When a Mathematician Orders a Slice of Pie

Probability Level 4

A mathematician regularly orders a slice of pie for dessert at a restaurant where he meets for dinner with friends. On any day, the restaurant usually has available two out of three kinds of pies, Apple, Blackberry, and Cherry, which they make fresh on the premises.

One evening, the waitress asks him what he would like to have, Blackberry or Cherry. The mathematician says he'll have the Blackberry. Then the waitress remembers, knowing that the mathematician often seems to prefer Apple over Blackberry, "Oh, by the way, we do also have Apple tonight. Would you like to have that?" The mathematician says,

"Well, in that case, I'll have the [...]."

As the mathematician explains to his friends, different people make the pies on different days at random, of variable quality. Over time, he's come up with a scoring system as follows:

Over $30$ days that any of the pies are made,

the Apple scores $\color{#D61F06} \large 3$ on all $30$ days;
the Blackberry scores $\color{#D61F06} \large 2$ on $18$ days, $\color{#D61F06} \large 4$ on $6$ days, and $\color{#D61F06} \large 6$ on $6$ days;
the Cherry scores $\color{#D61F06} \large 1$ on $16$ days, $\color{#D61F06} \large 5$ on $14$ days.

So, on any given day, he always orders the pie that is most likely to have the highest score among all that is available on that day. (See note.)

What pie did the mathematician order that evening?

Note: Not the highest expected score, but which pie is likely to have the highest score on that day. Imagine tossing two or three $30$ -sided dice and checking to see which die comes up with the highest number, which have numbers on the faces as follows:

Apple die: $3$ on all $30$ faces
Blackberry die: $2$ on $18$ faces, $4$ on $6$ faces, $6$ on $6$ faces
Cherry die: $1$ on $16$ faces, $5$ on $14$ faces

Whichever die comes up with the highest number wins. Whether there's $2$ or $3$ dice being played, the mathematician goes with the one that is most likely to win. $30$ sided dice are possible in which any side has equal probability of coming up, based on the Rhombic Triacontahedron

Comments about the solution and the mathematician's reasoning would be most welcome.

Apple Pie Blackberry Pie Cherry Pie

1 solution

Michael Mendrin
Jul 26, 2018

The mathematician ordered the Cherry pie because he is being logical. He just wants the pie that's the most likely to be the best among the three on that day. If there were only the Blackberry and Cherry pies available, his best bet is to go with the Blackberry. However, once it's known that the Apple is also available, then his best bet is to go with the Cherry.

Out of $225$ days, the scores of $($ Apple, Blackberry, Cherry $)$ pies are

$(3,2,1)$ in $72$ days
$(3,2,5)$ in $63$ days
$(3,4,1)$ in $24$ days
$(3,4,5)$ in $21$ days
$(3,6,1)$ in $24$ days
$(3,6,5)$ in $21$ days

so that the Cherry beats the rest on $63+21=84$ days, with a probability of $\dfrac{28}{75}$

Note that the Blackberry beats the Cherry on $72+24+24+21=141$ days, with a probability of $\dfrac{47}{75}$

For a more detailed explanation

When all three pies are available, then

The probability that Apple's $3$ beats both the Blackberry's $2$ and the Cherry's $1$ is

$\dfrac{30}{30} \dfrac{18}{30} \dfrac{16}{30} = \dfrac{24}{75}$

The probability that Blackberry's $4$ beats both the Apple's $3$ and the Cherry's $1$ , or the Blackberry's $6$ beats both the Apple's $3$ and the Cherry's $1$ or $5$ is

$\dfrac{6}{30} \dfrac{30}{30} \dfrac{16}{30} + \dfrac{6}{30} \dfrac{30}{30} \dfrac{30}{30} = \dfrac{23}{75}$

The probability that Cherry's $5$ beats both the Apple's $3$ and the Blackberry's $2$ or $4$ is

$\dfrac{14}{30} \dfrac{30}{30} \dfrac{18}{30} +\dfrac{14}{30} \dfrac{30}{30} \dfrac{6}{30}= \dfrac{28}{75}$

When there are only two pies available, then

The probability that Apple's $3$ beats the Blackberry's $2$ is

$\dfrac{30}{30} \dfrac{18}{30} = \dfrac{45}{75}$

The probability that Apple's $3$ beats the Cherry's $1$ is

$\dfrac{30}{30} \dfrac{16}{30} = \dfrac{40}{75}$

The probability that Blackberry's $2$ or $4$ or $6$ beats the Cherry's $1$ or the Blackberry's $6$ beats the Cherry's $5$ is

$\dfrac{30}{30} \dfrac{16}{30} + \dfrac{6}{30} \dfrac{14}{30} = \dfrac{47}{75}$

The expected scores of each pie

Apple:

$3$

Blackberry:

$2\cdot\dfrac{18}{30}+4\cdot \dfrac{6}{30}+6 \cdot \dfrac{6}{30}=\dfrac{16}{5} =3.2$

Cherry:

$1\cdot\dfrac{16}{30}+5\cdot \dfrac{14}{30}=\dfrac{43}{15} \approx 2.867$

so if the mathematician goes by the expected score, then he'd always order Blueberry whenever it's available, followed by Apple, then Cherry (the last if it's the only pie available). However, deciding this way does not mean the pie he has ordered will be most likely the best pie of the day, even if the Blackberrry will offer greater overall satisfaction over time. The Blackberry pie is kind of like a baseball hitter that overall turns in the best average performance even though he's less likely than others to be the star of the game.

Fun problem. As a potential follow-up, suppose all 3 are available half the month, while for the rest of the month only two are available, with each pairing being equally likely. The question would then be which pie does the mathematician likely have most often over the course of a month. I'm finding that the Apple would win out, followed by the Cherry and then Blackberry, (the latter two being very close). Sometimes consistent adequacy wins in the long haul. :)

Brian Charlesworth - 2 years, 10 months ago

This problem does have a strange parallel with quantum interference, which I've often wondered about, where the existence of a third alternative changes the original probabilities. I still don't know if there's any connection between the two, though, I kind of doubt it.

Michael Mendrin - 2 years, 10 months ago

Another variation would be that the mathematician brings with him the three dice, one each for the Apple, Blackberry, and Cherry. Then he throws either two or three depending on how many pies are available that day, and picks the one with the highest number showing. But I think the paradoxical effect would be diluted if we introduced this complication, i.e., harder to see.

Michael Mendrin - 2 years, 10 months ago

Would these dice differ for each of the different pies? Say each (balanced) die has 6 sides, with the Apple pie die having 3 on all sides, the Blackberry pie die having 2 on four sides with 4 and 6 each on one side, and the Cherry pie die having 1 and 5 each on three of the sides. This would closely match the scenario in your problem.

Brian Charlesworth - 2 years, 10 months ago

@Brian Charlesworth – Yes, this would require using non-standard numbering of die faces.

I wanted a situation where given only two pies available, then clearly $A>C$ , $A>B$ , and $B>C$ , and yet when all three are available, $C>A>B$ . With the dice and numbers you have suggested, we then have $A=C$ , and I wanted to make it obvious that the mathematician prefers Apple whenever it's available and there's only one other pie. I found it easier to adjust the probabilities the way I want them if I go with the idea that the mathematician is paying attention to the quality of the pies over a period of a 30 day month. Let me keep fooling with this.

Michael Mendrin - 2 years, 10 months ago

@Brian Charlesworth – On another note, non-transistive set of dices are possible, if each one is numbered differently, so that for example, die A usually beats die B, B usually beats die C, and C usually beats die A. Kind of a like a rock, scissors, paper version in dice form.

Here, I'm sort of borrowing that idea but this doesn't necessarily call for non-transistivity.

Michael Mendrin - 2 years, 10 months ago

This is a wonderful paradox.

Jeremy Galvagni - 2 years, 10 months ago

Thanks, Jeremy. This is actually an old problem that has been bothering me for years. I first read about it in one of Martin Gardner's books a long time ago, which unfortunately did not even provide or explain the mathematics of this that would give a plausible support of the mathematician's bizarre decision. I finally hit on the idea of using a plausible scoring system that the mathematician could have relied on in making that decision. Then it was a matter of finding the right numbers to make this work.

Whether or not the mathematician wants to decide which pie to order in this way depends on what he considers to be important to him, so one cannot really say that the mathematician is "being illogical" in deciding to switch to Cherry. But this was never explained in Martin Gardner's book. I wonder what could be other plausible reasons for the mathematician's decision.

I've used my experience in creating non-transitive dice in creating this scoring system.

Michael Mendrin - 2 years, 10 months ago

The idea seems sound to me. When matched two at a time A>B>C but in a three-way competition C wins. This is not like transitive dice, although computing the probabilities is similar.

The idea of choosing B over C, but then switching to C when A becomes available (even though A wins head-to-head) is amazing.

I don't recall ever seeing this in a Gardner book, but it's been a while. Kudos for turning it into a Brilliant problem.

Jeremy Galvagni - 2 years, 10 months ago

@Jeremy Galvagni – When I first read about it in that Gardner book, I wondered if there was any connection with quantum probabilities, where, in some experimental setups, given a choice between B or C, a photon goes to B, but given a choice of A, B, or C, it goes to C. in other words, the existence of another choice A interferes with the choice between B or C, even though there seems to be no way any "signal" could have gone from A to either B or C. The mere existence of a third choice somewhere else can alter the outcome. That's why I've always remembered this strange pie paradox.

I don't even remember the title of that book. All I remember is being frustrated with Gardner for not better explaining the strange pie paradox.

Michael Mendrin - 2 years, 10 months ago

@Michael Mendrin – A>B>C and C overall winning usually came up in the truel puzzles. But I don't know if the conditions of them going by turns + the options of shooting the ground helped in getting that kind of results. Has anyone found a truel puzzle where the 3 may shoot simultaneously?

Saya Suka - 7 months, 2 weeks ago

It would be funny to see the confusion on waitress' face after mathematician's respond. Or maybe even funnier to hear her saying something like: "Oh, by the way, we do also have Apple tonight. Would you like to have Cherry then?"

The mathematicians reasoning is logical as long as he sticks to his scoring system. Consider slightly changed scores:

Apple: score 3 each day
Blackberry: score 2 on 15 days, score 4 on 15 days
Cherry: score 1 on 19 days, score 5 on 11 days.

Again, mathematician would've chosen Cherry after previously being offered with an Apple pie. The significant advantage which Blackberry and Apple have over Cherry in head-to-head duel breaks once we introduce the third kind of pie because it mostly affects the probabilities of the favorites.

I come up with the following analogy: say you saved some money and you want to invest in business of one of the local bakeries. Before you make a final decision, you do a little research and find out following facts: $34\%$ of people in town only eat Cherry pies and other $66\%$ of people hate Cherry pies and only eat Apple or Blackberry pies, but have no specific preference for one of them. You already know that there's one bakery famous for its Cherry pies and one famous for its Apple pies. Naturally, you picked the second one, specialized for Apple pies because it has more customers. However, at the last moment, friend of yours says to you that a new bakery was opened recently and that they offer delicious Blackberry pie. Assuming the information you've collected are correct (especially those about no specific preference for either Apple pies or Blackberry pies), you conclude that it's better to invest in bakery with Cherry pie because it has a safe $34\%$ percent of the market share, while there are no indications that one of the other two will gain advantage over the other, which seems to be perfectly sound decision.

Uros Stojkovic - 2 years, 10 months ago

The question seems wrong, I don't know if I'm right, my answer would've been cherry too, but at the start if the question it has been written that on a particular day only two kinds of pies are available, if he asked Blackberry, and the waitress suggested Apple, only those two were available

A Former Brilliant Member - 9 months, 3 weeks ago

When a Mathematician Orders a Slice of Pie

1 solution

0 pending reports