Maths Problem

[Lul Thyme]

Ok what about this :
Suppose we have a coin or whatever with certain odds, but with don't know what the odds are.
Say it has P chance of landing heads, and 1-P chance of landing tails.
(like .75 and .25 or something)

Suppose me and you want to make an even bet.
Can you find a procedure using our coin we can make an even bet (.5 chance of winning each) without knowing what the odds for the coin actually are?


No spoilers please from those who know...
BTW this is usually attributed to von Neumann, one of the fathers of Computer Science, and Game Theory also.
And the trick is still used in many areas of Computer Science...

[Dauphin]

Heads I win, tails you lose.

Spoiler:

At a quick glance I would say you could set it so that we both toss the coin until only one of us gets a heads. In a kind of penalty shoot out fashion.

Not sure if that is what you had in mind though.

[Lul Thyme]

If you restart in case of equalities, that could work.

You didnt state it like the solution I was looking for, but after some thinking about it, they are both almost equivalent.

[Mercator]

Hmmm...

Toss the coin. Then toss the coin again. If the coin lands on the opposite side the second time it's a win (for the person who bet on the first toss correctly). If it lands on the same side you start over.

[Petek]

Here's a puzzle that's similar to the original one (since it involves a decision whether or not to switch from your original choice).

A wealthy friend places two envelopes in front of you. Your friend explains that one envelope contains twice as much money as the other one (but you don't know which is which). You may select one envelope and keep whatever money you find inside.

You select one of the envelopes at random, open it, and find $100. Your friend now tells you that you may, if you wish, change your mind and select the other envelope. Should you do so?

Common sense tells you that, since your original choice was made at random, there is no advantage to switching. However, what if you reason as follows: The envelope that you didn't choose either contains $50 or $200. If you choose the second envelope, your expected payout is one-half of $200 + $50, or $125. Based on that reasoning, it seems that you should switch.

The reasoning is absurd, since it implies that you improve your expectation by always switching to the second envelope (no matter which one you chose first). The puzzle is to explain the flaw in the reason given above for switching.

[Mr. Harley]

I'll stick with a simple double-blind study with four pages of statistics, thank you.

[Lul Thyme]

Originally posted by Mercator
Hmmm...

Toss the coin. Then toss the coin again. If the coin lands on the opposite side the second time it's a win (for the person who bet on the first toss correctly). If it lands on the same side you start over.

Yep thats what I was looking, again in a slightly different form, but Dauphins idea is equivalent I think.

[Lul Thyme]

Originally posted by Petek
Here's a puzzle that's similar to the original one (since it involves a decision whether or not to switch from your original choice).

A wealthy friend places two envelopes in front of you. Your friend explains that one envelope contains twice as much money as the other one (but you don't know which is which). You may select one envelope and keep whatever money you find inside.

You select one of the envelopes at random, open it, and find $100. Your friend now tells you that you may, if you wish, change your mind and select the other envelope. Should you do so?

Common sense tells you that, since your original choice was made at random, there is no advantage to switching. However, what if you reason as follows: The envelope that you didn't choose either contains $50 or $200. If you choose the second envelope, your expected payout is one-half of $200 + $50, or $125. Based on that reasoning, it seems that you should switch.

The reasoning is absurd, since it implies that you improve your expectation by always switching to the second envelope (no matter which one you chose first). The puzzle is to explain the flaw in the reason given above for switching.

This is a tricky one.The fallacy I view this way
There are 2 different possible situations :
The enveloppes had 50 and 100 and you chose the 100 one.
or The enveloppes had 200 and 100 and you chose the 100 one.

But by computing the averages, you assumed that these were somehow equiprobable, but in fact we have no knowledge of the probability distribution of the situations so we are not able to compute an average.

[KrazyHorse]

What LulThyme said.

If you are guaranteed the situations are equiprobable then you should switch.

Computing probabilities with 0 information as to the distribution of events is impossible.

[KrazyHorse]

Probabilities are what you do with incomplete, not 0 information.

[Dauphin]

But don't you know what the distribution could be. Its either $100, $50 or $200, $100. Isn't the fallacy that the two distributions you use in your expectation calc have different means and variances.

I may be oversimplifying, but couldn't you take a geometric mean, as opposed to an arithmetic mean, to get around this.
i,e ($200 * $50)^1/2 = $100 expectation if you switch. Therefore no gain or loss if you switch.

The same then works if you added in a third envelope of money with either 4 or .25 times as much money as you first pick.
i.e. ($400*$200*$50*$25)^1/4 = $100. Therefore no gain or loss if you switch.

Also, it would work if you changed the ratio.

Though I must admit, I feel as if I am using circular logic in thinking the maths of this one through.

[Petek]

Here's a link to an explanation of the problem:

http://www.maa.org/devlin/devlin_0708_04.html

For lots more information, Google on "two envelope paradox".

[Urban Ranger]

What KH and LT said. It's a mis-application of the Principle of Indifference.

[Lul Thyme]

Originally posted by Dauphin
But don't you know what the distribution could be. Its either $100, $50 or $200, $100. Isn't the fallacy that the two distributions you use in your expectation calc have different means and variances.

I may be oversimplifying, but couldn't you take a geometric mean, as opposed to an arithmetic mean, to get around this.
i,e ($200 * $50)^1/2 = $100 expectation if you switch. Therefore no gain or loss if you switch.

The same then works if you added in a third envelope of money with either 4 or .25 times as much money as you first pick.
i.e. ($400*$200*$50*$25)^1/4 = $100. Therefore no gain or loss if you switch.

Also, it would work if you changed the ratio.

Though I must admit, I feel as if I am using circular logic in thinking the maths of this one through.

but your calculations still assume that somehow you know something about how likely it was to be 100-200 vs 50-100....

[Snowflake]

Ok this is not a trick question. Just one that I didn't know how to do.

If you do a 3d6 12 times, and then throw away 6 lowest scores, what would be the expected value of total score?