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The Elliquian Herald & Post
Issue 74 ~ April thru July 2017

Wiki Blogs Dicebot

Author Topic: Math oddities  (Read 11106 times)

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Offline Pumpkin Seeds

Re: Math oddities
« Reply #50 on: June 16, 2011, 06:47:18 AM »
My head hurts.

Offline Oniya

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Re: Math oddities
« Reply #51 on: June 16, 2011, 09:12:46 AM »
This might help.   >:)

Tom Lehrer- New Math

Offline Kuroneko

Re: Math oddities
« Reply #52 on: June 16, 2011, 10:40:34 AM »
Math bad.

Offline ReanimateMagnus

Re: Math oddities
« Reply #53 on: June 16, 2011, 11:20:39 AM »


Offline MHaji

Re: Math oddities
« Reply #54 on: July 18, 2011, 05:53:45 PM »
For a serious math oddity, though, how about the St. Petersburg Paradox:

You pay a flat fee to enter a lottery where a fair coin is flipped repeatedly until a tails appears, thus ending the game.  The pot starts at $1, and the pot doubles every time a heads appears.  You win whatever is in the pot once the game ends.  For example, if the first flip is tails, then you win $1.  If you gets heads and then tails, you win $2.  Heads, heads, tails = $4.  Heads, heads, heads, tails = $8.  And so on.  So the question is how much should you be willing to pay to enter this lottery?

To determine this, you need to find the expected payout.  The expected value is found by taking the probability of each outcome, multiplying that probability by what you would win, and adding all the probability payoffs together.  In other words, there's a 1/2 chance that you will get tails on the first flip, thus getting $1.  There's a 1/4 chance that you will get heads on the first flip and tails on the second, thus getting $2.  The chance decreases as you continue the string of getting heads in a row, but that's compensated by doubling the pot.  So mathematically, that's expressed as the following:

E(x) = (1/2 x 1) + (1/4 x 2) + (1/8 x 4) + (1/16 x 8) + (1/32 x 16) + ...

Once you get far out there, the chances of getting that payoff are very miniscule, but it's compensated for by the enormous payoff.  As you can see from simplifying the math above, there's an interesting pattern:

E(x) = 1/2 + 1/2 + 1/2 + 1/2 + 1/2 + ...

This mathematical expression converges to infinity, meaning that if you keep adding 1/2 forever, so the total will keep growing without ever stopping.  So we see that the expected earnings of the St. Petersburg Paradox is an infinite amount.  Thus, you should be willing to spend everything you have on this game, because your probable earnings are infinite.  Would you be willing to risk everything?  Mathematically, it seems that you should!

Note: the math here is correct (unless I mistyped something), but obviously something doesn't seem right here because who do you know would actually be willing to risk everything?  There are answers to explain this paradox (some more mathematical than others), but it's nonetheless a cool mathematical oddity that I thought people would enjoy.

The problem isn't just that you risk everything - it's that expected payoff isn't the same thing as the expected usefulness of the money. To put it more simply, I would rather take $50 for sure than a 1/1,000,000,000,000,000 chance of $50,000,000,000,000,000. Even if that much money existed (and it pretty much has to exist for the infinite sum to genuinely be infinite), you'd be pretty crazy to lose $50 on such a tiny chance. Risk aversion exists for a reason - it keeps people alive.

The paradox also breaks if we assume a GIGANTIC, but finite, bankroll. Let's try it with "more money than exists in the entire world, even counting derivatives."

Suppose the game goes only up to a meager $10,000,000,000,000,000 maximum, or ten quadrillion dollars. That would be about 50 rounds, giving us an expected payoff of.... $25.

So if you pay more than $25 to play this, you'd better hope that you have a use for "more money than exists in any form."

Offline AndyZ

Re: Math oddities
« Reply #55 on: November 25, 2011, 02:13:40 AM »
I saw Mythbusters where they did the Monty Hall problem and did an E search to see if anyone else had talked about it.  Saw this thread, decided to jump in.  Hope nobody minds.

I read the thing about the Monty Hall problem and thought that they did a poor job of actually explaining it.  It helps to break things down, I think, case by case:

Spoiler: Click to Show/Hide
Let's say that there's 3 doors, doors 1 and 2 contain nothing but door 3 has a car.  The host knows this but the contestant does not.

If you pick door 1, the host will open up door 2 to reveal it as empty.  You'd have to switch to win.

If you pick door 2, the host will open up door 1 to reveal it as empty.  You'd have to switch to win.

If you pick door 3, the host will open up either door 1 or 2, but you should not switch.

Thus, not switching only lets you win 1/3 of the time, while switching lets you win 2/3 of the time.  It seems as though it would be a 50-50 shot, but this just isn't true.

With the St. Petersburg paradox, I'm reading it over and I think the problem is recursion.  You see, out of 100% possible solutions, you have a 50% chance to earn nothing at all, 25% to earn $1, 12.5% to earn $2, 6.25% to earn $4, and you can continue from there.

Were it truly E(x) = (1/2 x 1) + (1/4 x 2) + (1/8 x 4) + (1/16 x 8) + (1/32 x 16) + ..., then you'd need to have a 50% chance of earning a dollar, plus a 25% chance of earning another $2.  This means that the jump for two successive heads would be $3.  The next would be 1+2+4 to become 7, then 15, and so on.  I may be wrong on this, but that's how I see it.