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Probability

Alkatran

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I was reading the thread about the sports betting and was wondering about something.

Let's say I devise a game like so:

You pay me something. I flip a coin and you call it. If you get it right you win twice what you paid, but if you get it wrong you get nothing (double or nothing).

However, here's the twist: if you win I roll a dice and you call it: if you get it wrong you have to play again.

What is the average payout of playing this game?

Interestingly, it's -0.8 dollars.

It makes sense though: if you don't play again the average is 0: if you ALWAYS have to play again the average is -1 (since you play until you lose).

I bet I could make money off this!


"Hey, gimme a buck, I'll pay you a buck and a half if..." :D


Work (for 1$ initial bets):
(draws a tree)
payout = -1 + 2*1/2*1/6 + 4*(1/2*1/6)^2 + ...
= -1 + SUM[n = 1 to infinity](2^n/(2*6)^n)
= -1 - 1 + SUM[n = 0 to infinity]((1/6)^n)
= -2 + 1/(1 - 1/6) = -2 + 6/5 = -.8
General formula:
payout = 1 / (1 - p) - 2
where p is the probability of NOT replaying.

I KNOW I made a mistake because if p is greater than .5 you end up with way too much payout.
 
I did something similar in Stats recently. Game costs 2 dollars to play. You get 3 dollars if you win. After you pay, you draw a random card from a 52-card deck. If it's red or a jack, you win. On average a player loses 38 cents per play.
 
Well, now I end up with 0 average payout in all cases. I suppose that makes sense... :( Oh well.

I wrote a computer program to simulate it and it predicts losses when you get close to 0, but that could be related to the fact that it's pseudo random.


Coming up with another one...
 
Alright.

Same concept except now you can only win a dollar or lose a dollar and have to play again if the coin flip is called.

In this case the game is fair, but hitting 0 effectively ends the game. I'll work out the winnings for this one, and if they aren't 0 then it looks like there's two types of games: ones that are still fair when you continue to play and ones that aren't.
 
Oh my god this is difficult. I can tell that even when I find the formula it's going to be too complicated for me to solve the series! :(

So far I'm noticing connections to nCr... but it will break down later.
 
Well, I did part of it. If you constantly win and only stop if you lose the 'keep going' flip your payout is

-(x-4) / ([x-2]^2) - 1

Which is 0 at 0% chance of continuing and 2 at 100% chance of continuing. But, of course, that leaves only the infinitely many other places the game can end. :(
 
ReFLeX said:
I did something similar in Stats recently. Game costs 2 dollars to play. You get 3 dollars if you win. After you pay, you draw a random card from a 52-card deck. If it's red or a jack, you win. On average a player loses 38 cents per play.
Click to expand...

The slots on the casino are the worst against you as far as odds go and they are 20-30 percent against you. This would be in that range. :) Most of the casino games are not only heavily against you in odds, but allow for you to play badly thus making the odds even worse if you don't play smart, which most people do not.
 
I don't see why you're making this so complicated. Just say that the average payout is x. If you lose, that's it, you've lost 1. If you win, you get 1, but then you've got a 1/6 chance of having to play again. If you play again, you have an average payout of x. So, average payout is:
50% chance of -1 + 50% chance of +1 + 50% chance of 1/6 chance x=
-.5+.5+x/12=x/12

But x was defined as being the average payout, so x=x/12. Subtracting x/12 from both sides, 11x/12=0, so x=0. So the average payout is zero. See how a little algebra can make everything so much simpler?
 
Alkatran said:

You pay me something. I flip a coin and you call it. If you get it right you win twice what you paid, but if you get it wrong you get nothing (double or nothing).

However, here's the twist: if you win I roll a dice and you call it: if you get it wrong you have to play again.

What is the average payout of playing this game?
Click to expand...

Hi Alkatran,

Say you bet n dollars, and the probability of winning the coin flip is p, and the probability of winning the dice roll is q, then the expected winnings for one play of this game is

E(winnings) = p[q*2n+(1-q)*E(winnings)]+(1-p)*-n

The E(winnings) is on the right side of the equation also, because, as you know, if you don't win the dice roll, you have to play the game again.

Solving for E(winnings) gives

E(winnings) = (2npq+np-n)/(pq-p+1)

For p = .5 and q = 1/6, this gives

E(winnings) = (-4/7)*n
 
Re: Re: Probability

Originally posted by jzs
if you don't win the dice roll, you have to play the game again.
Click to expand...
Yes, you have to play the game again. But, you already called the coin correctly, so you get paid for winning the first game regardless of what happens later. At least, that's my interpretation. And Art Vandelay's. And apparently Alkatran's, because his latest answer is zero too.

BTW, if there's no dice roll but instead you always play the game again whenever you win the coin flip, the expectation is still zero, but Art's and your method doesn't work in that case.
 
Re: Re: Re: Probability

69dodge said:
But, you already called the coin correctly, so you get paid for winning the first game regardless of what happens later. At least, that's my interpretation. And Art Vandelay's. And apparently Alkatran's, because his latest answer is zero too.
Click to expand...


Well it wasn't mine. I interpreted the problem statement to mean that you only keep 2n if you win the die roll.


BTW, if there's no dice roll but instead you always play the game again whenever you win the coin flip, the expectation is still zero, but Art's and your method doesn't work in that case.
Click to expand...

If it is only a coin flip, I get E(winnings) = p*2n + (1-p)*-n = n/2, when p = .5, as I took the problem statement to mean that if you win the coin flip, you get double the money you spent to play, and if you lose, you just lose the money you spent to play.
 
Re: Re: Re: Probability

69dodge said:
BTW, if there's no dice roll but instead you always play the game again whenever you win the coin flip, the expectation is still zero, but Art's and your method doesn't work in that case.
Click to expand...
Other than not having the 1/6 term, I don't see how my method doesn't work.
Instead of x=-.5+.5+x/12, you have x=-.5+.5+x/2; x=x/2; x=0.
 
Vagabond said:
The slots on the casino are the worst against you as far as odds go and they are 20-30 percent against you. This would be in that range. :) Most of the casino games are not only heavily against you in odds, but allow for you to play badly thus making the odds even worse if you don't play smart, which most people do not.
Click to expand...

It is illegal in most US casinos to have machines with holds greater than 20-25% (Nevada is 25%). However, with that said... any hold is a losing one (for you) in the long term.

The only way to play a slot machine badly is to not play max coin. In this instance (less than max coin) it is easily possible to have holds greater than 20% if you play less than max coin (this is due to a factoring up of jackpots with max coin).

I consider a bad player anyone which goes to a casino expecting to leave with more than he started.
 
Re: Re: Re: Re: Probability

Originally posted by Art Vandelay
Other than not having the 1/6 term, I don't see how my method doesn't work.
Instead of x=-.5+.5+x/12, you have x=-.5+.5+x/2; x=x/2; x=0.
Click to expand...
Ha. You're right. I made a silly mistake and ended up with x = x instead of x = x/2. That's why I said it wouldn't work.

Ooh, that gives me an idea. Suppose if you win, you have to play again with double the money. Then, it really is x = x. So, what's the expectation? Well, it's just that famous gambling scheme, only backwards. You're guaranteed to lose the initial amount.
 
DangerousBeliefs said:
It is illegal in most US casinos to have machines with holds greater than 20-25% (Nevada is 25%). However, with that said... any hold is a losing one (for you) in the long term.

The only way to play a slot machine badly is to not play max coin. In this instance (less than max coin) it is easily possible to have holds greater than 20% if you play less than max coin (this is due to a factoring up of jackpots with max coin).

I consider a bad player anyone which goes to a casino expecting to leave with more than he started.
Click to expand...

In my opinion playing slots at all is "playing badly". :) I don't go to the casino all that often. I will go once or twice a year and play poker or craps. But, I expect to lose 50-60 bucks and this is enough for me to have some fun for a few hours. An expensive night of entertainment, but no more so than going to a ball game or show. The way I play I don't usually lose much if at all. But, playing craps exactly the same way everytime while it might be the best odds, it isn't the way to play that is the most fun.
 
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