joef1000 added a topic in ParadoxesAn infinite amount of money?So let's say I have a time machine, and I put £1,000 into a bank account (at time t=0). Now, if I don't touch it for a year (call this t=2), I'll have some interest on my original £1,000. Let's say the interest rate is 1% per year (works for any interest rate really).
Then at t=2, my balance is £1,010. So if I ask to withdraw all my money, so I now have £1,010 in my wallet.
Then I use my time machine to travel to the day after I deposited the original £1,000, call this t=1,and deposit half the money in my wallet, that is, £505.
I then go back to t=3, where I still have the other £505 in my wallet.
BUT... If I deposited the money, then at t=1 I had £1,505, instead of £1,000. So at t=2, adding 1% interest gives me £1,520.05, so THAT'S actually how much I withdraw. So I deposited half of that = £560.02 and kept £560.02 for myself at t=3 (I'll give the extra penny to charity)
But that's more money, which means it actually become more, and more, and more.
Does it converge, or do I have a truly infinite supply? Can we increase our final outcome if the initial parameters change?
Let's go to the maths!
Let M be the amount of money I have (since all my money at any given point is either in the bank or in my wallet, I only need one letter to denote this. M0 is my starting money at t=0, M1 is money at t=1, and so on. M0 = £1000
Let I be the interest rate, aka 0.01.
Let P be the proportion of my money I deposit at t=1 (the rest I keep at t=3, giving the remainder to charity), this = 0.5.
A. M1 = M0+P*M2
B. M2 = M1* 1 + I)
C. M3 = (1-P)*M2
Substituting A into B gives
M2 = (M0+P*M2)*(1+I)
M2/(1+I) = M0+P*M2
M2/(1+I)-P*M2 = M0
M2*(1/(1+I)) - M2*P = M0
M2 * (1/(1+I)-P) = M0
M2 = M0 / (1/(1+I) - P)
So the amount we end up with at then end (M3) is
------------- * M0
1/(1+I) - P
If 1-P > 1/(1+I)-P, we made a profit.
1 + I > 1
-> 1 / (1+I) < 1
-> 1 / (1+I) - P < 1-P
We made a profit.
How much profit? If it's more than I, this was worth it.
Let's plug in our values:
(1 - 0.5) / (1/1.01 - 0.5)
We (slightly more than) doubled the interest rate.
In fact, changing the interest rate with P = 0.5 doubles the interest rate
What about different values of P (putting depositing more at t=1)?
Lower values of P end up giving results close to 1.01, but higher values give more profit: for example, (1 - 0.9) / (1 / 1.01 - 0.9) 1.11, effectively bumping up the interest rate to 11%.
However, going too high gives negatives, so what's the highest point we can go to?
Answer: P = 1/1.01 = 100/101. Then our money goes to infinity. (as we divide by 0)
So for interest rate I, depositing (1/(1+I)) of the money at t=1 when in the past means you end up with an infinite amount.
Of course, due to rounding, you can't deposit exactly this amount, but you get more money each "iteration" so you can round it more accurately each time. And if you give the remainder of division after rounding to charity, you'd be helping infinitely too. (in most cases)
The only limit is the amount of money your wallet can carry at once as you travel through time.
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joef1000 added a topic in GamesThe wish gameIn this game, you have to make a wish, whilst corrupting the wish above you. Inspired by the same game on another forum here.
There are rules:
- No wishing for more wishes
- No wishing for more than one thing in a wish (although a wish can have as much detail as you like)
- No wishing to break these rules
- When you corrupt a wish the corruption must be related to the wish, for example if someone wishes for a cookie you can't say "Granted, but you die before receiving it" because that has nothing to do with the wish itself, you could, however, say "Granted but it's poisoned and you will die if you eat it".
- Make sure what you say
Person1: I wish for a sandwich.
Person2: Granted, but it's a dirt sandwich. I wish for a book.
Peerson3: Granted but it's boring. I wish for a 3DS.
...and so on.
Feel free to be as creative as you like in the wishes and corruptions!
I wish for a bar of chocolate.
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