Someone show me the numbers and % for the following situation

I want to know if the person is offering a fair deal

If I win 1 out of every 8 times and get paid 7 times my bet when I win, will I break even

I think explaining it with the coinflip concept is a would make it much easier because they are similar to all ins ( 50/50)
Basicly the game is : I am offering him a game where the odds are fair ( from my point of view ) and unfair ( from his point of view)
The game I offer is : You have to win 3 time in a row a coinflip in order to win my 7 pennies
if ever he lose one coinflip he must give me his 1 penny
Here is a site my friend showed me in order to argue with him
http://www.horseracing-and-bettingsystems.com/betting-odds-percentage-table.html
here are a couple screenshots from this site which proves my point



hopefully someone can help us with this debate as ive been hosting this game and though it was fair for quite a few weeks now
Am i wrong or is he wrong ?
Ty for reading and sorry for my bad english I am french lol

Well I figured it out, it's simple lmao


-1
-1
-1
-1
-1
-1
-1
+7
=
0

closed

In theory, but you’d need to use a “large” number of runs to achieve something close to this result. In essence the game is “fair” though.

Expected value = (87.5%*-1)+1*(12.5%*7) = 0

P(winning) = 1/8 = 12.5%, where you win 7x your bet
P(losing) = 1-(1/8) = 87.5%, where you lose your bet (-1x)

Its been awhile since ive done probability questions but at first glance it seems that the game is expected to yield nothing

This post was edited by cialda on Apr 20 2018 06:51am

The game is "unfair".



Fixed.

"get paid 7 times my bet when I win" means you've already lost your bet when winning 7 times your bet : +7 - 1 = +6 is your final gain in this case.



Situation is different here :

(1/8) chance of winning 7
(7/8) chance of losing 1

Average (mean, expected value) = 0, the game is "fair".



Even assuming the game is fair, whatever amount of money you are willing to invest, you will almost certainly be broke at a point.
That is to say : the probability of you being broke after some large number of deals is 100%.

@feanur because of the variation away from break-even? You'd be talking about a bigger number of games than one can play in several lifetimes at the rate of 1 penny per game, if someone is to invest a big sum

The average number of games before being broke does not exist.
Indeed, you could wait several lifetimes before being broke, or that could happen very quickly...

Haha this is quite true, I played a game where I would bet per 100k stacks against 50% odds, double-or-nothing kind of thing. I don't remember back then, but nowadays 3kk makes 6 bucks.
I lost 44kk despite these odds
Would only play a handful of games every day. Good thing I'm rich beyond human comprehension in that game

You lose 7 times. On the 8th bet, you win. But you would still have to place a bet on the 8th game. So you would break even, as long as you quit playing when you win on the 8th game.

You lose 70$ on the first 7 games, but on the 8th game, you place 10$, but you win 80$. Now you're at zero again.

Well, if you're paid 7 times your stake, the game would have to involve the rule that you get your original bet back. If not, it's a losing game.

This post was edited by Taurean on Apr 23 2018 11:40am