# Another gambling "paradox"

Hello everyone,

So I'm not a mathematician at all (i just watch some math videos from time to time) and maybe this question is obvious to answer.

Consider the case where you play a game of heads/tails. If you win, you win 10 dollars. If you lose, you lose 10 dollars. You can play this game as much as you want. If my understanding is correct, it is possible to leave this game with an arbitrary amount of money. You can play until you get many lucky rolls in a row and leave this game with say, 1000 dollars, even if on average you break even.

Now consider the same game, but when you win, you only make 1 dollar. You still lose ten dollars on loss. Can you expect to leave the game with 1000 dollars positive, or any other arbitrary positive amount, if you play long enough?

On one hand, you could argue that yes, just like in the first example, if you roll an infinite amount of times, you will eventually get lucky enough that you will be in the positive even for a game where you lose on average.

However you could also argue that the more you play trying to "get very lucky", the luckier you need to get to make all money that you lost back, making it impossible to make your money back.

Thank you

## 1 Answer

In neither of these games you will NOT win $1000 if you play it a large amount of times.

(I) Suppose that when you win, you win 10 dollars, and if you lose, you lose 10 dollars. Then the expectation of your win is

\[\frac{1}{2}\cdot 10-\frac{1}{2}\cdot 10=0,\]

which means that, on average, you are expected to win $0$ every time you play it.

(II) Suppose that when you win, you win 1 dollar, and if you lose, you lose 10 dollars. Then the expectation of your win is

\[\frac{1}{2}\cdot 1-\frac{1}{2}\cdot 10=-4.5,\]

which means that, on average, you are expected to lose $4.5$ dollars every time you play it.

It is possible to get lucky and win 1000 times in a row and leave with at least 1000 dollars in either of the cases above. But this is exteremely unlikely and even when it happens, if you play the game a large number of times, you are expected to leave with $0$ dollars in case (I), and lose 4.5 dollars every time you play in case (II). If you play the second game 1000,000 times, you are expected to lose 4,500,000 million dollars!

- 1 Answer
- 154 views
- Pro Bono

### Related Questions

- Calculating Dependant Probability of Multiple Events
- Probability of drawing a red ball, given that the first ball drawn was blue?
- Poisson process question
- What is the probability that the last person to board an airplane gets to sit in their assigned seat?
- Check if problems are correct
- Combinatorical Drawing
- applied probability
- Probability question

This is a time-consuming problem and I think it should with a bounty.