hand dealt results

[QFIT]

I'm sorry, but why do you keep making this claim? The odds are 50-50. There is no mathematical law that states that the results will be exactly 50-50. In fact, there are enormous branches of statistics and probability theory that say it won't be.

You used the word "given" five times in one post. Where is it "given?" What law of mathematics? Usually, people use this word when talking about the Bible. If you are talking science, you have to give a source or proof.

 

[aslan]

The "given" is simply that the game is 50/50. That's the known. That the game will have equal amounts of wins and losses. The game is not a 49.999% game or a 50.0001% game it is known that the game is 50/50. That's a given. It's inherent that it will eventually have equal amounts of wins and losses. It's not arguable, it's the "given". You're trying to argue that a 50/50 game is a 49.9999% game. It's inherentyly wrong. The given is that the game is 50/50. If you're results are any other then 50/50, you're game is biased and the results are no good. That's the "given". It's inherent that the game breaks even. It has to.

50/50 is about odds, not outcomes.

 

[JJR]

Carry it out a few trillion more trials. Eventually it has to break even or the game is biased. It has to, or the true odds for this biased game is 49.999%.

 

[Lonesome Gambler]

Carry it out a few trillion more trials. Eventually it has to break even or the game is biased. It has to, or the true odds for this biased game is 49.999%.

That's false. I think you're misunderstanding what they're telling you here. The odds of a trial are 50/50. It's equally likely that either outcome will come to pass. Over trillions of trials, it becomes—possibly counter-intuitively—even less likely that the number of each possible outcome are equal, even though the odds for either outcome on a single trial is still 50/50, since the amount of trials has increased. The law of large numbers does indeed tell us that we'll come closer and closer to our true expectation (breakeven) the larger the number of trials we repeat, but it's almost impossible that we'll be exactly at that figure.

To summarize, as has already been noted, you're confusing odds and outcomes. If you flip a fair coin a trillion times, it could be heads or tails each individual time, but the odds of it being 500,000,000,000 heads and 500,000,000,000 tails is almost nil. This does not change the odds of each outcome in any way.

 

[QFIT]

I fear that you do not understand the meaning of "given." When you state a problem, you have givens. Like a coin toss with an unbiased coin. You then use rules of logic and previously proved theorems to come to a conclusion. You are stating that the conclusion is a given. Frankly speaking, that is like a religious argument. You can't just claim something is true. You must demonstrate.

 

[k_c]

Carry it out a few trillion more trials. Eventually it has to break even or the game is biased. It has to, or the true odds for this biased game is 49.999%.

You can use binomial expansion to get an idea of where you might be headed for a large number of fair coin flips.

First of all the probability of an equal number of heads and tails is always 0 on an odd number of flips.

For 2 flips prob(equal ht) = 2/4
For 4 flips prob(equal ht) = 6/16
For 6 flips prob(equal ht) = 20/64
For 8 flips prob(equal ht) = 70/256

For a large even number of flips it appears prob(equal ht) is headed downward. Maybe it approaches zero or maybe another limit but it is continously getting lower. Since prob(equal ht) is always 0 for an odd number of flips the conclusion is that your chances for an equal number of heads/tails after a large number of flips is not good at best.

(h+t)^1 = h + t (prob equal h,t = 0)

(h+t)^2 = h^2 + 2*h*t + t^2 (prob equal h,t = 2/4)

(h+t)^3 = h^3 + 3*h^2*t + 3*h*t^2 + t^3 (prob equal h,t = 0/8)

(h+t)^4 = h^4 + 4*h^3*t + 6*h^2*t^2 + 4*h*t^3
+ t^4 (prob equal h,t = 6/16)

(h+t)^5 = h^5 + 5*h^4*t + 10*h^3*t^2 + 10*h^2*t^3
+ 5*h*t^4+t^5 (prob equal h,t = 0/32)

(h+t)^6 = h^6 + 6*h^5*t + 15*h^4*t^2 + 20*h^3*t^3 + 15*h^2*t^4
+ 6*h*t^5+t^6 (prob equal h,t = 20/64)

(h+t)^7 = h^7 + 7*h^6*t + 21*h^5*t^2 + 35*h^4*t^3 +35*h^3*t4
+ 21*h^2*t^5 + 7*h*t^6 + t^7 (prob equal h,t = 0/128)

(h+t)^8 = h^8 + 8*h^7*t + 28*h^6*t^2 + 56*h^5^t^3 + 70*h^4*t^4
+ 56*h^3^t^5 + 28*h^2*t^6 + 8*h*t^7 + t^8 (prob equal h,t = 70/256)

(h+t)^999999999 = ??????? (prob equal h,t = 0/(2^999999999))

(h+t)^1000000000 = ??????? (prob equal h,t = ?/(2^1000000000))

Edit: OK I went to this site and entered (h+t)^100
Chance of 50 heads and 50 tails in 100 flips is
100891344545564193334812497256 / 1267650600228229401496703205376 =~ .07959

I would guess that for a very large even number of flips prob(equal ht) approaches 0.
For an odd number of flips the prob is always 0.

 

[QFIT]

Yes, the probability over a billion hands is zero. The result is virtually never zero.

 

[aslan]

I would guess that for a very large even number of flips prob(equal ht) approaches 0.

Because in a very large number of flips the difference between h and t can grow so large that they may never even up again? I'm trying to understand in layman's terms.

 

[iCountNTrack]

You can use binomial expansion to get an idea of where you might be headed for a large number of fair coin flips.

First of all the probability of an equal number of heads and tails is always 0 on an odd number of flips.

For 2 flips prob(equal ht) = 2/4
For 4 flips prob(equal ht) = 6/16
For 6 flips prob(equal ht) = 20/64
For 8 flips prob(equal ht) = 70/256

For a large even number of flips it appears prob(equal ht) is headed downward. Maybe it approaches zero or maybe another limit but it is continously getting lower. Since prob(equal ht) is always 0 for an odd number of flips the conclusion is that your chances for an equal number of heads/tails after a large number of flips is not good at best.

(h+t)^1 = h + t (prob equal h,t = 0)

(h+t)^2 = h^2 + 2*h*t + t^2 (prob equal h,t = 2/4)

(h+t)^3 = h^3 + 3*h^2*t + 3*h*t^2 + t^3 (prob equal h,t = 0/8)

(h+t)^4 = h^4 + 4*h^3*t + 6*h^2*t^2 + 4*h*t^3
+ t^4 (prob equal h,t = 6/16)

(h+t)^5 = h^5 + 5*h^4*t + 10*h^3*t^2 + 10*h^2*t^3
+ 5*h*t^4+t^5 (prob equal h,t = 0/32)

(h+t)^6 = h^6 + 6*h^5*t + 15*h^4*t^2 + 20*h^3*t^3 + 15*h^2*t^4
+ 6*h*t^5+t^6 (prob equal h,t = 20/64)

(h+t)^7 = h^7 + 7*h^6*t + 21*h^5*t^2 + 35*h^4*t^3 +35*h^3*t4
+ 21*h^2*t^5 + 7*h*t^6 + t^7 (prob equal h,t = 0/128)

(h+t)^8 = h^8 + 8*h^7*t + 28*h^6*t^2 + 56*h^5^t^3 + 70*h^4*t^4
+ 56*h^3^t^5 + 28*h^2*t^6 + 8*h*t^7 + t^8 (prob equal h,t = 70/256)

(h+t)^999999999 = ??????? (prob equal h,t = 0/(2^999999999))

(h+t)^1000000000 = ??????? (prob equal h,t = ?/(2^1000000000))

Edit: OK I went to this site and entered (h+t)^100
Chance of 50 heads and 50 tails in 100 flips is
100891344545564193334812497256 / 1267650600228229401496703205376 =~ .07959

I would guess that for a very large even number of flips prob(equal ht) approaches 0.
For an odd number of flips the prob is always 0.

Using a calculator for very large numbers a colleague of mine coded, for 50 million flips the probability of getting equal heads and tails using the binomial formula is 0.000112

 

[JJR]

What are the chances of getting to break even or above break even? My guess is 50/50. I understand getting exactly equal numbers of heads and tails is uncommon, but what I really meant was at or above break even. I'm sure for a binomial distribution of a fair game of 50/50 half of the numbers are above break even and half are below break even.

I guess the answer we're looking for is how long will it take to return to break even once a trial has started. I guess a risk of ruin calculator for a 50/50 game that showed the number of trials to lose 1 unit (actually returning to 1 below break even) would do the trick, but like I already stated it's a "given" that the game will return to break even eventually. Which disproves the gamblers fallacy for multiple trials. Gamblers fallacy is only accurate for the next trial in a series. That is, if you lose 4 in a row you're odds of winning the next trial is still just 50/50. Not that a 50/50 game won't break even, because it's actually a 49.9999% game.

 

[QFIT]

What are the chances of getting to break even or above break even? My guess is 50/50. I understand getting exactly equal numbers of heads and tails is uncommon, but what I really meant was at or above break even. I'm sure for a binomial distribution of a fair game of 50/50 half of the numbers are above break even and half are below break even.

I guess the answer we're looking for is how long will it take to return to break even once a trial has started. I guess a risk of ruin calculator for a 50/50 game that showed the number of trials to lose 1 unit (actually returning to 1 below break even) would do the trick, but like I already stated it's a "given" that the game will return to break even eventually. Which disproves the gamblers fallacy for multiple trials. Gamblers fallacy is only accurate for the next trial in a series. That is, if you lose 4 in a row you're odds of winning the next trial is still just 50/50. Not that a 50/50 game won't break even, because it's actually a 49.9999% game.

Once again, you repeat that it is a "given" and that you have "disproved" Gambler's Fallacy. Why do you keep saying this? Why do you think you can just make a claim, and that serves as a "proof?"

A 50/50 game does not have to eventually break even. Odds are not a guarantee. There is no such law.

 

[Sonny]

Alright seriously, read the stickies already. Start with the one called The Gambler's Fallacy. It explains everything in perfect detail. The wikipedia link is also decent. This discussion is just going to continue in this death spiral until everyone gets up to speed. It looks like I have to close this thread in order to encourage people to do the research. If anyone has a legitimate question they can start a new thread.

-Sonny-

 

[iCountNTrack]

What are the chances of getting to break even or above break even? My guess is 50/50. I understand getting exactly equal numbers of heads and tails is uncommon, but what I really meant was at or above break even. I'm sure for a binomial distribution of a fair game of 50/50 half of the numbers are above break even and half are below break even.

I guess the answer we're looking for is how long will it take to return to break even once a trial has started. I guess a risk of ruin calculator for a 50/50 game that showed the number of trials to lose 1 unit (actually returning to 1 below break even) would do the trick, but like I already stated it's a "given" that the game will return to break even eventually. Which disproves the gamblers fallacy for multiple trials. Gamblers fallacy is only accurate for the next trial in a series. That is, if you lose 4 in a row you're odds of winning the next trial is still just 50/50. Not that a 50/50 game won't break even, because it's actually a 49.9999% game.

Okay so now you are convinced that the number if heads to tails dont have to be equal(and will be very unlikely as the number of flips increases) so we are getting somewhere
I ran a simulation for a coin flip, the random number generator i used was the Mersenne Twister, the results are summarized in the following table

We can see from the table that the number of heads was not equal for all the the different flip numbers. However what is more important to realize is that the larger the number of trials the closer the fraction of number of heads is to 0.5.
It is also important to realize that the number of heads in excess or in default from having an equal number of heads and tail does not matter as the sample size increases because the denominator (number of trials) will dilute it. For instance for 10 flips, even though we have an "excess" of only 1 heads, the fraction is still off by 10% from 0.5. While for instance for 1000000 flips while we have an "excess" of 243 heads, the fraction is off by 0.88%. So again it is all happening with the sample size (the denominator)