Game Question

[nsguy1350]

I made up this game out of nowhere. I have a question about it, and I want to see if anyone can solve it. I do not have the solution, and I haven't attempted to find one.

The game has the following rules:
You can "pass" or "play".
If you play, you win $1,000,000, but there is a 1% chance that you lose everything you've earned and that the game is over.
If you pass, you keep everything you have won.

For example, I could play twice, then pass. I have a chance of winning a million bucks twice, then keeping it, or winning once, then losing everything. I might even lose on the first play.

My question is: what (integer) number of times is the most efficient? What, if you had an infinite number of trials, would be the best number of times to play per game to ensure the most money?

Surely, it's not play once and pass, because that means that 98.01% of the time you would be winning $2000000, instead of 99% of the time where you win $1000000.
Of course, playing 1000 times per game isn't going to net you much money, either, because you're going to lose very often.

I'm sure this has a simple solution, but I want to see if anyone can figure this out.

[smeag]

OK, since you are not really taking out 1% chance every time you try, you are not losing or gaining any chance (theoretical explanation)
Mathematically, IDK
Can you please help me prepare for the math UIL?

[nsguy1350]

I don't think so, because if you try 100000 times, you will almost never win, but playing once everytime will net you $1000000 99% of the time, which will add up to more than continual playing. Sure, I'll help, but you have to help me too ;).

[chrishuff]

The people who can help you Mathematically solve this are people with a higher IQ than you.

Now if I had to answer that question, this is what I would say:

I don't care what the odds are, if I won the first game, I would get the **** out of it and take my $1,000,000. Gambling to try and get more money is just getting greedy. If I were to lose the first game, why in the world will I stop playing? I will just stop once I won $1,000,000 and be happy with it. XD

[nsguy1350]

You get no second chance (lose on first try = $0). I'm asking that if you had, say, a thousand tries (to make it more practical, for $1 each), and you needed as much money as possible, how many times would you play each time?
It definitely varies with number. If you and I play 10000 times, and for each game, you play only once (99% of time, get a million bucks!), and I play up to 10000 times (for a chance, incredibly small chance, to win a million bucks, have to dodge that bullet 10000 times each game to win), I would bet that you would whoop me. Every once in a while (very rarely), you'd get about 9900 x $1000000, and I'd get 10000 x $1000000 (unless I hit it more than once, incredibly unlikely). Obviously, you would win more than I would in the long run.

To give a similar, more manageable example, you have a 50% chance of losing when you play, with the same rules. If the win per play is $1, and you and I play 100 times, you'd get about 50 each time. If I try to play 100 times for each of the 100 games, I only have a 1/(2^100) chance of hitting the $100, which will definitely net me less money than you. This shows that there is some kind of variation, and I'm asking for which would be the most efficient (for the 50% chance one, I bet it's play once and get out for max).

This has to follow some simple formula, but I don't feel like figuring it out :D

[chrishuff]

I see you gambling in the future... :P

Well in that case, if I lose the 1st time, oh well, I would just move on.

The original game:
The mathematical expectation would be playing the game and winning 99 times, and expecting to lose one game after every 100 games. Of course this may not exactly happen, as with how probability goes.

[SumitG]

Hmm...to the above poster, I'm pretty sure that's not exactly what "expected value" implies. It's more like an "on average" thing where you're expected to lose once every 100 games.

If you play n times, you're expected to get (.99)^n*(1000000n) dollars. Now, if you want to maximize this, you just derive with respect to n and set the result to 0 to find the n value which produces the maximum. The solutions comes out to be something like ~99.5, implying that if you play about that number of times you have the highest expected value (dollar amount).

[nsguy1350]

I knew it was simple! I thought of making a predicted value function, taking the derivative, and then finding the maximum, but I was too lazy :D