I have searched numerous Yahtzee sites and might be missing the solution, but maximizing the Chance outcome doesn’t appear to be addressed. In order to maximize a Yahtzee Chance score, assuming no other options are available, how should it be played? After throwing the dice, do I only keep 6’s and toss the other dice, or keep 5’s and 6’s and toss the other dice, or keep 4’s, 5’s and 6’s and toss the other dice, etc? What is the best way to maximize the Chance score? Also, what score can I expect at the optimum or at the “best ways” to play the Chance?
I prefer cheating. I find it to be the most effective way to win.
Hey Surgoshan … I agree, never could play the game, maybe a better thread would have been “How to cheat at Yahtzee?”. I like that even more!
Generally, I end up filling in the Chance with a failed bid to score in another category, frex Four of a Kind, or when ending up with only two fives or sixes while going for their respective categories above the bonus box. Therefore, except in extraordinary circumstances, you can’t “go for” a good Chance score.
In those extraordinary circumstances, however:
Your expectation value for a single die roll is 3.5 (the sum of all the faces, divided by six). Keep all fours or above, and re-roll threes or below. If you’re feeling exceptionally lucky, re-roll fours and below on the second roll, but not on the third–but this has the unfortunate consequence of leaving you vulnerable to a bad second roll.
Do ya feel lucky, punk? 
LL
I usually play the game only waiting for my chance at the “Chance”! I think a full house is only 25 points. If the Chance outcome were greater than 25 points on average, not sure it is based on this question … and if the Chance were played correctly, wouldn’t I want to “go for” the Chance outcome and hope, by chance, for a full house outcome versus the opposite?
To maximize chance score, keep 5’s and 6’s after the first roll. After the second roll keep 4’s, 5’s and 6’s. This will give an average score of 23.33.
The reason is as follows:
With one roll remaining on a die, the expectation is 3.5, as explained by LazarusLong42 so you would keep 4, 5, or 6.
With two rolls remaining for a die, half the time you get 1, 2, or 3 (which would be rerolled with expectation of 3.5 on the final roll) and the other half the time you get 4, 5, or 6 which would be kept. So the expectation with two rolls remaining is:
(3.5 + 3.5 + 3.5 + 4 + 5 + 6)/6 = 4.25.
Since a total of 3 rolls are allowed per die, a 1, 2, 3, or 4 on the first roll has expectation of 4.25 since they would be rerolled up to 2 more times. 5’s and 6’s would be kept (expectation of 5 or 6). This gives an expected value of
(4.25 + 4.25 + 4.25 + 4.25 + 5 + 6)/6 = 4.667 per die,
or for 5 dice an expected total of 23.33.
Chance should be a throwaway score. If you can’t get anything 3 of a kind, full house, etc., put it in Chance. It should never be a targeted score.
Geez Major, if Manlob is correct, an average Chance of 23+ points is almost as good a 4 x 6’s (24 pts) and close to a full house at 25 pts? Dosen’t seem like a throw away to me.
Well, you should never be trying for a Chance throw, unless you’ve filled up everything else already. Of course if you’ve got mixed numbers that you can’t put in any other slot, Chance is a good place for it.
Through empirical testing (having the computer roll the Yahtzee dice millions of times), it appears that keeping 4s, 5s, and 6s is the best strategy for maximizing Chance.
My program asked for a threshhold value between 1 and 5. Any die that was less than or equal to that would be rethrown.
I kept the same threshhold value for both of the rethrows. Results would probably differ for different threshholds for each rethrow.
Here’s my results:
Threshhold Average
Value Score
========== =======
1 19.94
2 21.94
3 23.12
4 23.05
5 21.33
You might want to run that. “Keep 4, 5, 6” was my off-the-cuff answer until I thought for a second - it clearly yields the best expected value if you have one remaining throw, but if you had 1000 throws you would keep nothing but 6’s and almost certainly wind up eventually with 30. Having multiple rolls remaining gives you a higher expected value than 3.5 * number of dice if you are going to selectively retain some of the high dice on the rolls. I don’t know off the top of my head if two rolls remaining makes it profitable to reroll 4’s after the first roll, keeping 4, 5, 6 on the second roll. I don’t have time to calculate it right now - I suspect it does. If nobody else answers, I may later. There are other people on this board who can run the numbers.
After some more run-throughs with 2 threshholds, I’ve found that re-rolling 1s, 2s, 3s, and 4’s on the first re-roll, then just 1s, 2s, and 3s on the second gives the best average Chance score of 23.33.
Second Threshhold
1 2 3 4 5
F 1 | 19.94 21.40 21.88 21.37 19.92
i 2 | 21.53 21.94 22.50 21.94 20.27
r 3 | 22.30 22.91 23.12 22.52 20.64
s 4 | 22.22 23.07 23.33 23.04 20.97
t 5 | 21.31 22.36 22.71 22.36 21.33
After some more run-throughs with 2 threshholds, I’ve found that re-rolling 1s, 2s, 3s, and 4’s on the first re-roll, then just 1s, 2s, and 3s on the second gives the best average Chance score of 23.33.
Second Threshhold
1 2 3 4 5
F 1 | 19.94 21.40 21.88 21.37 19.92
i 2 | 21.53 21.94 22.50 21.94 20.27
r 3 | 22.30 22.91 23.12 22.52 20.64
s 4 | 22.22 23.07 23.33 23.04 20.97
t 5 | 21.31 22.36 22.71 22.36 21.33
Thanks. That’s what I suspected.
The method I gave for maximizing chance score isn’t always the best strategy, if your objective is to win the game.
Suppose you are last to play, and need 11 points in chance to tie your opponents, 12 or more to win. If your first roll is 11111 (a wasted Yatzee) and roll all dice again to get 33221, the strategy for highest score would say roll all dice again. However, that is not the way to maximize your chances of winning the game.
Hi Manlob, you obviously know the game … but my problem is this … if Chance is a 23+ point outcome on average … it seems better than 4 x 1’s, 2’s, 3’s, 4’s, 5’s and almost a full house, 25 pts … e.g. should I give up a 4 4 6 6 6 chance for a full house at 25 points vs 26 points Chance? The analysis here has really suprised me in terms of the closed form calculations you did, Yabob play strategies, and the Monte Carlo kind of analysis provided by AWB! An issue here is should one play to maximize the Chance and only accept the lower point results of 4 x 1’s, 2’s, 3’s, 4’s, etc., if the Chance is less than 23 pts or post the chance at 23+ pts?
You should put a 44666 in full house (or maybe 3 of a kind or 6’s depending on the situation). For full house you either get 25 points or no points, and if you are trying for it you have about a 37% chance of getting the points, for an expectation of around 9 points. When trying for a good chance score, you have close to a 90% chance of getting a score of 20 or more, and an expectation of 23.33.
If the last two things you have to fill are full house and chance, and you end a turn with 66666 (I assume this could be a full house), putting it full house and saving chance for last gives you an expectation of 25+23.33=48.33 points. If you put the 30 points in chance and try for full house on the last turn, your expectation is only 30+9=39 points.
As others have said, save chance for a failed attempt at something more difficult to get. However, there are a cases where you are better off putting a zero in something like Yatzee in order get a better chance score.