Natural 3 of a kind Yahtzee chances

firebus's picture

Typically what happens is that on your first roll you get 3 ones, and maybe a potentially useful singleton.

And you've already used the ones box.

So the question is, should you go for Yahtzee here? Or should you try to build off the singleton or just re-roll all five dice?

I suspect that if the probability is reasonably close to 1 in 6, it's probably your best chance at a Yahtzee in the game, and thus worth risking the Yahtzee box on.

There are 3 ways to get to Yahtzee from here, and we can sum their individual probabilities to get the overall chances.

  1. You could roll 2 ones on the next roll.
  2. You could roll 1 one on the next roll and 1 one on the final roll.
  3. You could roll no ones on the next roll and 2 ones on the final roll.

If I'm remembering my discrete math correctly (what are the odds?)

p(1) = 1/36
p(2) = 1/6 * 1/6 = 1/36
p(3) = 5/6 * 1/36 = 5/216

p(1) + p(2) + p(3) ~ 0.079. It's a little less than 1 in 12. Those odds are probably too crappy if you have something else useful you could try to do with the other 2 dice in the first roll. But if it's late in the game it might still be your best option.


Comments

good thinking

That's a great way to break the problem down.
I don't know probability that well, but I think your method will work but I don't think your answer is quite right. I suspect the probability of getting a yahtzee from 3 of a kind on the first roll is a little better than that.

first roll - 3 aces
p(0)- probability of rolling 0 aces on the next roll
p(1) - probability of rolling 2 aces on next throw
p(2) - probability of rolling 1 ace on the next throw and 1 ace on the 3rd throw
p(3) - probability of rolling 0 aces on the next throw and 2 aces on the final throw

p(1) = 1/6 * 1/6 = 1/36
p(0) = 5/6 * 5/6 = 25/36
p(2)= 1- p(1)- p(0)= 1 - 26/36 = 10/36
p(3) = p(0) * p(1) = 25/36 * 1/36 = 25/1296

p(1)+ p(2)+ p(3) = 121/1296 = ~9.34%
about a little bit better than 1 in 11 or
at a bit better than 1-10 odds

Well that is my stab at it anyway.


firebus's picture

the odds were low

yes, absolutely, my calculation of p(2) was naive.

thank you!!!


All depends on which boxes

All depends on which boxes are filled. You say "on your first roll", which could mean first of the game, but you indicate the ones are already taken.

The trick here is, wy are the ones already taken? The 1s should, most of the time, be the last box filled, unless you roll a 4+ in the 1s column. Any 4+ in 3,4,5,6 makes up the 1s, so the wise move is to never be in this situation by ALWAYS leaving the 1s for last.

Still, what if you fill in 4 1s the 2nd to last going for yahtzee, then get this... Well, then your point is valid. But, again, Yahtzee is worth more than 4 1s PLUS the bonus for filling the left side, so... Still weird to go for 1s. If my choice was completing the bonus or going for Yahtzee in the 2nd to last roll of the game... And you're giving me the 3 1s... Then yes, go for a yahtzee of 1s.

But, almost any other situation, go for 5 6s off 3 1s. Just throw them all in the crapper, and get something else. Odds be damned, there are no odds for how much luck fills your tendons.


firebus's picture

Rolls

You say "on your first roll", which could mean first of the game, but you indicate the ones are already taken.

Sorry, I meant first of the three rolls in a turn.

The 1s should, most of the time, be the last box filled

No. You should ditch a bad roll in the ones box early in the game, rather than take a low score in Chance or give up Yahtzee. This happens often.

It's also quite common to use the ones on a bad roll mid-game when you've got headroom on the bonus to absorb the loss.

I rarely fill the ones box last.


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