one or a six. Let's try this first with just one roll. Say you roll two dice, one after the other (I say that just because it helps me think about it, even though you're rolling them at the same time). If the first is a 1 or 6, you need the next to be a 2, and if the first is a 2, you need the next to be a 1 or 6. Thus, with only one roll, the probability is:

P = P(2) * (P(6) + P(1)) + (P(6) + P(1)) * P(2)

= 2 * P(2) * (P(6) + P(1))

= 2(1/6)(1/6 + 1/6)

= 1/9.

This is less than 11/36, but again, that's with only one roll. Now we generalize it to two rolls, taking into account that your decision for the second roll will change depending on what you get in the first (but assuming that you'll always go for a straight).

We know the odds of you getting a straight in one roll are 1/9, so we don't need to calculate that. What remains to be determined is what happens when you get a two but not a 1 or 6, a 1 or 6 but not a 2, or neither. We'll try the last one first.

The odds of you not getting a 2, a 6, or a 1 in one roll of two dice is:

P = (3/6)*(3/6) = 1/4,

since there is a 3/6 chance on each die of getting something besides those three. If that happens, you roll both again, and the probability of getting a straight then, we've established, is 1/9. (Now, you can see the just rolling the two dice twice in a row gives you a probability of 1/9 + 1/9 = 2/9 of getting that straight with either or both rolls. This is by itself only 8/36, though, so we gotta calculate further.)

So the probability of either getting a straight your first roll, or getting nothing helpful your first roll and getting a straight on the second, is

P = 1/9 + (1/4)*(1/9) = 5/36.

Now for if you get a 2 on your first roll.

The probability of getting a 2 but not a 1 or 6 is

P = P(2) * P(3, 4, or 5) * 2 + P(2) * P(2)

= (1/6) * (3/6) * 2 + (1/6)*(1/6)

= 7/36,

and the following probability of you getting a 1 or 6 on your next roll is

P = 2/6 = 1/3,

so the total probability of you getting a 2 on your first and a 1 or 6 on your next is

P = (7/36) * (1/3) = 7/108.

Now for the probability of you getting a 1 or 6 on your first roll but not a 2, and getting a 2 on your next roll.

The probability of getting a 1 or 6 but no 2 is

P = P(1) * P(not 1 or 2) * 2 + P(6) * P(not 2 or 6) * 2 + P(6) * P(6) + P(1) * P(1)

= (1/6) * (4/6) * 2 + (1/6) * (4/6) * 2 + (1/6) * (1/6) + (1/6) * (1/6)

= 8/36 + 8/36 + 1/36 + 1/36

= 1/2.

The probability of then getting a 2 with one die is 1/6, so the probability of doing both is

P = (1/2) * (1/6) = 1/12.

Adding up all these probabilities, you have that the total probability of getting a straight if you roll two dice on your second roll instead of just one, is

P = 5/36 + 7/108 + 1/12

= 15/108 + 7/108 + 9/108

= 31/108,

which is less than 11/36, which equals 33/108.

So your odds are a bit less going for a straight with two dice than they would be if you went with the one die. Then again, if you roll two you have an additional small possibility of getting two sixes or something, which may also help a bit. But if you gotta have a straight, go with the one die..

Hope this helps with something (or is at least mildly entertaining).

I've checked this, and it seems okay, again, assuming that you always go for a straight, and keep any ones, twos, or sixes you get on your second roll if you roll two dice.

Have fun. ^_^

Edit: Oops, I overlapped something there. The probability of getting a 1 or a 6 but no two in one roll is not 1/2. The calculation I used counts the possibilities (1,6) and (6,1) twice. So basically you subtract 2/36 from that, which means that the actual probability is 16/36. This changes the final probability to:

P = 5/36 + 7/108 + (16/36) * (1/6)

= 15/108 + 7/108 + 8/108

= 30/108

= 10/36.

So the probability of getting a straight if you roll two is 10/36, not 31/108. My bad. But either way, you wanna go with the one die.

Answer 2

Your best choice in this situation (first roll) depends on the value of the fifth die.

If it is a 1, you should roll the 1's and go for the open-ended straight. You have less of a chance of getting the large straight than if you rolled a single 1, but your chance of getting the small straight is very good. Or, you might have a good pair or even trips going into the next roll.

If it is a 2, you have a large straight and can score it.

If it is 3 through 5, you should keep the pair and roll the other three dice. Keeping the pair vs going outside to the straight is marginally better for 3's (less than 1 pt of expected value) and significantly better for 5's (over 2 pts of expected value)

If it is a 6, you should roll the 1. You can keep your small straight while trying for the large straight, and that's a good play.

This all assumes trying to maximize your expected value in solitaire yahtzee. It is a complex game and so the results are difficult to prove through math, but there is a site which has a computer program to develop expected values for all combinations of filled/unfilled boxes. And then work from there to compute the optimum strategy based on dice rolls.

There is a web-interface to the program here:

http://www-set.win.tue.nl/~wstomv/misc/yahtzee/osyp.php3

If it is a 2-player game, then the decision is even more difficult. When behind, you are often willing to trade off expected value for high variance (go for the bigger roll), while when ahead conservative play is generally better.

I noticed most of the folks here are analyzing the situation where it is the last turn of the game and the large straight is the only box available. That is a simpler situation, and much easier to analyze algebraically.

Answer 3

I have a 1,3,4,5 on the board on the first roll

If you keep all and throw the die to roll a 2 in two attempts

P(a 2 on 1st roll) = 1/6
P (Not a 2 1st but a 2 2nd) = 5/6*1/6

So P (rolling a 2) in two rolls = 1/6 + 5/36
= 1/6

If you just keep the 3, 4, 5

P(Rolling (2 and 6) or (1 and 2) on 1st roll) = 2/36 + 2/36
= 1/9

P(Rolling just a 2 and on 1st roll without a 1 or 6) = 2 * 1/6 * 4/6
= 2/9
Then P(Rolling 1 or 6 on subsequent roll) = 1/3
So P(completing straight this way) = 2/9 * 1/3
= 2/27

P (No 2 on first roll and rolling (2 and 6) or (1 and 2) on 2nd roll)= 5/6 * 1/9
=5/54

So P(gutshot straight in 2 rolls) = 1/9 + 2/27 + 5/54
= (6 + 4 + 5) /54
= 15/54
= 5/18 >1/6 (in fact its almost 1/3)

Answer 4

I think you're better off going for the open-ended, but let's prove it.
P(2) = 1/6 + 5/6*1/6 = 6/36 + 5/36 = 11/36

The other path is quite a lot more complicated:
Favorable outcomes are:

[1,2] [2,1] [2,6] [6,2] -->4/36 desired result complete

(2,2) (2,3) (2,4) (2,5) (3,2) (4,2) (5,2) --> 7/36 open ended
(1,1) (1,3) (1,4) (1,5)
(1,6) (3,1) (3,6) (4,1)
(4,6) (5,1) (5,6) (6,1)
(6,3) (6,4) (6,5) (6,6) -->16/36 "inside straight"

x3,3x x3,4x x3,4x x4,3x x4,4x -->9/36 must roll both dice
x4,5x x5,3x x5,4x x5,5x ..... ..... ..... .. again

Taking the last first, now only 4/36 possibilities are favorable, -->9/36*4/36 = 36/1296 (+ 44/1296 = 157/1296)
With the middle 23, it's again decision time. You already have 1 favorable number. if you re-roll both dice your probability of success is (4/36)(23/36 = 92/1276

(2,2) (2,3) (2,4) (2,5) (3,2) (4,2) (5,2) --> 7/36 open ended -- here rolling 1 die gives you P = (2/6) while rolling 2 gives only P = 4/36 = 1/9

With the remaining 16/36, you have 1 number, but only one number available. Rolling 1 die gives P = 1/6. Rolling both again gives P = 4/36 = 1/9 again.

Totaling all these up, you have P(success) =
4/36 + (7/36)(2/6) + (16/36)(1/6) + (9/36)(4/36) =
(144 + 84 + 96 + 36)/1296 = 360/1296 = 10/36

Nope, slightly better off (1/36) trying to fill the inside straight!

Answer 5

if you role just the other die two times, one of the two times would require a 2 , each time there is a 5 out of 6 chance of not getting 2
so chances of winning is 1-(5/6 * 5/6) = 1-25/36 or 11/36

if you role the 1 and the extra die your chances of winning include
(1,2), (2,1), (2,6),(6,2)
Thus there are 4 possible winning roles of 36 possibilities.

Thus rolling 1 die would have significantly better odds!

Answer 6

Wal starts to go downhill when he adds 1/6 + 5/36 and gets 1/6. The answer goes downhill from there, but at least he spells "roll" correctly.

Answer 7

probability on one die of getting 123456 is 1/6 so you have a 1 outta 6th chance of gettin a 2 if it was me in this situation i would pick the one up that i didnt need and role 4 a 2 and if i didnt get the 2 i would pick up the 1 and the other die i didnt want and roll them.

Answer 8

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Answer 9

It totally depends on what your fifth die is. I actually created a computer program in C++ that can be adapted for this purpose...