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OK, so here is a real-world problem that I am actually facing, right now.

I ordered ten almond trees from a nursery: 6 Texas Mission (TM), and 4 Hall's Hardy (HH) which are pollinators for each other.

I have two plots that the trees are to go in, one with 6 holes and one with 4. So in the 6 hole plot I need 4 TM and 2 HH in between the TM. In the 4 hole plot I need 2 TM and 2 HH.

I absolutely must have pollinators near each other (in the same plot) or they will not bear fruit and all is for naught.

The problem is ... THE VENDOR DID NOT LABEL THE TREES!!! And it's looking like there's no way to tell them apart until they bloom which is too late. So ... if I just plant the trees at random in the holes, what is the probability of the following:

enter image description here

I really do appreciate any help I can get here.

Kerry Thomas
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  • An interesting problem! +1 Start by figuring out the probability of your nightmare situation happening. – A rural reader Mar 09 '23 at 18:20
  • Note to Math.SE: Silent down-voters are frowned upon (at least by me) for real-world mathematics problems. At least explain to OP what the problems are! – Brian Tung Mar 09 '23 at 18:21
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    OP: Do you want that specific arrangement in Case $1$—HH in the middle on the six-plot, and along that specific diagonal in the four-plot? Or just any two on the six-plot and any two in the four-plot? If the latter, I think the probability is $3/7$ of a successful planting. – Brian Tung Mar 09 '23 at 18:23
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    If it's the exact specific arrangement, the probability is $1/210$; if the diagonal can go the other way, it's $1/105$. ¶ The nightmare is also $1/210$, I think. – Brian Tung Mar 09 '23 at 18:28
  • OP: It's possible, incidentally, that this looks like a homework problem, I'm afraid—that may explain some down-votes. I don't believe it is, but it's certainly true that there are homework problems in combinatorics and probability that look exactly like this. – Brian Tung Mar 09 '23 at 18:35
  • I swear ... this is not a homework problem ... I really am facing this. This why I figured that this would be right up y'all's alley – Kerry Thomas Mar 09 '23 at 19:33
  • @BrianTung yes sir, I would like the HH in the middle in the 6-plot, and really in the 4 plot they are close enough to be in just about any configuration (in my yard it's a zigzag line.) They just need to be close enough for the bees to go from one to the other – Kerry Thomas Mar 09 '23 at 19:35
  • BTW the nursery did finally call me back and tell me how they packed them. I'm a little skeptical (how can they remember that?), but they seem like honest folks so I'm going to trust them (not much choice). It was an interesting problem. – Kerry Thomas Mar 09 '23 at 19:40

1 Answers1

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To begin, note that there are $\binom{10}{4} = 210$ total configurations: pick where the $4$ TM go.

Nightmare: Note that there is only one configuration in which this happens: all $6$ TM go into the $6$-hole pot and all $4$ TM go into the $4$-hole pot. The probability this happens is therefore $\frac{1}{210}$, which is quite small.

Perfect: Also only one configuration, since it has to be "exactly like" what is depicted. So also $\frac{1}{210}$: not very likely.

Acceptable: All configurations that are not perfect and not nightmare: this is $1 - \frac{1}{210} - \frac{1}{210} = \frac{208}{210} = \frac{104}{105}$.

K. Jiang
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