We play a game, you flip a coin, and you keep fliping untill you get tail

>How much would you pay
$2 because I won't risk you taking anything from me.

Your post is stupid.

-1 /12 $

calculate both

2nd option.

You'd calculate the probability density as a function of n. Its a negative binomial so pretty easy to do. Then find out what chances you are comfortable.

>still a 50% chance of losing 2$
Try again

Expected gain is infinity, so I'd bet all my bank account.

The dollar symbol goes in front of the number, brainlets.

>not reading the problem statement

"keep flipping until you get tail"

n dollars, payable upon loss

That user should bet all of his money because he probably never gets tail

expected value for number of successes 0.5/(1-0.5) = 1

if you play the game an infinite amount of times you should always pay less than 2 dollars

What? Expected value is +infinity, no way to lose, worst case scenario is +2$.

you're a dumbass.

>In probability theory and statistics, the negative binomial distribution is a discrete probability distribution of the number of successes in a sequence of independent and identically distributed Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs.
the expected value is [math]\frac{rp}{1-p}[/math] where r is the number of failures that stops the trial (1 failure) and p=0.5. the expected value is 1.

if, mathematically, you only expect to win on average 1 game, you should bet less than you expect to win from one game.

You must be mentally challenged, so I'll explain slowly.
The gain GROW the longer you flip, meaning the gain is NOT CONSTANT.
The expected value is therefor NOT:
[math]\sum 0.5^n=\frac{0.5}{1-0.5}=1 [/math]
but instead is:
[math] \sum 0.5^n\times 2^n=\sum 1=+\infty[/math]

>if, mathematically, you only expect to win on average 1 game, you should bet less than you expect to win from one game.
Also this makes no sense, you can only play one game, and you can't lose it.

The dealer ask you 10$ to play 1 round, do you agree ?

I'm describing expected value in the context of a probability distribution, not in terms of expected returns.

Whatever the expected return is on any given nth throw is doesn't matter if on average you only win one game.

You still don't make any sense, but I have a cousin with down syndrome so I'm used to translate gibberish.
>I'm describing expected value in the context of a probability distribution, not in terms of expected returns.
Are you trying to say "the expected rank of the first tails" ? then it's not 1 (one), since one is the absolute minimum.
It's value is :
[math]\sum n\times0.5^n=\frac{0.5}{(0.5-1)^{2}}=2[/math]

But why would this matter? If we bet on you rolling a 6 with a dice, and I multiply your bet by 10 if you do, will you not play because on average you'll lose more often than not ?

Nothing. It must be a scam.

At least 1 assumption made in the description of the game does not apply to reality. Therefore the game cannot be implemented.

Nice bait though.
Counter question: What physical properties have to be changed to create a fictional world in which anyone would want to play this game for all the money they own?

What rules would have to be added to make a lot of people in our reality want to play this game for n-1 $?

> ?
DUDE PHONEPOSTERS AREN'T ALL UNDERAGED'S BRINGING THE LEVEL OF CONVERSATION DOWN YOU'RE JUST ALL ABOUT MUH SEKRIT CLUB

>Are you trying to say "the expected rank of the first tails"
No, I'm not. The expected value is 1 because that's the expected value of successes.

>If we bet on you rolling a 6 with a dice, and I multiply your bet by 10 if you do, will you not play because on average you'll lose more often than not ?
You're describing a different situation there. My answer to that question has no relevance to OP's scenario.

I once did 120 damage with this little guy