"Coin-flipping" situation

gazarsgo · #1 04-20-2005, 07:51 PM

What kind of EV are you looking at in a situation like this:
You are an 80% favorite every round.
You must pay $5 to play.
Whether you win or lose, you get $1. If you lose, you have to pay another $5 to continue playing.

Is this +EV, -EV, or breakeven? If it is profitable, what size bankroll do you need to avoid going broke with 95%, 99%, and 99.9% confidence?

Intuitively, this seems like it should be a breakeven proposition, but I can't quite wrap my head around it...

mannika · #2 04-20-2005, 08:09 PM

It's breakeven. 80% of the time you win $1, 20% of the time you lose $4.

Popinjay · #3 04-20-2005, 08:10 PM

1 * .8 = .8 EV
-4 * .2 = -.8 EV

Total EV of one flip in this game = 0

However because you have to pay $5 to start playing it is a -EV game.

gazarsgo · #4 04-20-2005, 09:12 PM

[ QUOTE ]
However because you have to pay $5 to start playing it is a -EV game.

[/ QUOTE ]

I guess I did not explain this clearly. It does not matter if you win or lose, you are getting $1 for playing a round. You cannot play another round if you lose without spending another $5.

gazarsgo · #5 04-20-2005, 09:19 PM

[ QUOTE ]
It's breakeven. 80% of the time you win $1, 20% of the time you lose $4.

[/ QUOTE ]

Can you explain your logic? Am I just overthinking this?
100% of the time you win at least $1 (net -$4).
80% of the time you win $2 (net -$3), by winning the first round.
Winning twice is .8 * .8 = 64% of the time, which you win at least $3 and net -$2.
Winning three times is .8^3 = 51.2% of the time, in which you win at least $4 and net -$1
winning four times is .8^4 = 40.96% of the time, in which you win at least $5 and breakeven.
winning five times is .8^5 = 32.768% of the time, where you profit.

Right?
I know you profit 1/3rd of the time...but do you lose more often than you break even?

mannika · #6 04-20-2005, 10:10 PM

[ QUOTE ]
However because you have to pay $5 to start playing it is a -EV game.

[/ QUOTE ]

This is what I thought at first, but this is not correct, because you can choose to stop playing when you lose for the first time, thereby negating the fact that you have to pony up $5 to begin with.

mannika · #7 04-20-2005, 10:13 PM

[ QUOTE ]
[ QUOTE ]
It's breakeven. 80% of the time you win $1, 20% of the time you lose $4.

[/ QUOTE ]

Can you explain your logic? Am I just overthinking this?
100% of the time you win at least $1 (net -$4).
80% of the time you win $2 (net -$3), by winning the first round.
Winning twice is .8 * .8 = 64% of the time, which you win at least $3 and net -$2.
Winning three times is .8^3 = 51.2% of the time, in which you win at least $4 and net -$1
winning four times is .8^4 = 40.96% of the time, in which you win at least $5 and breakeven.
winning five times is .8^5 = 32.768% of the time, where you profit.

Right?
I know you profit 1/3rd of the time...but do you lose more often than you break even?

[/ QUOTE ]
Sure, you may profit only 1/3rd of the time, but you stand to gain more when you profit than when you lose. If you take the summation of (0.8 + 0.8^2 + 0.8^3 ... + 0.8^n), you come out with:

(0.8 - 0.8^infinity)/(1-0.8)
= (0.8/0.2) = 4

Therefore, the mean amount you get for winning is $4, and you still get your dollar the last time you lose, so $4+$1 = $5, therefore breakeven.

Hojglad · #8 04-21-2005, 07:00 AM

[ QUOTE ]
[ QUOTE ]
However because you have to pay $5 to start playing it is a -EV game.

[/ QUOTE ]

This is what I thought at first, but this is not correct, because you can choose to stop playing when you lose for the first time, thereby negating the fact that you have to pony up $5 to begin with.

[/ QUOTE ]

This game definitely carries a negative expectation. There is no way you can play it so that you end up a winner. The fact that you have to pay 5 dollars up front to play it shows you this.

P(win 5 in a row to recoup your loss) = 0.32768
P(lose any of the first 5) = 0.67232

Keep in mind that in order to break even, you HAVE to win at least 5 in a row. I wrote some code to simulate playing this game 100 million times and the net result was that, starting with a bankroll of 1 billion dollars, you'd end up losing about 20 million.

Come on, the probability that you lose one of the first 5 times is almost twice that you win the first 5 times in a row to recoup your starting fee. Combine that with an empirical test that shows you lose money playing this game in the long run, and there is no way that this game carries a positive expectation.

Hojglad · #9 04-21-2005, 08:04 AM

Nevermind, I completely forgot about the stipulation that you win a dollar whether or you win or lose. This game still has a negative expectation, however.

EDIT: When I added the new stipulation into my code and simulated 1 billion trials, the player wound up about 25,000 dollars behind.

mannika · #10 04-21-2005, 09:06 AM

[ QUOTE ]
Nevermind, I completely forgot about the stipulation that you win a dollar whether or you win or lose. This game still has a negative expectation, however.

EDIT: When I added the new stipulation into my code and simulated 1 billion trials, the player wound up about 25,000 dollars behind.

[/ QUOTE ]

I have no idea how you're getting this number, when I run my simulation I only get a -EV sample mean of $5, and that's only because the simulation tells them to play the game up to a certain number of plays, not to play the game until they lose for the last time (if this was the case, EV would be $0).

How many times are you running this simulation, and what is the standard deviation of the sample mean? If you are running it a billion times, the standard deviation of the final amount won is likely very high, and may account for the $25000 loss you are seeing.