Author |
Topic: HARD: ENVELOPE GAMBLE (Read 9810 times) |
|
Padzok
Junior Member
Gender: 
Posts: 104
|
 |
Re: HARD: ENVELOPE GAMBLE
« Reply #100 on: Sep 14th, 2004, 11:02am » |
 Quote  Modify
|
Quote:
There are 1000 pairs of envelopes, the kth pair being {2k-1, 2k} (suppose for simplicity and to save my putting in so many superscripts, the actual contents of each envelope are a piece of card with a number between 0 and 1000 on (inclusive), representing an actual value of 2 to that power) |
|
on Sep 9th, 2004, 2:15pm, rmsgrey wrote:
So in practice, I pick an abitrary large integer and decide in advance that I'll switch for anything less than that and accept that moving my cut-off higher would improve my expected winnings (provided the probabilities are such that (on the log2 scale) P(K)/P(K+2)<2) |
|
I agree of course that swapping for all (k-1) for k=1 to k=1000 is the best strategy, so long as we stick when we actually do see 21000 on the card.
But is that not a feature of this particular series? Namely 21000 dwarfs the second highest number in the series (2999 which appears twice).
In other words, is it not the very fact that we have chosen a tactic which guarantees a win for 21000 which produces the benefit,and not the fact we swap for every other number?
Indeed, according to my arithmetic (which is usually wrong) if we swapped for every single number, then the loss from 21000 to 2999 is so great that it means the entire strategy shows a net deficit.
I do think the idea of taking an arbitrary cut-off is promising. 
But I think that higher is not necessarily better. 
For a finite series of the type rmsgrey describes, where we did not know the upper limit of k, then the most important thing is that our cut off is somewhere below the upper limit of k.
I think that will always show a profit (again my arithmetic could be wrong) for this particular type of series. But if the cut off is even slightly too high, then we would expect a loss over a long enough time period.
Is it agreed this type of series is a very special case, though? If I was playing this gameshow, my strategy would not be based on the assumption that the show used such a series from which to draw its envelopes.
{PS...I would still gamble on a swap though...  }
|
|
 IP Logged |
|
|
|
rmsgrey
Uberpuzzler
    
Gender:
Posts: 2873
|
|
Re: HARD: ENVELOPE GAMBLE
« Reply #101 on: Sep 15th, 2004, 6:40am » |
Quote Modify
|
For the finite case, always swapping has to break even - for instance, with the 1000 pairs of envelopes, if you play 2000 times, and (by some freak of circumstance) get each possible outcome exactly once, you'd get the same set of 2000 outcomes as if you never swapped - each value between 0 and 1000 twice, and the extremes once each.
For a more general distribution, the role of the unique end values is taken by the set of values less than twice the minimum value and the set of values more than half the maximum.
In general, playing with a cut-off means you win big on the cut-off value, and break even elsewhere in the long run, so for any given distribution, your winnings are solely determined by the cut-off value. With infinite distributions with unbounded expectation, the higher the cut-off value, the greater the expected winnings, but it's unclear what happens when the cut-off actually reaches infinity.
|
|
IP Logged |
|
|
|
Nasta
Guest
|
|
Re: HARD: ENVELOPE GAMBLE
« Reply #102 on: Sep 3rd, 2005, 5:38pm » |
Quote Modify
 Remove
|
nice problem! i read all the pages and was very intrigued by some posts, esp these of rmsgray. he had some really deep insight that is going right in the heart of the problem - which is - THE FORMULATION of the riddle, and THE ASSUMPTION of the player in it. Lets use 200 and 400 as the "real world" numbers. two cases, equally likely that may happen are - he gets 200, thinks i can get 400 x 0.5 + 100 x 0.5. Or he gets 400 and thinks - i will get 800 x 0.5 + 200 x 0.5 and obviously in both cases i win if i switch. truth is however that he cannot ever IN FACT get 800 or 100, and the imaginary expected value won by getting 800(which cannot happen in fact) is bigger than the imaginary expected value brought by getting 100(which cannot be achieved either). the 800 brings +400(equal to the sum in his envelope), while 100 brings +50(0.25 of the sum in his envelope). The bonus comes just because one of the assumed figures u supposedly get is twice as higher than the higher real value of the actual envelopes, and the other is twice as lower as the lower of the envelopes u get. i think that is what rmsgray meant and i hope i got it a little clearer(if not a lot more confusing :lol: ) for you now
|
|
IP Logged |
|
|
|
Nasta
Newbie
Gender:
Posts: 9
|
|
Re: HARD: ENVELOPE GAMBLE
« Reply #103 on: Sep 3rd, 2005, 5:48pm » |
Quote Modify
|
This is why the reasoning will fail in the long run - despite the player's thoughts bringing 500 EV from 400 = X and 250 EV from 200 = X, so 375 on average he will actually gain only 300, these 25% are only imaginary.
props to rmsgray for solving the problem 
p.s. no reason not to change though, both cases lead to same result, so u may swap or not as u wish...
|
| « Last Edit: Sep 3rd, 2005, 8:22pm by Nasta » |
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print