Odds of winning Orchard Game? I recently started playing the Orchard Game with my sons.  It is a simple children's game but I cannot figure out how to calculate the probability that you win.  Here are the setup/rules:


*

*There are 4 pieces of 4 different fruits (4 plums (blue), 4 green apples (green), 4 red apples (red) and 4 pears (yellow)).

*There is a bird

*There is a basket 

*All the fruit starts outside the basket

*There are no teams, you rotate rolling the dice but everyone is on the same team

*On the dice there are 4 colors - one for each fruit - a basket and a bird

*If you roll a color of the fruit, you place one piece of that fruit in the basket.  If all four pieces of that fruit are in the basket, nothing happens

*If you roll the basket you can pick any one piece of any fruit you want and put it into the basket

*If you roll the bird the bird moves forward one space


Win/loss
You win if you get all the pieces of fruit into the basket before the bird moves 5 spaces.  If the bird moves 5 spaces before you get all the fruit into the basket you lose.
What are the odds that you win?  The basket along with rolling a color when there is nothing of that color left complicates the calculation and it is beyond abilities.
Thanks!
 A: Commentor Parcly Taxel is correct to point out that the strategy used when rolling a basket matters. I propose that the strategy that maximizes the chance of winning is to choose a type of fruit that has the least number already in the basket. I don't see an easy way to prove this is optimal, but, intuitively, it is most likely to delay getting 4 copies of fruits for as long as possible which means more rolls will be helpful. I will proceed at first assuming that this strategy is being used.
I suggest computing this probability using dynamic programming. We can consider the state of the game as an ordered quintuple $(b, x_1, x_2, x_3, x_4)$ where $b$ represents the number of spaces the bird has moved, and $x_i$ represents the number of the $i$th fruit in the basket. Then, for each game state, we can define $P(b, x_1, x_2, x_3, x_4)$ to be the probability of winning from that state.
We have base cases that are trivial to compute:


*

*$P(5, x_1, x_2, x_3, x_4) = 0$

*$P(b, 4, 4, 4, 4) = 1$ (assuming $b < 5$)


Let $S$ be the strategy function that represents the transformation of a game state under our given strategy. Then, setting aside for the moment the case of a "dud" roll of a fruit that already has 4, we can express the probability for more complicated states in terms of their successors:
$P(b, x_1, x_2, x_3, x_4) = \frac{1}{6}P(b+1, x_1, x_2, x_3, x_4) +  \frac{1}{6}P(b, x_1 + 1, x_2, x_3, x_4) +  \frac{1}{6}P(b, x_1, x_2 + 1, x_3, x_4) +  \frac{1}{6}P(b, x_1, x_2, x_3 + 1, x_4) +  \frac{1}{6}P(b, x_1, x_2, x_3, x_4 + 1) +  \frac{1}{6}P(S(b, x_1, x_2, x_3, x_4))$
A "dud" roll would just mean that one of the terms on the right-hand-side of such an equation would be the same as the term on the left-hand-side. Due to the difference of coefficients, this can easily be resolved by algebraically moving all such terms to the left-hand-side. It turns out to be equivalent just to average over all the non-dud rolls. (We can equivalently reformulate the game as rerolling any such dud roll, and then the chance of each non-dud is the same.)
For example: $P(4,4,4,4,3) = \frac{2}{3}P(4,4,4,4,4) + \frac{1}{3}P(5,4,4,4,3) = \frac{2}{3}\cdot 1 + \frac{1}{3}\cdot 0 = \frac{2}{3}$
There is some symmetry that can be taken advantage of here, due to the fact that the different fruits and equivalent to each other. However, this still leaves hundreds of equations to solve. This is possible to do by hand, but instead, I wrote a computer program to compute an approximate result.
Using the (claimed-to-be) optimal strategy, the chance of winning is: $P(0,0,0,0,0) \approx 63.136\%$
It is also interesting to ask how much the strategy matters. I believe that the pessimal strategy would be to choose a fruit of a type that already has as many as possible in the basket. Using this strategy I get: $P(0,0,0,0,0) \approx 55.494\%$
Also, I noticed that the version of the rules on the page you linked to describes a different variation of the game:


*

*There are 10 of each fruit

*The bird needs 9 steps to cause the game to be lost

*Rolling the basket allows choosing any two fruits


For this version of the game I get:


*

*With optimal strategy, $P(\cdot) \approx 68.405\%$

*With pessimal strategy, $P(\cdot) \approx 53.240\%$

A: I recently started playing this game with my son, and had the same question! I ended up writing a story about it which ends by showcasing win probabilities for a few different strategies, and for different variations of the game.
Here are the parameters I end up adjusting:


*

*The number of fruits per color (4 in First Orchard, 10 in Orchard Game)

*The number of raven rolls required for a loss (5 in First Orchard, 9 in Orchard Game)

*The number of fruit you can remove when you roll the fruit basket (1 in First Orchard, 2 in Orchard Game)

*The number of fruit colors (4 in both games).


Fortunately, I was able to duplicate the probabilities that tehtmi reported. 
Here's the link if you're curious. Enjoy! https://mattlane.us/stories/harvesting-wins/
