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

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We play a game, you flip a coin, and you keep fliping untill you get tail.
If you get tail on 1st flip you gain 2$
if you get tail on 2nd flip you gain 4$
if you get tail on 3rd flip you gain 8$
...
if you get tail on nth flip you gain 2^n $

How much would you pay to participate in a round of such game ?
>>
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>>8919391
This method is better and works great on roulette. Profit.
>>
>>8919397
this has nothing to do with roulette or martingale.
>>
>>8919391
Well my expected winnings are 1+1+1+...

So I'd say it's worth about -1/2$
>>
I recognize that you are trying to steal a dollar from me and don't play your Jewish games
>>
>>8919406
you wouldn't pay even 2$ or less to play this game ?
>>
>>8919391
This looks suspiciously like a homework thread.

The "math" answer is less than the expected value of the series.

I'd pay $3 and no more.
>>
>>8919412
So you'd bet all your money to play a single round ?
>>
slightly less than a £1 maybe 98/99p
>>
>>8919391
wait, it's a new bet per flip?

or one single bet and then flip until i win?
>>
>>8919391
>How much would you pay
$2 because I won't risk you taking anything from me.
>>
>>8919446
Your post is stupid.
>>
-1 /12 $
>>
>>8919451
calculate both
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>>8919451
2nd option.
>>
>>8919391
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.
>>
>>8919452
>still a 50% chance of losing 2$
Try again
>>
>>8919391
Expected gain is infinity, so I'd bet all my bank account.
>>
The dollar symbol goes in front of the number, brainlets.
>>
>>8919733
>not reading the problem statement

"keep flipping until you get tail"
>>
>>8919391
n dollars, payable upon loss
>>
>>8919816
That anon 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
>>
>>8919857
What? Expected value is +infinity, no way to lose, worst case scenario is +2$.
>>
>>8920087
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]
>>
>>8921198
>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.
>>
>>8919405
>>8919406
>>8919412
>>8919452
>>8919466
>>8919466
>>8919857


The dealer ask you 10$ to play 1 round, do you agree ?
>>
>>8921235
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.
>>
>>8921243
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 ?
>>
>>8919391
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 $?
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>>8919446
> ?
DUDE PHONEPOSTERS AREN'T ALL UNDERAGED'S BRINGING THE LEVEL OF CONVERSATION DOWN YOU'RE JUST ALL ABOUT MUH SEKRIT CLUB
>>
>>8921291
>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.
>>
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I once did 120 damage with this little guy
Thread posts: 35
Thread images: 3


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