__Decision__

An archive of questions and answers that may be of interest to puzzle enthusiasts.
__Question 1 - allais:__

The Allais Paradox involves the choice between two alternatives:

A. 89% chance of an unknown amount

10% chance of $1 million

1% chance of $1 million

B. 89% chance of an unknown amount (the same amount as in A)

10% chance of $2.5 million

1% chance of nothing

What is the rational choice? Does this choice remain the same if the
unknown amount is $1 million? If it is nothing?
Show Answer

This is "Allais' Paradox".Which choice is rational depends upon the subjective value of money. Many people are risk averse, and prefer the better chance of $1 million of option A. This choice is firm when the unknown amount is $1 million, but seems to waver as the amount falls to nothing. In the latter case, the risk averse person favors B because there is not much difference between 10% and 11%, but there is a big difference between $1 million and $2.5 million.

Thus the choice between A and B depends upon the unknown amount, even though it is the same unknown amount independent of the choice. This violates the "independence axiom" that rational choice between two alternatives should depend only upon how those two alternatives differ.

However, if the amounts involved in the problem are reduced to tens of dollars instead of millions of dollars, people's behavior tends to fall back in line with the axioms of rational choice. People tend to choose option B regardless of the unknown amount. Perhaps when presented with such huge numbers, people begin to calculate qualitatively. For example, if the unknown amount is $1 million the options are:

A. a fortune, guaranteed

B. a fortune, almost guaranteed

a tiny chance of nothingThen the choice of A is rational. However, if the unknown amount is nothing, the options are:

A. small chance of a fortune ($1 million)

large chance of nothing

B. small chance of a larger fortune ($2.5 million)

large chance of nothingIn this case, the choice of B is rational. The Allais Paradox then results from the limited ability to rationally calculate with such unusual quantities. The brain is not a calculator and rational calculations may rely on things like training, experience, and analogy, none of which would be help in this case. This hypothesis could be tested by studying the correlation between paradoxical behavior and "unusualness" of the amounts involved.

If this explanation is correct, then the Paradox amounts to little more than the observation that the brain is an imperfect rational engine.

__Question 2 - division:__

N-Person Fair Division

If two people want to divide a pie but do not trust each other, they can still ensure that each gets a fair share by using the technique that one person cuts and the other person chooses. Generalize this technique to more than two people. Take care to ensure that no one can be cheated by a coalition of the others. Show Answer

N-Person Fair DivisionNumber the people from 1 to N. Person 1 cuts off a piece of the pie. Person 2 can either diminish the size of the cut off piece or pass. The same for persons 3 through N. The last person to touch the piece must take it and is removed from the process. Repeat this procedure with the remaining N - 1 people, until everyone has a piece.

References:

Luce and Raiffa, "Games and Decisions", Wiley, 1957, p. 366

Kenneth Rebman, "How To Get (At Least) A Fair Share of the Cake",

in Mathematical Plums, Ross Honsberger, ed, Dolciani Mathematical

Expostions Number 4, published by the MAA.There is a cute result in combinatorics called the Marriage Theorem.

A village has n men and n women, such that for all 0 < k <= n and for any set of k men there are at least k women, each of whom is in love with at least one of the k men. All of the men are in love with all of the women :-}. The theorem asserts that there is a way to arrange the village into n monogamous couplings.The Marriage Theorem can be applied to the Fair Pie-Cutting Problem.

One player cuts the pie into n pieces. Each of the players labels some non-null subset of the pieces as acceptable to him. For reasons given below he should "accept" each piece of size > 1/n, not just the best piece(s). The pie-cutter is required to "accept" all of the pieces.

Given a set S of players let S' denote the set of pie-pieces acceptable to at least one player in S. Let t be the size of the largest set (T) of players satisfying |T| > |T'|. If there is no such set, the Marriage Theorem can be applied directly. Since the pie-cutter accepts every piece we know that t < n.

Choose |T| - |T'| pieces at random from outside T', glue them together with the pieces in T' and let the players in T repeat the game with this smaller (t/n)-size pie. This is fair since they all rejected the other n-t pieces, so they believe this pie is larger than t/n.

The remaining n-t players can each be assigned one of the remaining n-t pie-pieces without further ado due to the Marriage Theorem. (Otherwise the set T above was not maximal.)

The problem of getting not just a fair solution, but an envy-free solution, is not solved. A reference to this problem:

David Gale, "Dividing a Cake," in Mathematical Entertainments,

Mathematical Intelligencer, Vol. 15, No. 1, Winter 1993, p. 50,

contains references to work by Steven Breams and Alan Taylor.

__Question 3 - dowry:__

Sultan's Dowry

A sultan has granted a commoner a chance to marry one of his hundred daughters. The commoner will be presented the daughters one at a time. When a daughter is presented, the commoner will be told the daughter's dowry. The commoner has only one chance to accept or reject each daughter; he cannot return to a previously rejected daughter. The sultan's catch is that the commoner may only marry the daughter with the highest dowry. What is the commoner's best strategy assuming he knows nothing about the distribution of dowries? Show Answer

Since the commoner knows nothing about the distribution of the dowries, the best strategy is to wait until a certain number of daughters have been presented then pick the highest dowry thereafter. The exact number to skip is determined by the condition that the odds that the highest dowry has already been seen is just greater than the odds that it remains to be seen AND THAT IF IT IS SEEN IT WILL BE PICKED. This amounts to finding the smallest x such that:

x/n > x/n * (1/(x+1) + ... + 1/(n-1)).Working out the math for n=100 and calculating the probability gives: The commoner should wait until he has seen 37 of the daughters, then pick the first daughter with a dowry that is bigger than any preceding dowry. With this strategy, his odds of choosing the daughter with the highest dowry are surprisingly high: about 37%.

(cf. F. Mosteller, "Fifty Challenging Problems in Probability with Solutions",

Addison-Wesley, 1965, #47; "Mathematical Plums", edited by Ross Honsberger, pp. 104-110)Here's a twist on the sultan's dowry problem I hope hasn't been posted yet. I became interested in an iterated version of this problem, which goes as follows:

There's a long line of suitors outside the sultan's palace, and one by one they come in. If a suitor gets the right girl, he marries her, as before. Unfortunately (for the suitor, at least), if he doesn't, he gets his head lopped off, and the next suitor comes in.

Anyway, the first few dozen guys all know their probability theory, so they know that the best strategy is to skip the first 37 girls, and then pick the first girl who is the best up to that point. Alas, each one assumes that he's the only one who knows that strategy, so one by one, these few dozen guys get their heads lopped off.

After the 49th head has just rolled down the hill, and the sultan's vizier has just cried out, "Next!" the next guy in line says, "This isn't working out. We might all be doing the same thing. It doesn't hurt any of us to tell the rest what strategy we'll be using, so that none of us sets out to pick the same girl over and over again. I might as well just tell you, though, that I'm going to get her! I know this great strategy where you skip the first 37 blah blah blah..." Naturally, a few moments later, head number 50 comes rolling down.

"Next!" cries the vizier.

Well, suitor number 51 is in a quandary. He's all set to skip 37, etc, etc, except now he knows, that's not the right strategy. But he doesn't know if the last guy skipped the right girl because she was in the first 37, or if he didn't meet her yet because he stopped too early.

QUESTION 1: What is his best strategy?

ANSWER 1: His best strategy is:

"Skip the first 14. Take the first girl in [15,37] who is better

than the first 14. If there isn't one, take the SECOND girl in [38,100]

who is the best up to that point."Unfortunately, head number 51 rolls down the hill. "Next!" cries the vizier, who is getting a little hoarse, and wishes he didn't have this job.

QUESTION 2: What is suitor number 52's best strategy?

ANSWER 2: His best strategy is:

"Skip the first 5. Take the first girl in [6,14] who is better than the first 5. If there isn't one, take the SECOND girl in [15,37] who is the best up to that point. If there isn't one, take the THIRD girl in [38,100] who is the best up to that point."By the end of the day, of course, a wedding will be set, won't it?

MORE QUESTIONS: If each suitor uses the best strategy at that point, how many suitors will it take before the right girl is certain to be found? Does each succeeding suitor always have a better chance of winning than the preceding one?

SPECULATION: The last strategy is "Pick the last girl." Its probability of success is 1. And it is strategy number 100. (The corresponding conditions hold for 3, 4, and 5 daughters.)

Does anyone have any observations on this one?

byron elbows

(mail to brian@cs.ucla.edu)

__Question 4 - envelope:__

Someone has prepared two envelopes containing money. One contains twice as
much money as the other. You have decided to pick one envelope, but then the
following argument occurs to you: Suppose my chosen envelope contains $X,
then the other envelope either contains $X/2 or $2X. Both cases are
equally likely, so my expectation if I take the other envelope is
.5 * $X/2 + .5 * $2X = $1.25X, which is higher than my current $X, so I
should change my mind and take the other envelope. But then I can apply the
argument all over again. Something is wrong here! Where did I go wrong?

In a variant of this problem, you are allowed to peek into the envelope you chose before finally settling on it. Suppose that when you peek you see $100. Should you switch now? Show Answer

Let's follow the argument carefully, substituting real numbers for variables, to see where we went wrong. In the following, we will assume the envelopes contain $100 and $200. We will consider the two equally likely cases separately, then average the results.First, take the case that X=$100.

"I have $100 in my hand. If I exchange I get $200. The value of the exchange is $200. The value from not exchanging is $100. Therefore, I gain $100 by exchanging."

Second, take the case that X=$200.

"I have $200 in my hand. If I exchange I get $100. The value of the exchange is $100. The value from not exchanging is $200. Therefore, I lose $100 by exchanging."

Now, averaging the two cases, I see that the expected gain is zero.

So where is the slip up? In one case, switching gets X/2 ($100), in the other case, switching gets 2X ($200), but X is different in the two cases, and I can't simply average the two different X's to get 1.25X. I can average the two numbers ($100 and $200) to get $150, the expected value of switching, which is also the expected value of not switching, but I cannot under any circumstances average X/2 and 2X.

This is a classic case of confusing variables with constants.

OK, so let's consider the case in which I looked into the envelope and found that it contained $100. This pins down what X is: a constant.

Now the argument is that the odds of $50 is .5 and the odds of $200 is .5, so the expected value of switching is $125, so we should switch. However, the only way the odds of $50 could be .5 and the odds of $200 could be .5 is if all integer values are equally likely. But any probability distribution that is finite and equal for all integers would sum to infinity, not one as it must to be a probability distribution. Thus, the assumption of equal likelihood for all integer values is self-contradictory, and leads to the invalid proof that you should always switch. This is reminiscent of the plethora of proofs that 0=1; they always involve some illegitimate assumption, such as the validity of division by zero.

Limiting the maximum value in the envelopes removes the self-contradiction and the argument for switching. Let's see how this works.

Suppose all amounts up to $1 trillion were equally likely to be found in the first envelope, and all amounts beyond that would never appear. Then for small amounts one should indeed switch, but not for amounts above $500 billion. The strategy of always switching would pay off for most reasonable amounts but would lead to disastrous losses for large amounts, and the two would balance each other out.

For those who would prefer to see this worked out in detail: Assume the smaller envelope is uniform on [$0,$M], for some value of $M. What is the expectation value of always switching? A quarter of the time $100 >= $M (i.e. 50% chance $X is in [$M/2,$M] and 50% chance the larger envelope is chosen). In this case the expected switching gain is -$50 (a loss). Thus overall the always switch policy has an expected (relative to $100) gain of (3/4)*$50 + (1/4)*(-$50) = $25.

However the expected absolute gain (in terms of M) is:

/ M

| g f(g) dg, [ where f(g) = (1/2)*Uniform[0,M)(g) +

/-M (1/2)*Uniform(-M,0](g). ]

= 0. QED.OK, so always switching is not the optimal switching strategy. Surely there must be some strategy that takes advantage of the fact that we looked into the envelope and we know something we did not know before we looked.

Well, if we know the maximum value $M that can be in the smaller envelope, then the optimal decision criterion is to switch if $100 < $M, otherwise stick. The reason for the stick case is straightforward. The reason for the switch case is due to the pdf of the smaller envelope being twice as high as that of the larger envelope over the range [0,$M). That is, the expected gain in switching is (2/3)*$100 + (1/3)*(-$50) = $50.

What if we do not know the maximum value of the pdf? You can exploit the "test value" technique to improve your chances. The trick here is to pick a test value T. If the amount in the envelope is less than the test value, switch; if it is more, do not. This works in that if T happens to be in the range [M,2M] you will make the correct decision. Therefore, assuming the unknown pdf is uniform on [0,M], you are slightly better off with this technique.

Of course, the pdf may not even be uniform, so the "test value" technique may not offer much of an advantage. If you are allowed to play the game repeatedly, you can estimate the pdf, but that is another story...

__Question 5 - exchange:__

At one time, the Canadian and US dollars were discounted by 10 cents on
each side of the border (i.e., a Canadian dollar was worth 90 US cents
in the US, and a US dollar was worth 90 Canadian cents in Canada). A
man walks into a bar on the US side of the border, orders 10 US cents
worth of beer, pays with a US dollar and receives a Canadian dollar in
change. He then walks across the border to Canada, orders 10 Canadian
cents worth of beer, pays with a Canadian dollar and receives a US
dollar in change. He continues this throughout the day, and ends up
dead drunk with the original dollar in his pocket.

Who pays for the drinks? Show Answer

The man paid for all the drinks. But, you say, he ended up with the same amount of money that he started with! However, as he transported Canadian dollars into Canada and US dollars into the US, he performed "economic work" by moving the currency to a location where it was in greater demand (and thus valued higher). The earnings from this work were spent on the drinks.

Note that he can only continue to do this until the Canadian bar runs out of US dollars, or the US bar runs out of Canadian dollars, i.e., until he runs out of "work" to do.

__Question 6 - high.or.low:__

I pick two numbers, randomly, and tell you one of them. You are supposed
to guess whether this is the lower or higher one of the two numbers I
picked. Can you come up with a method of guessing that does better than
picking the response "low" or "high" randomly (i.e. probability to guess
right > .5) ?
Show Answer

Pick any cumulative probability function P(x) such that a > b ==> P(a) > P(b). Now if the number shown is y, guess "low" with probability P(y) and "high" with probability 1-P(y). This strategy yields a probability of > 1/2 of winning since the probability of being correct is 1/2*( (1-P(a)) + P(b) ) = 1/2 + (P(b)-P(a)), which is > 1/2 by assumption.

__Question 7 - monty.hall:__

You are a participant on "Let's Make a Deal." Monty Hall shows you
three closed doors. He tells you that two of the closed doors have a
goat behind them and that one of the doors has a new car behind it.
You pick one door, but before you open it, Monty opens one of the two
remaining doors and shows that it hides a goat. He then offers you a
chance to switch doors with the remaining closed door. Is it to your
advantage to do so?
Show Answer

Under reasonable assumptions about Monty Hall's motivation, your chance of picking the car doubles when you switch.The problem is confusing for two reasons: first, there are hidden assumptions about Monty's motivation that cloud the issue; and second, novice probability students do not see that the opening of the door gave them any new information.

Monty can have one of three basic motives:

1. He randomly opens doors.

2. He always opens the door he knows contains nothing.

3. He only opens a door when the contestant has picked the grand prize.These result in very different strategies:

1. No improvement when switching.

2. Double your odds by switching.

3. Don't switch!Most people, myself included, think that (2) is the intended interpretation of Monty's motive. Interviews with Monty Hall indicate that he did indeed try to lure the contestant who had picked the car with cash incentives to switch. However, if Monty always adopted this strategy, contestants would soon learn never to switch, so one presumes that occasionally Monty offered another door even when the contestant had picked a goat. At any rate, analyzing the problem with this strategy is difficult, since it requires knowing something about Monty's probability of bluffing.

A good way to see that Monty is giving you information by opening doors that he knows are valueless is to increase the number of doors from three to 100. If there are 100 doors, and Monty shows that 98 of them are valueless, isn't it pretty clear that the chance the prize is behind the remaining door is 99/100?

The original Monty Hall problem (and solution) appears to be due to Steve Selvin, and appears in American Statistician, Feb 1975, V. 29, No. 1, p. 67 under the title ``A Problem in Probability.'' It should be of no surprise to readers of this group that he received several letters contesting the accuracy of his solution, so he responded two issues later (American Statistician, Aug 1975, V. 29, No. 3, p. 134). However, the principles that underlie the problem date back at least to the fifties, and probably are timeless. See the references below for details.

Reference (too numerous to mention, but these contain bibliographies):

Leonard Gillman, "The Car and the Goats", AMM 99:1 (Jan 1992), p. 3

Ed Barbeau, "The Problem of the Car and Goats", CMJ 24:2 (Mar 1993), p. 149

The second reference contains a list of equivalent or related problems.

__Question 8 - newcomb:__

Newcomb's Problem

A being put one thousand dollars in box A and either zero or one million
dollars in box B and presents you with two choices:

(1) Open box B only.

(2) Open both box A and B.

The being put money in box B only if it predicted you will choose option (1).

The being put nothing in box B if it predicted you will do anything other than

choose option (1) (including choosing option (2), flipping a coin, etc.).

Assuming that you have never known the being to be wrong in predicting your actions, which option should you choose to maximize the amount of money you get? Show Answer

This is "Newcomb's Paradox".You are presented with two boxes: one certainly contains $1000 and the other might contain $1 million. You can either take one box or both. You cannot change what is in the boxes. Therefore, to maximize your gain you should take both boxes.

However, it might be argued that you can change the probability that the $1 million is there. Since there is no way to change whether the million is in the box or not, what does it mean that you can change the probability that the million is in the box? It means that your choice is correlated with the state of the box.

Events which proceed from a common cause are correlated. My mental states lead to my choice and, very probably, to the state of the box. Therefore my choice and the state of the box are highly correlated. In this sense, my choice changes the "probability" that the money is in the box. However, since your choice cannot change the state of the box, this correlation is irrelevant.

The following argument might be made: your expected gain if you take both boxes is (nearly) $1000, whereas your expected gain if you take one box is (nearly) $1 million, therefore you should take one box. However, this argument is fallacious. In order to compute the expected gain, one would use the formulas:

E(take one) = $0 * P(predict take both | take one) + $1,000,000 * P(predict take one | take one)

E(take both) = $1,000 * P(predict take both | take both) + $1,001,000 * P(predict take one | take both)

While you are given that P(do X | predict X) is high, it is not given that P(predict X | do X) is high. Indeed, specifying that P(predict X | do X) is high would be equivalent to specifying that the being could use magic (or reverse causality) to fill the boxes. Therefore, the expected gain from either action cannot be determined from the information given.

__Question 9 - prisoners:__

Three prisoners on death row are told that one of them has been chosen
at random for execution the next day, but the other two are to be
freed. One privately begs the warden to at least tell him the name of
one other prisoner who will be freed. The warden relents: 'Susie will
go free.' Horrified, the first prisoner says that because he is now
one of only two remaining prisoners at risk, his chances of execution
have risen from one-third to one-half! Should the warden have kept his
mouth shut?
Show Answer

Each prisoner had an equal chance of being the one chosen to be executed. So we have three cases:

Prisoner executed: A B C Probability of this case: 1/3 1/3 1/3

Now, if A is to be executed, the warden will randomly choose either B or C, and tell A that name. When B or C is the one to be executed, there is only one prisoner other than A who will not be executed, and the warden will always give that name. So now we have:

Prisoner executed: A A B C Name given to A: B C C B Probability: 1/6 1/6 1/3 1/3

We can calculate all this without knowing the warden's answer. When he tells us B will not be executed, we eliminate the middle two choices above. Now, among the two remaining cases, C is twice as likely as A to be the one executed. Thus, the probability that A will be executed is still 1/3, and C's chances are 2/3.

__Question 10 - red:__

I show you a shuffled deck of standard playing cards, one card at a
time. At any point before I run out of cards, you must say "RED!".
If the next card I show is red (i.e. diamonds or hearts), you win. We
assume I the "dealer" don't have any control over what the order of
cards is.

The question is, what's the best strategy, and what is your probability of winning ? Show Answer

If a deck has n cards, r red and b black, the best strategy wins with a probability of r/n. Thus, you can say "red" on the first card, the last card, or any other card you wish. Proof by induction on n. The statement is clearly true for one-card decks. Suppose it is true for n-card decks, and add a red card. I will even allow a nondeterministic strategy, meaning you say "red" on the first card with probability p. With probability 1-p, you watch the first card go by, and then apply the "optimal" strategy to the remaining n-card deck, since you now know its composition.

The odds of winning are therefore: p * (r+1)/(n+1) + (1-p) * ((r+1)/(n+1) * r/n + b/(n+1) * (r+1)/n).

After some algebra, this becomes (r+1)/(n+1) as expected.

Adding a black card yields: p * r/(n+1) + (1-p) * (r/(n+1) * (r-1)/n + (b+1)/(n+1) * r/n).

This becomes r/(n+1) as expected.

__Question 11 - rotating.table:__

Four glasses are placed upside down in the four corners of a square
rotating table. You wish to turn them all in the same direction,
either all up or all down. You may do so by grasping any two glasses
and, optionally, turning either over. There are two catches: you are
blindfolded and the table is spun after each time you touch the
glasses. Assuming that a bell rings when you have all the glasses up,
how do you do it?
pt type="text/javascript">
function reveal(a){
var e=document.getElementById(a);
if(!e)return true;
if(e.style.display=="none"){
e.style.display="block"
} else {
e.style.display="none"
}
return true;
}
Show Answer

1. Turn two adjacent glasses up.

2. Turn two diagonal glasses up.

3. Pull out two diagonal glasses. If one is down, turn it up and you're done. If not, turn one down and replace.

4. Take two adjacent glasses. Invert them both.

5. Take two diagonal glasses. Invert them both.References

"Probing the Rotating Table"

W. T. Laaser and L. Ramshaw

_The Mathematical Gardner_,

Wadsworth International, Belmont CA 1981.... we will see that such a procedure exists if and only if the parameters k and n satisfy the inequality k >= (1-1/p)n, where p is the largest prime factor of n.

The paper mentions (without discussing) two other generalizations: more than two orientations of the glasses (Graham and Diaconis) and more symmetries in the table, e.g. those of a cube (Kim).

__Question 12 - stpetersburg:__

What should you be willing to pay to play a game in which the payoff is
calculated as follows: a coin is flipped until it comes up heads on the
nth toss and the payoff is set at 2^n dollars?
Show Answer

Classical decision theory says that you should be willing to pay any amount up to the expected value of the wager. Let's calculate the expected value: The probability of winning at step n is 2^-n, and the payoff at step n is 2^n, so the sum of the products of the probabilities and the payoffs is:E = sum over n (2^-n * 2^n) = sum over n (1) = infinity

So you should be willing to pay any amount to play this game. This is called the "St. Petersburg Paradox."

The classical solution to this problem was given by Bernoulli. He noted that people's desire for money is not linear in the amount of money involved. In other words, people do not desire $2 million twice as much as they desire $1 million. Suppose, for example, that people's desire for money is a logarithmic function of the amount of money. Then the expected VALUE of the game is:

E = sum over n (2^-n * C * log(2^n)) = sum over n (2^-n * C' * n) = C''

Here the C's are constants that depend upon the risk aversion of the player, but at least the expected value is finite. However, it turns out that these constants are usually much higher than people are really willing to pay to play, and in fact it can be shown that any non-bounded utility function (map from amount of money to value of money) is prey to a generalization of the St. Petersburg paradox. So the classical solution of Bernoulli is only part of the story.

The rest of the story lies in the observation that bankrolls are always finite, and this dramatically reduces the amount you are willing to bet in the St. Petersburg game.

To figure out what would be a fair value to charge for playing the game we must know the bank's resources. Assume that the bank has 1 million dollars (1*K*K = 2^20). I cannot possibly win more than $1 million whether I toss 20 tails in a row or 2000.

Therefore my expected amount of winning is

E = sum n up to 20 (2^-n * 2^n) = sum n up to 20 (1) = $20

and my expected value of winning is

E = sum n up to 20 (2^-n * C * log(2^n)) = some small number

This is much more in keeping with what people would really pay to play the game.

Incidentally, T.C. Fry suggested this change to the problem in 1928 (see W.W.R. Ball, Mathematical Recreations and Essays, N.Y.: Macmillan, 1960, pp. 44-45).

The problem remains interesting when modified in this way, for the following reason. For a particular value of the bank's resources, let

e denote the expected value of the player's winnings; and let p denote the probability that the player profits from the game, assuming the price of getting into the game is 0.8e (20% discount).

Note that the expected value of the player's profit is 0.2e. Now let's vary the bank's resources and observe how e and p change. It will be seen that as e (and hence the expected value of the profit) increases, p diminishes. The more the game is to the player's advantage in terms of expected value of profit, the less likely it is that the player will come away with any profit at all. This is mildly counterintuitive.

__Question 13 - truel:__

A, B, and C are to fight a three-cornered pistol duel. All know that
A's chance of hitting his target is 0.3, C's is 0.5, and B never misses.
They are to fire at their choice of target in succession in the order
A, B, C, cyclically (but a hit man loses further turns and is no longer
shot at) until only one man is left. What should A's strategy be?
Show Answer

This is problem 20 in Mosteller _Fifty Challenging Problems in Probability_ and it also appears (with an almost identical solution) on page 82 in Larsen & Marx _An Introduction to Probability and Its Applications_.Here's Mosteller's solution:

A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly {I disagree; this is not at all clear!} shoots the more dangerous C first, and A gets one shot at B with probability 0.3 of succeeding. If he misses this time, the less said the better. On the other hand, suppose A hits B. Then C and A shoot alternately until one hits. A's chance of winning is (.5)(.3) + (.5)^2(.7)(.3) + (.5)^3(.7)^2(.3) + ... . Each term corresponds to a sequence of misses by both C and A ending with a final hit by A. Summing the geometric series we get ... 3/13 < 3/10. Thus hitting B and finishing off with C has less probability of winning for A than just missing the first shot. So A fires his first shot into the ground and then tries to hit B with his next shot. C is out of luck.

As much as I respect Mosteller, I have some serious problems with this solution. If we allow the option of firing into the ground, then if all fire into the ground with every shot, each will survive with probability 1. Now, the argument could be made that a certain strategy for X that both allows them to survive with probability 1 *and* gives less than a probability of survival of less than 1 for at least one of their foes would be preferred by X. However, if X pulls the trigger and actually hits someone what would the remaining person, say Y, do? If P(X hits)=1, clearly Y must try to hit X, since X firing at Y with intent to hit dominates any other strategy for X. If P(X hits)<1 and X fires at Y with intent to hit, then P(Y survives)<1 (since X could have hit Y). Thus, Y must insure that X can not follow this strategy by shooting back at X (thus insuring that P(X survives)<1). Therefore, I would conclude that the ideal strategy for all three players, assuming that they are rational and value survival above killing their enemies, would be to keep firing into the ground. If they don't value survival above killing their enemies (which is the only a priori assumption that I feel can be safely made in the absence of more information), then the problem can't be solved unless the function each player is trying to maximize is explicitly given.

-- clong@remus.rutgers.edu (Chris Long)

OK - I'll have a go at this.

How about the payoff function being 1 if you win the "duel" (i.e. if at some point you are still standing and both the others have been shot) and 0 otherwise? This should ensure that an infinite sequence of deliberate misses is not to anyone's advantage. Furthermore, I don't think simple survival makes a realistic payoff function, since people with such a payoff function would not get involved in the fight in the first place!

[ I.e. I am presupposing a form of irrationality on the part of the fighters: they're only interested in survival if they win the duel. Come to think of it, this may be quite rational - spending the rest of my life firing a gun into the ground would be a very unattractive proposition to me :-)]

Now, denote each position in the game by the list of people left standing, in the order in which they get their turns (so the initial position is (A,B,C), and the position after A misses the first shot (B,C,A)). We need to know the value of each possible position for each person.

By definition:

valA(A) = 1 valB(A) = 0 valC(A) = 0

valA(B) = 0 valB(B) = 1 valC(B) = 0

valA(C) = 0 valB(C) = 0 valC(C) = 1Consider the two player position (X,Y). An infinite sequence of misses has value zero to both players, and each player can ensure a positive payoff by trying to shoot the other player. So both players deliberately missing is a sub-optimal result for both players. The question is then whether both players should try to shoot the other first, or whether one should let the other take the first shot. Since having the first shot is always an advantage, given that some real shots are going to be fired, both players should try to shoot the other first. It is then easy to establish that:

valA(A,B) = 3/10 valB(A,B) = 7/10 valC(A,B) = 0

valA(B,A) = 0 valB(B,A) = 1 valC(B,A) = 0

valA(B,C) = 0 valB(B,C) = 1 valC(B,C) = 0

valA(C,B) = 0 valB(C,B) = 5/10 valC(C,B) = 5/10

valA(C,A) = 3/13 valB(C,A) = 0 valC(C,A) = 10/13

valA(A,C) = 6/13 valB(A,C) = 0 valC(A,C) = 7/13Now for the three player positions (A,B,C), (B,C,A) and (C,A,B). Again, the fact that an infinite sequence of misses is sub-optimal for all three players means that at least one player is going to decide to fire. However, it is less clear than in the 2 player case that any particular player is going to fire. In the 2 player case, each player knew that *if* it was sub-optimal for him to fire, then it was optimal for the other player to fire *at him* and that he would be at a disadvantage in the ensuing duel because of not having got the first shot. This is not necessarily true in the 3 player case.

Consider the payoff to A in the position (A,B,C). If he shoots at B, his expected payoff is:

0.3*valA(C,A) + 0.7*valA(B,C,A) = 9/130 + 0.7*valA(B,C,A)

If he shoots at C, his expected payoff is:

0.3*valA(B,A) + 0.7*valA(B,C,A) = 0.7*valA(B,C,A)

And if he deliberately misses, his expected payoff is:

valA(B,C,A)

Since he tries to maximise his payoff, we can immediately eliminate shooting at C as a strategy - it is strictly dominated by shooting at B. So A's expected payoff is:

valA(A,B,C) = MAX(valA(B,C,A), 9/130 + 0.7*valA(B,C,A))

A similar argument shows that C's expected payoffs in the (C,A,B) position are:

For shooting at A: 0.5*valC(A,B,C)

For shooting at B: 35/130 + 0.5*valC(A,B,C)

For missing: valC(A,B,C)So C either shoots at B or deliberately misses, and:

valC(C,A,B) = MAX(valC(A,B,C), 35/130 + 0.5*valC(A,B,C))

Each player can obtain a positive expected payoff by shooting at one of the other players, and it is known that an infinite sequence of misses will result in a zero payoff for all players. So it is known that some player's strategy must involve shooting at another player rather than deliberately missing.

Now look at this from the point of view of player B. He knows that *if* it is sub-optimal for him to shoot at another player, then it is optimal for at least one of the other players to shoot. He also knows that if the other players choose to shoot, they will shoot *at him*. If he deliberately misses, therefore, the best that he can hope for is that they miss him and he is presented with the same situation again. This is clearly less good for him than getting his shot in first. So in position (B,C,A), he must shoot at another player rather than deliberately miss.

B's expected payoffs are:

For shooting at A: valB(C,B) = 5/10

For shooting at C: valB(A,B) = 7/10So in position (B,C,A), B shoots at C for an expected payoff of 7/10. This gives us:

valA(B,C,A) = 3/10 valB(B,C,A) = 7/10 valC(B,C,A) = 0

So valA(A,B,C) = MAX(3/10, 9/130 + 21/100) = 3/10, and A's best strategy is position (A,B,C) is to deliberately miss, giving us:

valA(A,B,C) = 3/10 valB(A,B,C) = 7/10 valC(A,B,C) = 0

And finally, valC(C,A,B) = MAX(0, 35/130 + 0) = 7/26, and C's best strategy in position (C,A,B) is to shoot at B, giving us:

valA(C,A,B) = 57/260 valB(C,A,B) = 133/260 valC(C,A,B) = 7/26

I suspect that, with this payoff function, all positions with 3 players can be resolved. For each player, we can establish that if their correct strategy is to fire at another player, then it is to fire at whichever of the other players is more dangerous. The most dangerous of the three players then finds that he has nothing to lose by firing at the second most dangerous.

Questions:

(a) In the general case, what are the optimal strategies for the other two players, possibly as functions of the hit probabilities and the cyclic order of the three players?

(b) What happens in the 4 or more player case?

-- David Seal

In article <1993Mar25.022459.10269@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:

The Good, the Bad, and the Ugly are standing at three equidistant

"P" "Q" "R" -- allow me these alternate names.

points around a very large circle, about to fight a three-way duel to see who gets the treasure. They all know that the Good hits with probability p=.9, the Bad hits with probability q=.7, and the Ugly hits with probability r=.5.Yes, I know this sounds like decision/truel from the rec.puzzles archive. But here's the difference:

At some instant, all three will fire simultaneously, each at a target of his choice. Then any who survive that round fire simultaneously again, until at most one remains. Note that there are then four possible outcomes: the Good wins, the Bad wins, the Ugly wins, or all are killed.

A multi-round multi-person game can get complicated if implicit alliances are formed or the players deduce each other's strategies. For simplicity let's disallow communication and assume the players forget who shot at whom after each round.

Now the questions:

These are not easy questions, even with the simplifying assumptions I've made.

1. What is each shooter's strategy?

Each player has two possible strategies so there are eight cases to consider; unfortunately none of the players has a strictly dominant strategy:

P aims at Q aims at R aims at P survival Q survival R survival Noone lives Q P P 0.0649 0.0355 0.7991 0.1005 Q P Q 0.1371 0.0146 0.6966 0.1517 * Q R P 0.3946 0.0444 0.1470 0.4140 Q R Q 0.8221 0.0026 0.0152 0.1601 R P P 0.0381 0.8221 0.0152 0.1246 * R P Q 0.1824 0.3443 0.0426 0.4307 R R P 0.1371 0.5342 0.0027 0.3260 R R Q 0.6367 0.0355 0.0008 0.3270 (The similarity of, say, the 4th and 5th lines here looks wrong: the intermediate expressions are quite different. I can't explain *why* P_survival(q,r,q) = Q_survival(r,p,p) = 0.8221 but I *have* double-checked this result.)

If I *know* who my opponents are going to aim at, I should shoot at the better shooter if they're both aiming at me or neither is aiming at me. Otherwise I should aim at whoever is *not* aiming at me. There are two equilibrium points (marked "*" above):

Good aims at Bad; Bad aims at Ugly; Ugly aims at Good.

and

Good aims at Ugly; Bad aims at Good; Ugly aims at Bad. Here, unlike for zero-sum two-person games, the equilibria are *not* equivalent and "solution", if any, may lie elsewhere. Perhaps a game-theorist lurking in r.p can offer a better comment.Note that the probability all three shooters die is highest at the equilibria! This seems rather paradoxical and rather sad :-(

2. Who is most likely to survive?

Good, Bad, or Ugly, depending on the strategies.

3. Who is least likely to survive?

Bad or Ugly, depending on the strategies.

4. Can you change p, q, and r under the constraint p > q > r so that the answers to questions 2 and 3 are reversed? Which of the six possible permutations of the three shooters is a possible ordering of probability of survival under the constraint p > q > r?

Yes. Of the six possible survival-probability orderings, five can be obtained readily:

p q r P_surv Q_surv R_surv Order 0.255 0.25 0.01 0.408 0.413 0.172 Q P R 0.26 0.25 0.01 0.412 0.406 0.173 P Q R 0.75 0.25 0.01 0.675 0.076 0.242 P R Q 0.505 0.50 0.01 0.325 0.324 0.344 R P Q 0.505 0.50 0.02 0.314 0.320 0.353 R Q P Unlike the p=.9, q=.7, r=.5 case we are given, the five cases in this table *do* have simple pure solutions: in each case p shoots at q, while q and r each shoot at p. (I've found no case with a "simple pure" solution other than this "obvious" p aims at q, q aims at p, r aims at p choice.)

5. Are there any value of p, q, and r for which it is ever in the interest of one of the shooters to fire into the ground?

No. It can't hurt to shoot at one's stronger opponent. This is the easiest of the questions ... but it's still not easy enough for me to construct an elegant proof in English.

-- David Karr (karr@cs.cornell.edu)

Speaking of decision/truel, I recall a *very* interesting analysis (I *might* have seen it here in rec.puzzles) suggesting that the N-person "truel" (N-uel?) has a Cooperative Solution (ceasefire) if and only if N = 3. But I don't see this in the FAQL; anyone care to repost it?

-- James Allen

In article <1993Apr1.123404.18039@vax5.cit.cornell.edu> mkt@vax5.cit.cornell.edu writes:

In article <1993Mar25.022459.10269@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes: [...]

5. Are there any value of p, q, and r for which it is ever in the interest of one of the shooters to fire into the ground?Yes, p=1, q=1, r=1. The only way for one to survive is to have the other two shoot at eachother. Shooting at anyone has no effect on ones personal survival.

I assume by "has no effect on" you mean "does not improve."

If all follow the same logic, they will keep shooting into the ground and thus all live.

I was very pleased by this answer but I had to think about it. First of all, it assumes that continuing the fight forever has a positive value for each shooter. My preferred assumption is that it doesn't. But even if each shooter is simply trying to maximize his probability of never being shot, I wonder about the "has no effect" statement.

Suppose that in round 1 the Good fires into the ground and the Bad shoots at the Good. Then the Ugly lives if he shoots the Bad and dies if he does anything else. (The Bad will surely shoot at the Ugly if he can in round 2, since this dominates any other strategy.) So it definitely makes a difference to the Ugly in this case to shoot at the Bad.

But all this is under the assumption that no shooter can tell what the others are about to do until after all have shot. This isn't entirely unreasonable--we can certainly set up a game that plays this way--but suppose we assume instead:

All three start out with guns initially holstered.

Each one is a blindingly fast shot: he can grab his holstered gun, aim, and fire in 0.6 second.

A shooter can redirect his unholstered gun at a different target and fire in just 0.4 second.

The reaction time of each shooter is just 0.2 second. That is, any decision he makes to act can be based only on the actions of the other two up to 0.2 second before he initiates his own action.

The bullets travel between shooters in less than 0.1 second and stop any further action when they hit.Then I *think* the conclusion holds for p=q=r=1: The best strategy is to wait for someone else to grab for their gun, then shoot that person, therefore nobody will shoot at anyone. At least I haven't yet thought of a case in which you improve your survival by shooting at anyone. Of course this is only good if you don't mind waiting around the circle forever.

-- David Karr (karr@cs.cornell.edu)

In article <1993Apr5.210749.2657@cs.cornell.edu>,

karr@cs.cornell.edu (David Karr) writes:

In article <1993Apr1.123404.18039@vax5.cit.cornell.edu> mkt@vax5.cit.cornell.edu writes:

In article <1993Mar25.022459.10269@cs.cornell.edu>, karr@cs.cornell.edu (David Karr) writes:

[...]

5. Are there any value of p, q, and r for which it is ever in the interest of one of the shooters to fire into the ground?Yes, p=1, q=1, r=1. The only way for one to survive is to have the other two shoot at eachother. Shooting at anyone has no effect on ones personal survival.

I assume by "has no effect on" you mean "does not improve."

If all follow the same logic, they will keep shooting into the ground and thus all live.

I was very pleased by this answer but I had to think about it. First of all, it assumes that continuing the fight forever has a positive value for each shooter. My preferred assumption is that it doesn't. But even if each shooter is simply trying to maximize his probability of never being shot, I wonder about the "has no effect" statement.

Suppose that in round 1 the Good fires into the ground and the Bad shoots at the Good. Then the Ugly lives if he shoots the Bad and dies if he does anything else. (The Bad will surely shoot at the Ugly if he can in round 2, since this dominates any other strategy.) So it definitely makes a difference to the Ugly in this case to shoot at the Bad.

Here's where the clincher comes in! If we "assume" the object of the game is to survive, and that there exists _one_ unique method for survival, then all the shooters will behave in the same fashion. Obviously the above case will not hold. How do we distinguish between the good, the bad and the ugly? If the command is "Shoot" then all will shoot and somebody is going to wind up lucky (Prob that it is you is 1/3). If the command is "No Shoot", then all will fire into the ground (or just give up and go home--no sense waitin' around wastin' time, ya know)...

But wait, you cry! What if there exists _more than one_ solution for optimal survival. Then what? Will the Good the Bad and the Ugly each randomly decide between "Shoot" and "No Shoot" with .5 probability? If this is true, then is it in your best interest to shoot someone? If it is, then we arrive back at square one: since we assume all shooters are geniouses, then all will shoot-- arriving at an optimal solution of "Shooting". If the answer is "No Shooting", we arrive at an optimal solution of "No Shooting". If there is no effect on your personal survival, then do we analyze this with another .5 probability between the chances of soemone shooting or not shooting? If the answer to this is "Shoot" then we arrive at square one: all will Shoot; if no, then all will withold. If there is no effect, then we arrive at another .5 probability... Obviously you can see the recursion of this process.

Perhaps this would be easier to discuss if we let p=1, q=1, r=0. Obviously, in terms of survival, shooting at the ugly would be wasting a shot. Thus we have made a complex problem more simple but retaining the essence of the paradox:

If there are two gunmen who shoot and think with perfect logic and are kept inside a room and are allowed to shoot at discrete time intervals without being able to "see" what your opponent will do, what will happen?

Let's say that you are one of the gunmen (the Good). You reason "My probability to survive the next round is independent on whether or not I fire at him." So you say to yourself, "Fire at the opponent! I'll get to stop playing this blasted game." But then you realize that your opponent will also think the same way...so you might think that you might as well not shoot. But if your opponent thinks that way, then you know that: 1. You can survive the next round. 2. You can shoot him if you wish on this round (if you like). So you say to yourself, "Fire at the opponent!". But you know the opponent thinks the same way so... you're dead. But really, you say. Is there a way of "knowing" what the opponent thinks? Of course not. You reason that you can know your probability of shooting your opponent (either 1 or 0). You reason that your opponent has a variable probability of shooting you. Thus from your perspective, p=1 and r<1. We already discussed this case and said "Shoot!". But wait you cry! What if the opponent figures this out too: p<1, r=1? Sorry, you're both dead. 'nuff said! This applies to the p=r=q=1 case as well.

But all this is under the assumption that no shooter can tell what the others are about to do until after all have shot.

Ay, there's the rub!

This isn't entirely unreasonable--we can certainly set up a game that plays this way--but suppose we assume instead:

All three start out with guns initially holstered. Each one is a blindingly fast shot: he can grab his holstered gun, aim, and fire in 0.6 second.

A shooter can redirect his unholstered gun at a different target and fire in just 0.4 second.

The reaction time of each shooter is just 0.2 second. That is, any decision he makes to act can be based only on the actions of the other two up to 0.2 second before he initiates his own action.

The bullets travel between shooters in less than 0.1 second and stop any further action when they hit.Then I *think* the conclusion holds for p=q=r=1: The best strategy is to wait for someone else to grab for their gun, then shoot that person, therefore nobody will shoot at anyone. At least I haven't yet thought of a case in which you improve your survival by shooting at anyone. Of course this is only good if you don't mind waiting around the circle forever.

Hmmn...alternate ploy:

0.0 You begin to unholster your gun

0.2 Opponents begin unholstering guns. You aim into the ground for .2 sec.

0.4 Opponents are unholstered you are unholstered. They note you aren't aiming at them. They haven't aimed at anyone yet.What happens now? I'll have to think about it, but I haven't seen anything fundamentally different between this and the above case yet.

More ideas to consider:

You begin unholstering your gun but only for .1 sec (you place it by .2 )

You begin unholstering your gun but only for .09 sec (you place it by .19)You start to aim for .1 sec and then stop aiming.

You start to aim for .1 sec and then turn and aim at another.

You start to aim for .09 sec and then stop aiming (or aim at another)-Greg

Looking at the answer for decision/truel, I came across the following:

Each player can obtain a positive expected payoff by shooting at one of the other players, and it is known that an infinite sequence of misses will result in a zero payoff for all players. So it is known that some player's strategy must involve shooting at another player rather than deliberately missing.

This may be true but it's not obvious to me. For example, suppose A, B, and C are passengers in a lifeboat in a storm. If they all stay aboard, the lifeboat is certain to sink eventually, taking all three to the bottom with it. If anyone jumps overboard, the two remaining in the boat are guaranteed to survive, while the person who jumped has a 1% chance of survival.

It seems to me the lifeboat satisfies the quoted conditions, in the sense that if nobody jumps then the payoff for all is zero, and the payoff for jumping is 0.01 which is positive. But it is not clear to me that the three shouldn't just all sit still until someone goes nuts and jumps overboard despite everything, for this strategy gives a 67% chance of survival (assuming everyone is equally likely to "crack" first) vs. only 1% for jumping by choice. Even if there is a wave about to swamp the boat, I'd wonder if the situation wouldn't just reduce to a game of "chicken," with each person waiting until the last minute and jumping only if it seems the other two have decided to sink with the boat if you don't jump.

On the other hand, this situation is set up so it is always worse to be the first person to jump. In the truel I don't think this is true, but only because of the asymmetry of the odds, and to determine whether *anyone* shoots, it is easiest to proceed directly to considering B's point of view.

Whenever it is B's turn to shoot, B can divide the possible courses of action into four possibilities (actually there are seven, but three are ruled out a priori by obvious optimizations of each individual's strategy):

Nobody ever shoots (expected value 0)

A shoots first (at B, expected value <= .7)

C shoots first (at B, expected value <= .5)

B shoots first (at C, expected value .7)In fact the value of "A shoots first" is strictly less than .7 because in case A misses, the same four possibilities recur, and all have expected payoff < 1.

So the value of "B shoots first" uniquely maximizes B's value function, ergo B will always shoot as soon as possible.

The rest of the analysis then follows as in the archive.

-- David Karr (karr@cs.cornell.edu)

Looking at the answer for decision/truel, I came across the following:

Each player can obtain a positive expected payoff by shooting at one of the other players, and it is known that an infinite sequence of misses will result in a zero payoff for all players. So it is known that some player's strategy must involve shooting at another player rather than deliberately missing.

This may be true but it's not obvious to me. For example, suppose A, B, and C are passengers in a lifeboat in a storm. If they all stay aboard, the lifeboat is certain to sink eventually, taking all three to the bottom with it. If anyone jumps overboard, the two remaining in the boat are guaranteed to survive, while the person who jumped has a 1% chance of survival.

It seems to me the lifeboat satisfies the quoted conditions, in the sense that if nobody jumps then the payoff for all is zero, and the payoff for jumping is 0.01 which is positive. But it is not clear to me that the three shouldn't just all sit still until someone goes nuts and jumps overboard despite everything, for this strategy gives a 67% chance of survival (assuming everyone is equally likely to "crack" first) vs. only 1% for jumping by choice. ...

Yes and no. Yes in the sense that if you treat the game as a psychological one, the best strategy is as you say. But treating it as a mathematical game, you've got to adhere to your strategy and you've got to assume that others will adhere to theirs.

I.e. as a mathematical game, "Don't jump at all" and "Don't jump unless I crack" are different strategies, and the first one is often (not always) superior - e.g. if I take "Don't jump at all" and the others take "Don't jump unless I crack", I'm certain to survive and the others each have a 50.5% chance, which is better from my point of view than a 67% chance of survival for all of us. As a psychological game, some of the mathematical strategies may simply not be available - i.e. you cannot control what you will do if you crack, and so we commonly use "Don't jump" to mean "Don't jump unless I crack", since "Don't jump at all" is not an available strategy for most real humans. But for mathematical analysis, the problem has to tell you what strategies you are not allowed to take.

What the argument above shows is that "Don't jump at all" is not a stable strategy, in the sense that if everyone takes it, it is in everyone's interest to change strategy. I.e. it shows that someone will jump eventually, even if it's only the result of someone actually having taken "Don't jump unless I crack".

Applied to the truel, the argument above *does* show that someone's strategy will involve shooting at another player: the strategy "Don't shoot at all" is unstable in exactly the same way as "Don't jump at all" was. But I agree it allows for a lot of leeway about how and when the deadlock gets broken, and your argument showing that it is always in B's interest to shoot is more satisfactory.

David Seal