BS: An amusing little problem

Subject: BS: An amusing little problem
From: BK Lick
Date: 10 Feb 13 - 04:11 AM

Suppose I present you with two personal checks made out to you and offer to gift you with whichever one of them you choose.

I place the checks face down before you, labeled A and B on their backs, and stipulate that their values are 2ⁿ and 2ⁿ⁺¹ dollars for some integer value of n.

I invite you to turn one of them face up and then choose one of them to keep as a gift from me.

Suppose that the check you turn face up has value K. Then the other check is worth either 2K or K/2 and therefore by choosing the face-down check instead of the one you've revealed, you stand to gain K dollars at the risk of losing only K/2. Thus it appears that whatever the value of the check you turn over, it is to your advantage to choose the other.

Suppose now that the game is changed and you must make your choice without seeing the value of either check. Say you choose check A. Whatever its value, by the reasoning above it's to your advantage to change your mind and choose check B. But then, of course, you should change back to check A, and so forth -- but this is surely absurd.What's going on here? Why should the act of revealing the value of one check make it advisable to choose the other?

—BK

Subject: RE: BS: An amusing little problem
From: SPB-Cooperator
Date: 10 Feb 13 - 05:59 AM

I haven't read up on game theory since te early 80s

Subject: RE: BS: An amusing little problem
From: DMcG
Date: 10 Feb 13 - 06:20 AM

Its an interesting to which the answer is be no means too obvious. It is, I hope, clear that if you played the game sufficient times the average return per game would be the same whether you always switched or never switched. Equally it is clear that you could write two cheques like that and make the offer. If is also true that on an individual game by choosing the face-down check instead of the one you've revealed, you stand to gain K dollars at the risk of losing only K/2. So the flaw must be in the assertion that it is therefore sensible to switch. And it is, since we agreed that overall switching brings no benefit. SO we have a mismatch in our understanding of potential benefit for a single game and the impact that has on out our probable benefit.

Subject: RE: BS: An amusing little problem
From: DMcG
Date: 10 Feb 13 - 06:21 AM

Sorry for the typos!

Subject: RE: BS: An amusing little problem
From: kendall
Date: 10 Feb 13 - 08:20 AM

I wouldn't place any value on either check from a stranger.

Subject: RE: BS: An amusing little problem
From: SINSULL
Date: 10 Feb 13 - 09:58 AM

I'm from NYC. I'd take both checks and leeave.
SINS

Subject: RE: BS: An amusing little problem
From: GUEST,Grishka
Date: 10 Feb 13 - 02:52 PM

My feeling (not knowledge) is that if you know nothing about the probability of any particular n or K being chosen, you cannot say anything about the probability of the value of the other check. If however this probability is known (or you can make a reasonable guess judging by the issuer's generosity), it cannot be equal for all those infinitly many possible n. From your K, you can then calculate the probability for the other check being of double value, which will rarely be 50%.

If my K is high, I would keep that check. If it is very low, I may take the other one.

BK, do you know the correct answer? Do we have mathematicians here?

Subject: RE: BS: An amusing little problem
From: DMcG
Date: 10 Feb 13 - 04:19 PM

Yes, me for one. But as you say if it took place in real life there's a complicating psychological factor. If you chose £4 you may well swap because it is not a significant amount. If the amount was £32,768 you would probably keep it because losing half that is significant. However, if it was £20,000,000 you might well swap again because £10,000,000 is more than most people are likely to have in one go it is worth the risk again

Subject: RE: BS: An amusing little problem
From: Joe_F
Date: 10 Feb 13 - 06:24 PM

I agree with Grishka. The fallacy lies in the superficially plausible assumption that it is equally probable that the revealed check will turn out to be the smaller or the larger one. Consider the problem faced by the person who writes the checks. Even if he has infinite resources, it is impossible for him to choose a sum randomly from that infinite set, for then the probability of any particular value would have to be zero. In the real world, of course, he has finite resources, and that fact makes certain outcomes impossible. You might guess that he is only a billionaire; then if the visible check is for $600,000,000, it must be the larger one. In general, the larger the value of the visible check, the greater the likelihood that it is the larger one.

DMcG is also right in pointing out the complication that (in decision-theoretic terms) the value of money is not linear in money. However, that is not the real source of the difficulty.

Subject: RE: BS: An amusing little problem
From: GUEST,leeneia
Date: 10 Feb 13 - 06:58 PM

What if n's a negative number?

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 10 Feb 13 - 07:25 PM

Yes, insert 0 ≤ n after for some integer value of n.

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 10 Feb 13 - 07:47 PM

Some good insights are presented above, focusing on the infinite number of possibilities. Consider a finite formulation of the game, say stipulating that 0 ≤ n ≤ 4 resulting in only ten possibilities -- then it is advantageous to choose the unrevealed check in every case except when the revealed check is worth $32.

Subject: RE: BS: An amusing little problem
From: Rapparee
Date: 10 Feb 13 - 09:01 PM

If n is less than or equal to 0, then K is either $0 or a negative number (e.g., $2^-1 is -$2, no? It's a lousy gift that takes away my money! And $2^-1+1 is zero bucks. You running a three card monte game or something?

Subject: RE: BS: An amusing little problem
From: GUEST,999
Date: 11 Feb 13 - 02:47 AM

Examples of integers which increase in both directions simultaneously

-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5

The original statement

2n and 2n+1

Statistically, there are more possibilities to have the amount of the cheque increased when the integer value is 0 or any positive number by a margin of 6 to 5, so take the cheque that has not been turned over.

However, as stated earlier, if the first cheque is a worthwhile value take it and sight unseen burn the second cheque.

Subject: RE: BS: An amusing little problem
From: Rumncoke
Date: 11 Feb 13 - 01:32 PM

Just take either one, say thank you and put it in your pocket - whatever it is you're better off than when you met the strange person who wants to perplex you, and you haven't spent time being disconcerted or uncertain of your own judgement.

Subject: RE: BS: An amusing little problem
From: McGrath of Harlow
Date: 11 Feb 13 - 08:09 PM

Suppose that the check you turn face up has value K. Then the other check is worth either 2K or K/2 and therefore by choosing the face-down check instead of the one you've revealed, you stand to gain K dollars at the risk of losing only K/2. Thus it appears that whatever the value of the check you turn over, it is to your advantage to choose the other.

I can't see the problem - or rather I can't see how that last sentence stands up. If the other cheque is in fact worth half the cheque you have turned up it would not be to your advantage to choose it.

Subject: RE: BS: An amusing little problem
From: gnu
Date: 11 Feb 13 - 08:38 PM

Monte Hall thread all over again with a slight twist. (Perhaps this comes as a clarification of that thread's basic analyses?)

One choice, two options. No matter how many times you get to "change your choice", it will never change your odds as there is only ONE choice between two options no matter how many times one changes that choice.

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 12 Feb 13 - 02:47 AM

Joe_F, Grishka, and DMcG have all touched on problems introduced by allowing the set of all possible checks to be unbounded. It may serve to clarify matters somewhat to consider a small, finite case; let's say that 0 ≤ n ≤ 4. Then the set of possible values for Checks A and B consists of the following ten combinations:

   A   B
  
   1   2
   2   1
   2   4
   4   2
   4   8
   8   4
   8  16
  16   8
  16  32
  32  16

Let's assume that I randomly choose one of these ten combinations to present to you for your choice (using a random number generator or a table of pseudo-random numbers) so that the probability of each combination is 0.1.

In that randomly chosen combination:
If Check A is worth 1 then Check B is worth twice as much.
If Check A is worth 32 then Check B is worth half as much.
If Check A is worth 2, 4. 8. or 16 then the probability that Check B is worth twice as much is 0.5 (as is the probability that Check B is worth half as much).

In the absence of any other information it seems clear that choosing Check A, you will on average receive (1+2+2+4+4+8+8+16+16+32) / 10 = 9.3 dollars. Likewise, choosing Check B, you will on average receive (2+1+4+2+8+4+16+8+32+16) / 10 = 9.3 dollars.

However, if you are allowed to see the value of Check A before choosing and you then choose Check A if it's worth $32 and choose Check B in every other case, you will on average receive (2+1+4+2+8+4+16+8+32+32) / 10 = 10.9 dollars.

In general, for 0 ≤ n ≤ N, it's straightforward to calculate that you will on average receive [3(2^N+1 - 1) ] / 2(N + 1) in the absence of other information and [3(2^N+1 - 1) + 2^N] / 2(N + 1) if you can first see Check A and then choose Check B unless Check A has the value 2^N+1.

Subject: RE: BS: An amusing little problem
From: GUEST,BobL
Date: 12 Feb 13 - 04:19 AM

Rapparee, a negative exponent doesn't result in a negative value!

2² = 4
2¹ = 2
2⁰ = 1
2^-1 = 0.5
2^-2 = 0.25

Which presumably sets the lower bound for N as -2, with cheques made out for 50 cents and 25 cents.

Subject: RE: BS: An amusing little problem
From: Mysha
Date: 12 Feb 13 - 07:09 AM

Hi,

The cheques are k and 2k, for a strict positive value of k. Since you don't know the value of k, actually looking at the cheque has no influence whatsoever.

You pick one of the cheques:
- 50% chance to pick k
- 50% chance to pick 2k.

If you were to switch, they continue as:
- 50% chance to pick k, and then switch to 2k, winning k
- 50% chance to pick 2k, and then switch to k, losing k

As you can see, the winning and losing values of switching are equal, so switching is not worth the bother (but does no harm either).


How is this different from the initial approach? There, the extremes of two separate cases are compared:
Case 1: Choose one of 2k and k
- 50% chance to pick k, and then switch to 2k, winning k
- 50% chance to pick 2k, and then switch to k, losing k

Case 2: Choose one of k and 0.5k
- 50% chance to pick 0.5k, and then switch to k, winning 0.5k
- 50% chance to pick k, and then switch to 0.5k, losing 0.5k

This is incorrectly combined to
- 50% chance to pick k, and then switch to 2k, winning k
- 50% chance to pick k, and then switch to 0.5k, losing 0.5k
by considering the read value k in both cases. However, the knowledge of the problem shows these to be separate cases.


It's that knowledge that makes this different from the quiz master who says:
- You! have won! k! But!! You may open! the Joker! box.! If you do?, the amount is either halved?? or doubled!!!
If the winner could reasonably assume the chances to half and double are equal, then indeed it would be advantagous to open the box.


Nice one; really had to think this one through.

Bye,
                                                                Mysha

Subject: RE: BS: An amusing little problem
From: Pete Jennings
Date: 12 Feb 13 - 07:14 AM

Send me the cheques and I'll let you know the answer.

Subject: RE: BS: An amusing little problem
From: Donuel
Date: 12 Feb 13 - 10:26 AM

You never said if those checks are mine to begin with.

There is no advantage to any choice.

Subject: RE: BS: An amusing little problem
From: McGrath of Harlow
Date: 12 Feb 13 - 12:15 PM

Can BK Lick, or someone, explain why exchanging a cheque you have for one half its value "is to your advantage" in the scenario outlined?

Subject: RE: BS: An amusing little problem
From: catspaw49
Date: 12 Feb 13 - 01:08 PM

If you were in Chicago though, unlike New York nowadays, they would take BOTH checks, shoot your ass and the 14 innocent bystanders as well.


Spaw

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 12 Feb 13 - 06:27 PM

Mysha, you've put your finger exactly on the perplexing aspect of this situation -- but, alas, you have not explained away the mystery as you cheerily seem to believe you have. As I concluded my original post: What's going on here? Why should the act of revealing the value of one check make it advisable to choose the other?

Consider the small, finite case scenario I described above. Say I've randomly generated the number five and accordingly presented to you two checks, face down, labeled A and B. worth $4 and $8, respectively. You turn over Check A and reveal that its value is $4. Now you know that Check B is worth either $2 or $8 with equal probability of 1/2. So the decision to choose B instead of taking the $4 is equivalent to placing a bet with me on the choice of a coin flip where you put up $2 but I put up $8. Clearly that decision would be favorable to you.

Subject: RE: BS: An amusing little problem
From: Bobert
Date: 12 Feb 13 - 07:10 PM

Bad news, BK...

I just got off the phone with yer banker and he says yer checking account is over-drawn...

Hated to have to tell you this...

B~

Subject: RE: BS: An amusing little problem
From: McGrath of Harlow
Date: 12 Feb 13 - 08:02 PM

Clearly that decision would be favourable to you

No it wouldn't. No bet is favourable if you lose. It might well be a rational bet if the odds are good and you can afford to lose, but that is a completely different thing. And the crucial thing isn't ever how much you might win, but how much you might lose.

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 12 Feb 13 - 11:52 PM

I find it hard to believe that McGrath actually means to say that a bet on the outcome of a coin-flip, in which he will collect $8 for winning and pay only $2 for losing, would not be favourable to him.

Subject: RE: BS: An amusing little problem
From: GUEST,Grishka
Date: 13 Feb 13 - 06:03 AM

McGrath does have a point: for a person, the value of winning or losing money may not be the same as its face value. If you proverbially "put your shirt on a horse", i.e. bet with your entire existence, you are being unreasonable ("addicted") no matter what the chances. Conversely, if you need a medicine for your life but only own a fraction of its price, an unfavourable bet may be reasonable if no better one is available.

Subject: RE: BS: An amusing little problem
From: GUEST,Grishka
Date: 13 Feb 13 - 06:07 AM

PS: The situations I mentioned are not as uncommon as you may think. The unfavourable bet is called "health insurance".

Subject: RE: BS: An amusing little problem
From: TheSnail
Date: 13 Feb 13 - 06:35 AM

My chances of gaining or losing if I switch are not determined by my knowledge or, more to the point, lack of knowledge, of the values of the cheques but by their actual values. In the above example, the cheques have values $4 and $8. If I have chosen the $4 cheque, I don't know whether the other cheque is $2 or $8 but if I switch I will not get the $2 cheque because it does not exist. I will get the $8 cheque with absolute certainty and be $4 better off.

If I had first chosen the $8 cheque, I would not know whether the other cheque was $4 or $16 but if I switch, I will not get the £16 cheque because it does not exist. I will get the $4 cheque with absolute certainty and be $4 worse off.

The way the problem is stated turns "I don't know whether the other cheque is half or double the one I've got" into "There is an equal chance that the other cheque is half or double the one I've got". Sorry, not so.

Incidentally, the 2ⁿ and 2ⁿ⁺¹ business is just a smoke screen to add further confusion. The cheques don't have to have values of powers of 2. It is sufficient that one is twice the other. $5 and $10 would have worked just as well.

Subject: RE: BS: An amusing little problem
From: McGrath of Harlow
Date: 13 Feb 13 - 06:45 AM

A losing bet is never a good bet, however tempting it might have looked in advance.

Tempting is not the same as, to use the language of the OP "to your advantage".

Subject: RE: BS: An amusing little problem
From: Nigel Parsons
Date: 13 Feb 13 - 06:55 AM

"A bird in the hand is worth two in the bush"

"A ring on the finger's worth two on the phone!"

Subject: RE: BS: An amusing little problem
From: TheSnail
Date: 13 Feb 13 - 07:04 AM

"A mole on the arm is worth two on the leg"

Subject: RE: BS: An amusing little problem
From: GUEST,Grishka
Date: 13 Feb 13 - 08:24 AM

Here is another one, for those who know about probability:

In an imaginary TV show, there is a large horizontal clock face with an hour hand mounted on a ball bearing. The hand is pushed so that it rotates many times. When at last it comes to a standstill, its position in terms of hours past midnight (larger than zero, up to 12) is measured exactly, including the fractional part. The "candidate" wins one dollar divided by that position number: 1/12 dollar for 12 o'clock, 1/6 dollar for 6 o'clock, 60 dollars for 1 minute past midnight, etc. Of course, the amount is rounded to full cents. The TV company asks an insurance company to take the risk. What would be a fair insurance premium? Or, in other words: how much would it be worth to be that candidate (strictly by face value, disregarding the usefulness for a particular person)?

(Courtesy of a friend of mine who minored in maths.)

The connection to the OP: Assume the TV company operates this oracle to determine an amount N, invisible for you, but closely watched by an impartial jury. Then the company writes one check for N and another one for twice N. The rest is as in the OP.

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 14 Feb 13 - 07:32 AM

Let me try one more time to help poor Snail grasp the situation: here's what's happened, in slow motion -- watch closely.

I've generated a random number between 1 and 10. You don't know what the number is. Let's suppose for the sake of this discussion that it's 5, but for all you know it could have been any number between 1 and 10.

Accordingly, I chose the fifth combination from the list of ten possible combinations and wrote Check A for $4 and Check B for $8, but for all you know each check could be worth 1, 2, 4, 8, 16, or 32 dollars.

Now you turn over Check A and observe it's worth $4. So you can determine that I must have chosen either the fourth combination (with Check B worth $2) or the fifth combination (with Check B worth $8) and they are equally likely, but you don't know which it was. Remember, we just chose the number 5 for the sake of this discussion.

So indeed it is so: there is an equal chance that Check B is half or double Check A.

Hope that helps. (And why, by the way, would I wish to employ a smoke screen to add further confusion? There is no attempt at deception involved here.)

Subject: RE: BS: An amusing little problem
From: TheSnail
Date: 14 Feb 13 - 02:58 PM

there is an equal chance that Check B is half or double Check A.

No, there is not. Cheque B is $8. It has no chance of being $2.

On the other hand, my best guess, given the limited information that I have, is that there is an equal chance that cheque B is half or double cheque A. These odds are a reflection of my uncertainty about the situation; they do not magically invoke the possibility that cheque B might actually be $2. It isn't, it's $8. The outcome, when I finally settle on one of the cheques, is not determined by my ignorance but by the actual facts. If I switch, I will get $8 not $2.

And why, by the way, would I wish to employ a smoke screen to add further confusion?. It's fairly standard in this sort of puzzle. Part of the skill in solving them is to identify the relevant facts.

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 14 Feb 13 - 04:22 PM

Ah, my molluscan friend, you are being so perversely unreasonable.

Imagine we place a coin into an opaque box which we then shake vigorously.
Would you not then agree that there is an equal chance that the coin is lying heads or tails? The position you're taking is equivalent to saying in this situation: "No, there is not. Whichever way the coin lies, it has no chance of lying the other way."

And you have assumed incorrectly that this is a puzzle to be skillfully solved.

Subject: RE: BS: An amusing little problem
From: TheSnail
Date: 14 Feb 13 - 06:17 PM

The position you're taking is equivalent to saying in this situation: "No, there is not. Whichever way the coin lies, it has no chance of lying the other way."

On the contrary, I am saying that it can ONLY be heads or tails. You seem to be saying that it has an equal chance of being legs.

In your example there are two cheques, one for $4 and one for $8. You know this because you wrote them. The $4 cheque is exposed and you are insisting that the unexposed cheque has an equal chance of being $2 or $8. How can that be? Where has this $2 cheque come from?

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 14 Feb 13 - 06:46 PM

No, there are not just two checks -- there are initially ten possibilities, all of them being equally likely. When Check A is revealed to be for $4 eight of the possibilities are ruled out, and only two are left; one possibility is that Check B is for $2, the other is that Check B is for $8. Each of these remaining possibilities is equally likely.

Try tilting your head slightly when you look at the situation.

Subject: RE: BS: An amusing little problem
From: TheSnail
Date: 14 Feb 13 - 07:36 PM

Previously from you -

Say I've randomly generated the number five and accordingly presented to you two checks, face down, labeled A and B. worth $4 and $8, respectively.

At the time that I am making my choices there are two cheques. One of them is worth $4 and the other is worth $8. Neither of them is worth $2.

Subject: RE: BS: An amusing little problem
From: McGrath of Harlow
Date: 14 Feb 13 - 08:47 PM

Basically it's a restatement of Schlesinger's cat thought experiment.   And I've tended to agree with Schlesinger himself, who viewed this as a reduction ad absurdem.

"The position you're taking is equivalent to saying in this situation: "No, there is not. Whichever way the coin lies, it has no chance of lying the other way." Precisely so. Not in this universe anyway. "Has" is not to be confused with "had".

Subject: RE: BS: An amusing little problem
From: Jeri
Date: 14 Feb 13 - 08:57 PM

I KNOW it's possible for a cat to be owned by two people simultaneously, or so THEY think. But is it possible that those two individuals actually are one person named "Schrödinger"?

I obviously know nothing about quantum psychics.

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 14 Feb 13 - 10:22 PM

"At the time that I am making my choices there are two cheques. One of them is worth $4 and the other is worth $8. Neither of them is worth $2."

I give up -- I fear that understanding this matter is beyond The Snail's ability.

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 14 Feb 13 - 10:29 PM

Yes, Erwin Rudolf Josef Alexander Schrödinger -- his celebrated thought experiment is charmingly portrayed here:
Schrödinger's Cat

Subject: RE: BS: An amusing little problem
From: TheSnail
Date: 15 Feb 13 - 05:27 AM

That's not fair! I thought this was a probability question not a quantum mechanics question. What, pray tell, is the equivalent of the atomic decay and the vial of poison, that may or may not have killed Schrödinger's cat, that can transform a cheque made out to $8 into one made out to $2?

Subject: RE: BS: An amusing little problem
From: GUEST,999
Date: 15 Feb 13 - 09:02 AM

Q: "What, pray tell, is the equivalent of the atomic decay and the vial of poison, that may or may not have killed Schrödinger's cat, that can transform a cheque made out to $8 into one made out to $2?"


A: Lysergic acid diethylamide

Subject: RE: BS: An amusing little problem
From: IanC
Date: 15 Feb 13 - 10:02 AM

This is essentially a variant of Zeno's "Achilles and the tortoise" paradox. Like theo original, It's based on a false premise - as pointed out by TheSnail.

:-)

Subject: RE: BS: An amusing little problem
From: BK Lick
Date: 15 Feb 13 - 11:52 PM

There are a few of you following this thread that seem genuinely interested but are completely missing the point of the original post. There is no trick here, no puzzle to be solved -- just a very easy to follow probabilistic analysis that leads to a surprisingly counter-intuitive result.

I seem to have led TheSnail astray by writing "Say I've randomly generated the number five and accordingly presented to you two checks, face down, labeled A and B, worth $4 and $8, respectively." That was intended only as an illustrative case. Please forget about it and take a fresh look at what I'm saying here.

You see, if you focus on the one case [A = 4, B = 8] you are considering only one possible event from the following ten-member "event-space":
[A = 1, B = 2] [A = 2, B = 4] [A = 4, B = 8] [A = 8, B = 16] [A = 16, B = 32]
[A = 2, B = 1] [A = 4, B = 2] [A = 8, B = 4] [A = 16, B = 8] [A = 32, B = 16]
if we can use this shorthand notation instead of talking about checks and dollars.

Now let's consider a sequence of "trials." In each trial I randomly choose to present to you for your choice one of these ten combinations (using a random number generator or a table of pseudo-random numbers) so that the probability of each combination is 1 in 10.

In any one trial you are presented with some combination [A = a, B = b] where, with equal probability, either a = 2b or a = b/2.

Then, over a succession of 100 trials, on average, each one of the ten possible combinations would occur ten times. Over those 100 trials, on average, A = 1 would occur 10 times; A = 32 would occur 10 times; and A = 2, A = 4, A =8, and A = 16 would each occur 20 times. (The last sentence remains true with B substituted for A.)

Clearly, in any given trial when both the values of A and B are unknown there is no advantage in choosing A or in choosing B.

However, if in a given trial you can learn the value of A, that new information changes the situation. For then
if A = 1, then B = 2 -- you should obviously choose B
if A = 2, then B = 1 or B = 4 with equal probability -- you should obviously choose B
if A = 4, then B = 2 or B = 8 with equal probability -- you should obviously choose B
if A = 8, then B = 4 or B = 16 with equal probability -- you should obviously choose B
if A = 16, then B = 8 or B = 32 with equal probability -- you should obviously choose B
if A = 32, then B = 16 -- you should obviously choose A

Note that if the value of A is revealed in a succession of trials, 90% of the time (whenever A ≠ 32) it's to your advantage to choose B.

Now consider a slightly larger example where the check values are bounded by 2¹⁰ = 1024 instead of 32; the same reasoning demonstrates that if the value of A is revealed in a succession of trials, 95% of the time (whenever A ≠ 1024) it's to your advantage to choose B.

In general, when the check values are bounded by 2^N the same reasoning demonstrates that if the value of A is revealed in a succession of trials, (100 - 50/N)% of the time (whenever A ≠ 2^N) it's to your advantage to choose B.

In the infinite case where there is no bound on the check values the same reasoning leads to the conclusion that it's always to your advantage to choose B, no matter what value of A is revealed.

Only in the infinite case is the result perplexing. Then one can ask: What's going on here? Why should the act of revealing the value of one check make it advisable to choose the other?

Subject: RE: BS: An amusing little problem
From: GUEST,999
Date: 16 Feb 13 - 02:05 AM

Good Keriste: If maths were that easy everyone would be doing them!

If I may, the premise is faulty.

The premise says that there are two choices. In fact there are three.



That's one small toke for man, and one GIANT toke for . . . . .