Yahtzee: If one gets a bonus Yahtzee, must he take a zero elsewhere so each player has 13 turns?

Answer 1

No. You can only count Yahtzee once - but you can use the extra one as a large straight, small straight, or full house (with the corresponding amount of points), as well as 3 of a kind, 4 of a kind, and Chance of course.

Answer 2

So far I can tell you that the answer should be in the ball park of 0.0463 ± error. This is based on a program that I wrote which simulates the Yahtzee random and decision making process. I doubt there is an analytical approach other than brute force. Then again maybe there is (see Zo Maar's answer below). Let's consider each case one by one. Case 1: yahtzee in one roll Case 2: 4 of kind in first roll Case 3: 3 of kind in first roll (including full house) Case 4: 2 of kind (but not three of kind, includes 2 pairs) Case 5: 5 distinct numbers on first roll We can work out the probability of getting a Yahtzee given each of these cases. P(case 1) = 6/6^5 P(case 2) = 6*5*(1/6)^4 * (5/6) P(case 3) = 6 * 5C3 * (1/6)^3 * (5/6)^2 P(case 5) = 6! / 6^5 = 720 / 6^5 P(case 4) = 1 - ∑ P(other cases) = 25/36 Clearly case 4 is tough to work out otherwise. Y - event that you eventually get a yahtzee P(Y | case 1) = 1 Case 2 At this point you hold the number, X, that occurs 4 times, and roll the last die until you get X. P(Y | case 2) = 1/6 + 5/6 * 1/6 = 11/6^2 Case 3 Again you hold the number, X, occurring 3 times P(Y | case 3) = 121 / 6^4 Case 4 This is where it gets nasty. Let's assume that if you get more than one pair, then you hold the larger one (it doesn't really matter). The only way you'll change that hold is if you get 3 of something else in the 2nd roll. You can use a probability tree to get hte prob of a yahtzee once you have 2 of a kind in the first round. P(Y | case 4) = 167/5832 *** see edit below Case 5 At this point, it makes no sense holding anything. You might as well roll all the dice again. You can make use of the above results. Now it’s like trying to get a yahtzee in 2 rolls rather than three. P(Y | case 5) = 5279 / (54*6^5) *** see edit below Finally P(Yahtzee) = ∑ P(Y | case i) * P(case i) = 0.0457 *** see edit below ***EDIT*** Zo Maar is right. I've checked over my work and indeed P(Y | case 4) = 113/3888 P(Y | case 5) = 221/17496 hence P(Y) = 0.04603 And this result is more consistent with my earlier computational result. Hmmm, Markov chains. The good stuff I never learned as an engineer.

Answer 3

If you get a bonus Yahtzee you score it as normal if it is on the top section of the score board. Example: if you have a Yahtzee already and roll five 3's, you can take a 15 in the 3's section as well as the bonus Yahtzee. If the top section is already filled, the bonus Yahtzee acts as a wild score for the bottom section. So if you have five 3's again after filling in your score for the 3's already, you can use the Yahtzee as a wild and fill in a score of 40 for the large straight, 30 for the small straight, etc. No matter what you do, you still get the bonus 100 points for every Yahtzee past the first.

Answer 4

if you have a sutable blank open you must take it like a yahtzee in sixes is also sixes,3 of a kind,4 of a kind, full house,or chance.if nothing is open you take max points anywhere you want to and also get 100 points credet fot extra yahtzee.

Answer 5

I believe that if you get another yahtzee you can take anything else on the sheet and get the points for it. you do not take a 0.

Answer 6

no i got it for christmas and i play it until they only need yahtzee bonuses then put 0 in if they dont get them
if it is all stored in a little red box think

hope u find it useful

Answer 7

I don't think so!

Answer 8

yes, that is the way I have played it