News1 min ago
Sacks
22 Answers
After Socks and Sex here's the last one from me for a while as work beckons. (and I need to buy some new light bulbs and socks and knock on doors to see whether a boy or girl answers)
I have two sacks, and inside each I have put some money. In fact, one sack contains twice as much money as the other.
I'll blindfold you and let you select one sack, which you can have after the game is over. But as soon as you select one, I offer you the option to switch sacks. Should you switch?
You reason as follows: My sack has $x, and with probability 1/2 the other sack has either $x/2 or $2x dollars. Thus the expected value of the other sack is (1/2)($x/2) + (1/2)($2x) which is $1.25x. This is greater than the $x in my current sack. Therefore I should switch sacks...
But if you do switch, a similar argument would instruct you to switch back... and therefore keep switching! What's going on here? Is there a flaw in the reasoning?
I have two sacks, and inside each I have put some money. In fact, one sack contains twice as much money as the other.
I'll blindfold you and let you select one sack, which you can have after the game is over. But as soon as you select one, I offer you the option to switch sacks. Should you switch?
You reason as follows: My sack has $x, and with probability 1/2 the other sack has either $x/2 or $2x dollars. Thus the expected value of the other sack is (1/2)($x/2) + (1/2)($2x) which is $1.25x. This is greater than the $x in my current sack. Therefore I should switch sacks...
But if you do switch, a similar argument would instruct you to switch back... and therefore keep switching! What's going on here? Is there a flaw in the reasoning?
Answers
Best Answer
No best answer has yet been selected by factor-fiction. Once a best answer has been selected, it will be shown here.
For more on marking an answer as the "Best Answer", please visit our FAQ.
Common sense tells you to infinitely switch isn't helping, and that here is always a 50% chance of doing better or doing worse; so the issue here is not whether it is a good idea to switch (one doesn't know) but what the flaw is.
I think BA may be on the right lines, your expected other sack value can not be 1.25x. It is either 0.5x or 2x. Plus each time you swop you redefine x since what you are holding is always considered 100% and the other sack judged in comparison.
I'll give it some thought when I have time.
I think BA may be on the right lines, your expected other sack value can not be 1.25x. It is either 0.5x or 2x. Plus each time you swop you redefine x since what you are holding is always considered 100% and the other sack judged in comparison.
I'll give it some thought when I have time.
Expected payoffs don't have to equal any actual amount that can be won, though, so that's not the flaw in itself. The game could be that you get either half, or double, the amount you are already given based on a coin toss. Then, after 100 tries you could get double the amount 50 times, and half fifty times. The sum of this is then indeed 50*2x + 50*x/2 = 125x, so that each game you won an average of 1.25x.
It's not unlike the claim that each family has an average of 2.4 children (or whatever the figure is) -- obviously no family ever has that amount and you would lose every bet you made on a family having that many children, but the figure is still correct.
It's not unlike the claim that each family has an average of 2.4 children (or whatever the figure is) -- obviously no family ever has that amount and you would lose every bet you made on a family having that many children, but the figure is still correct.
That's still not a flaw because it's the "expected" pay-off. If 50% of the time you get double your starting amount, and 50% of the time you get half, and the starting amount is the same, then running enough trials will mean that switching increases your winnings to 1.25 times your starting amount on average.
The flaw (possibly) is what I've highlighted. The 1.25 figure is correct in certain conditions, namely that each time you play the game the amount in the envelope you receive is the same. But those conditions are up for dispute, and I think this is where the problem lies, because the full parameters of the game haven't been laid out.
The flaw (possibly) is what I've highlighted. The 1.25 figure is correct in certain conditions, namely that each time you play the game the amount in the envelope you receive is the same. But those conditions are up for dispute, and I think this is where the problem lies, because the full parameters of the game haven't been laid out.
I suspect Derren may have pointed out that first swop is risking 50% of the value but gives you the possibility of gaining 100% so arguably worth it since it is a 50/50 chance either way.
Long term though you can only reverse what you achieved, so the rest of any calculation, and any average (rather than expected) gain/loss, is irrelevant. The 125% average stays the same throughout it isn't cumulative.
Long term though you can only reverse what you achieved, so the rest of any calculation, and any average (rather than expected) gain/loss, is irrelevant. The 125% average stays the same throughout it isn't cumulative.
Can anyone explain what is wrong with my reasoning?
Invisibly label the two sacks A and B. A has £100 in, B has £200. I don't know which is which.
I pick A and stick with A. I get £100.
Pick A and switch to B. I get £200.
Pick B and stick with B. I get £200
Pick B and switch to A. I get £100.
These are all of the possibilities, and I get the higher amount in 2/4 cases. A 1/2 chance of winning the higher amount, the same as the chance of choosing A or B, so I have the same chance of doing well no matter what. I understand Jim's and others' reasoning, but can't see why mine doesn't work the same as theirs.
Invisibly label the two sacks A and B. A has £100 in, B has £200. I don't know which is which.
I pick A and stick with A. I get £100.
Pick A and switch to B. I get £200.
Pick B and stick with B. I get £200
Pick B and switch to A. I get £100.
These are all of the possibilities, and I get the higher amount in 2/4 cases. A 1/2 chance of winning the higher amount, the same as the chance of choosing A or B, so I have the same chance of doing well no matter what. I understand Jim's and others' reasoning, but can't see why mine doesn't work the same as theirs.
No, that looks fine to me Clover. Do you think it isn't mutually ok with the other comments ?
I think it is just another way to look at it. You do have a 50% chance of getting the higher amount when you stop swopping.
You also can argue that a single swop might be worthwhile considering the possible loss/gain situation.
And it is also true that continual swopping isn't cumulative but simply undoes what went before: which is clearer if one doesn't consider each subsequent swop in isolation, reevaluating the amount in your hand to be 100% each time, but accepting you are hold either 50% or 200%.
I think it is just another way to look at it. You do have a 50% chance of getting the higher amount when you stop swopping.
You also can argue that a single swop might be worthwhile considering the possible loss/gain situation.
And it is also true that continual swopping isn't cumulative but simply undoes what went before: which is clearer if one doesn't consider each subsequent swop in isolation, reevaluating the amount in your hand to be 100% each time, but accepting you are hold either 50% or 200%.
Then try spending the cash in your local supermarket. When you come to the checkout you choose a queue. There is a holdup at the till because somebody's credit card has gone "tilt". You change queues and the same thing happens. You watch in dismay as your original queue is now flowing smoothly. Do you swap queues again?
I should just mention that I'm not arguing that the 1.25x reasoning is correct exactly -- I think it depends on the problem -- but only that it's not automatically wrong because of the number 1.25. Weird numbers can well appear in averages, eg the expected pay-off if you get £1 for a head and nothing for a tail is 50p per coin toss, even though on no single game will you ever win that amount. It's just the average pay-off based on how the problem is formulated. In the same way, if the sack problem is set up in a particular way then switching could well lead to an expected pay-off of 1.25x compared to sticking.
Latest Posts
Quizzes & Puzzles2 mins ago
News4 mins ago
Quizzes & Puzzles4 mins ago
Inflation Increases By 35% - Minister...
News16 mins ago
