[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 530 KB, 1200x1600, Coins.jpg [View same] [iqdb] [saucenao] [google]
5650766 No.5650766 [Reply] [Original]

I have a fun game. A coin-flipping game. I begin my game by flipping the coin once.

Every time a flip comes up heads, my score increases by one.

Every time a flip comes up tails, I flip the coin two more times.

My minimum score is one. My maximum score is infinite. But what is my average score?

>> No.5650777

>>5650766
If you continue indefinitely, then your score is going to be infinite, with a margin of error of infinite.

>> No.5650784

1 + 4 = 5

>> No.5650790

If I understand you correctly, there's a nonzero chance of the game never terminating.

>> No.5650799

>>5650777

But continuing indefinitely is only one of the possible outcomes.

You can flip heads, resulting in a score of 1 and game over.

Or you can flip tails, giving you two more flips, which both come up heads. Score of two, game over.

Or you can flip tails, then heads and tails, then heads twice, giving you a score of three and game over.

Or...

etc.

An infinite result is possible, but highly unlikely. What's the average?

>> No.5650805

>>5650790
Never mind, worked it out, chance of termination is actually 100%.

>> No.5650810

>>5650799
I think that even though the chance of infinite score is very low, the infinite score is more than enough to throw off the infinitesimally small probability.

Conversely, and I'm just spit-balling here, infinite score * infinitesimally small chance = 1, so average equals one.

>> No.5650814

>>5650799
It's just speculation

>> No.5650817

>>5650790

Right. But does that make it impossible to calculate the average score of the games that DO terminate? I mean, 50% of the outcomes will be 1, and 12.5% of the outcomes will be 2. But I don't have the math to figure the average.

>> No.5650821

>>5650805
>>5650817

Right, sorry, there's a 100% chance of eventual termination, but no theoretical maximum score.

>> No.5650824

>>5650821
No, there's not. Saying that termination is inevitable is saying that flipping tails eventually becomes impossible.

>> No.5650826

Why are you guys so afraid of doing the math yourself?
Like taking a piece of paper and a pen and trying to work it out?

>> No.5650828

>>5650766
I don't understand your game.
What happens if tails comes up; then heads come up? what decides what happens next?

>> No.5650831

>>5650821
Yes, but it is something you have to check. If the rules instead said you flipped three coins, there would be a roughly 38% chance of nontermination.

>>5650817
You can show that the average score is infinite pretty easily. Suppose the mean score is finite and equal to x. If you flip tails, your expected score is 1. If you flip heads, your expected score is 2x since each coin gives rise to an independent version of the original game. So you get x = (1+2x)/2 which has no solution.

>> No.5650834

>>5650828

If tails comes up, you get two more flips.

If one of them is heads, your score increases by one. You still have another flip, which could come up heads (+1 score) or tails (get two more flips.)

Naturally, the game ends when you run out of flips.

>> No.5650838

>>5650831
>So you get x = (1+2x)/2 which has no solution.

I see. Hmm...

But, conversely, it's so easy to calculate the odds of any individual outcome. You've got a 50% chance of getting a score of 1, a 12.5% chance of getting a score of 2, a 3.25% chance of getting a score of 3... It's not possible to work it out from that direction? Or does this function have a mode but no mean?

>> No.5650837

>>5650766
The expected value of a coinflip, E, is 1*1/2 (1 pt for heads) + 2*E*1/2 (2 flips for tails). That is, E = .5 + E. The expected value (an average score, in the standard sense) is then infinite.

>> No.5650840

For the game to end, you need to flip a number of heads equal to twice the number of tails plus one.

The game therefore always ends after exactly 3n+1 flips for some integer n, and the probability that this happens for a given n is simply <span class="math">\frac{ 1}{2^{3n+1}} {3n+1 \choose n}[/spoiler], and the score you then get is 2n+1.

The expected value is therefore <div class="math">\sum_{n=0}^{ \infty} (2n+1)\frac{ 1}{2^{3n+1}} {3n+1 \choose n}</div>, which wolframalpha estimates to be around 10.

>> No.5650843

>>5650840
(By the way, this is obviously false, the purpose of the post is to get you guys to figure out what is false in that statement)

>> No.5650845

You can drop the one point aspect and just try to solve for the number of rounds you will go. If you know the number of rounds, you can solve for the number of points you got(total points is (number of rounds+1)/2

Each round there is a 50% chance of getting 2 turns, in other words each round you are expected to get 1 turn. So carrying forward you are expected to get an infinite number of turns on average. Plug that back into equation 1 and you get infinite points.

So answer is INFINITE POINTS

>> No.5650848

>>5650838
>You've got a 50% chance of getting a score of 1, a 12.5% chance of getting a score of 2, a 3.25% chance of getting a score of 3... It's not possible to work it out from that direction?
Should be possible; it will just be harder. You will get an infinite sum that approaches infinity.

>Or does this function have a mode but no mean?
The distribution has a mode of 0 and a mean of <span class="math">\infty[/spoiler].

>> No.5650852

>>5650840
>>5650843
For one thing,
>you need to flip a number of heads equal to twice the number of tails plus one.
isn't even true for the game where you get a single tail followed by two heads.

>> No.5650854

>>5650837
>>5650831
these

>> No.5650856

>>5650848

Mode of 1, you mean? 0's not a possible outcome.

Hmm. So, translating this to real-world terms... if you were in Vegas, and this was one of the games on offer, with every "heads" winning you one dollar, how much would a chance to play be worth?

>> No.5650860

>>5650856
Not much.

https://en.wikipedia.org/wiki/St._Petersburg_paradox

>> No.5650864

>>5650860
>not much
>just unscientific and irrational blattering about "thought process and feelings"
I'd pay anything to play this game

>> No.5650866

>>5650860

Ohh, learning.

See, I was lying in bed trying to sleep and thinking about math (specifically game mechanics for tabletop RPGs, I'm from /tg/) when the coin game occured to me, and now I have both the mathematical answer, the practical answer, and I know more about decision theory.

Thanks, /sci/. You're a good board.

>> No.5650867

reminds me of the collatz conjecture

>> No.5650871

>>5650856
Yeah, sorry, I meant 1.

>> No.5650873

>>5650864

You would pay all your money plus whatever you could get from loan sharks in order to play a game with a 68.25% chance of returning $3 or less?

The average score is, technically, infinite. But it's also possible to calculate the odds of winning less than you paid to play, and those odds go up the more you pay.

>> No.5650960

If I were to roll three tails in a row, do I have four rolls remaining, or two? That is to say, is it cumulative?