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Fair coins tend to land on the same side they started

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Re: Fair coins tend to land on the same side they started

#221

Post by thih9 »

098799 wrote: post #84
ChocMontePy wrote: post #69

Earlier quoted context omitted.

If anyone wants to test it, someone wrote a short code that simulates doing that 100,000 times: https://www.techiedelight.com/generate-fair-results-biased-c... The coin is biased to come up TAILS 80% of the time, but using Von Neumann's method in the program I got HEADS 50.035%, TAILS 49.965%.

Why would you test it? Probability of two heads: p*p Probability of two tails: (1-p)*(1-p) Probability of head followed by tails: p*(1-p) Probability of tails followed by heads: (1-p)*p It's not difficult to notice that if you remove the first two, the last two form a 50/50 distribution

> Why would you test it?

Why not?

The fact that you can show something with a mathematical equation doesn’t make other demonstrations any less cool.

Re: Fair coins tend to land on the same side they started

#222

Post by joshuamorton »

matsemann wrote: post #219
cstrahan wrote: post #201

Earlier quoted context omitted.

> that's what the host does in the other scenario to me. Is it, though? It seems apparent that, after the first guess, the host opens all but the last two doors, which just so happens to be 1 door. To check the math: Start with the $NUM_DOORS open doors. Now open all but the last two. So that’s $NUM_DOORS-2, which is 3-2, which equals 1 open door.

My interpretation of the host opening a single door and asking if you want to switch is equally valid when you expand it to 100 doors.

Sure, it could be that the host only opens one door, or it could be that he opens all but one door. In every case, however, it is better to switch. The all-but-one example is hyperbolic but still follows precisely the same mathematical rules. Your interpretation is valid, but so is the all-but-one example, and they all lead to the same result, it's just more obvious when you open nearly all the doors.

Re: Fair coins tend to land on the same side they started

#223

Post by aqme28 »

toxik wrote: post #117
098799 wrote: post #84

Earlier quoted context omitted.

Why would you test it? Probability of two heads: p*p Probability of two tails: (1-p)*(1-p) Probability of head followed by tails: p*(1-p) Probability of tails followed by heads: (1-p)*p It's not difficult to notice that if you remove the first two, the last two form a 50/50 distribution

I think about it this way p(th) = p(t) p(h) p(ht) = p(h) p(t) Hence p(th) = p(ht) regardless of coin imbalance as long as both events actually will happen. QED.

Yeah this is simpler. You're just throwing away every pair that isn't a TH or HT

Re: Fair coins tend to land on the same side they started

#224

Post by lcnPylGDnU4H9OF »

cantrevealname wrote: post #128
098799 wrote: post #84

Earlier quoted context omitted.

Why would you test it? Probability of two heads: p*p Probability of two tails: (1-p)*(1-p) Probability of head followed by tails: p*(1-p) Probability of tails followed by heads: (1-p)*p It's not difficult to notice that if you remove the first two, the last two form a 50/50 distribution

> Why would you test it? I recall conversations on Usenet decades ago about the Monty Hall problem[1] in which people gave elementary proofs that probabilities don't change by opening a door. Even from mathematicians and statisticians. People were very insistent that the analytical solution was simple and obvious and that switching doors didn't change anything. The only thing that changed some people's minds was a pr…

Something that makes this a lot more intuitive is to increase the number of doors. If there's 100 doors, the host will open all remaining doors except 1, and they will never open the door with the car behind it, then one has a 1% chance of winning the car if they don't switch doors and it would happen only because they initially chose the door with the car.

Re: Fair coins tend to land on the same side they started

#225

Post by tshaddox »

cantrevealname wrote: post #128
098799 wrote: post #84

Earlier quoted context omitted.

Why would you test it? Probability of two heads: p*p Probability of two tails: (1-p)*(1-p) Probability of head followed by tails: p*(1-p) Probability of tails followed by heads: (1-p)*p It's not difficult to notice that if you remove the first two, the last two form a 50/50 distribution

> Why would you test it? I recall conversations on Usenet decades ago about the Monty Hall problem[1] in which people gave elementary proofs that probabilities don't change by opening a door. Even from mathematicians and statisticians. People were very insistent that the analytical solution was simple and obvious and that switching doors didn't change anything. The only thing that changed some people's minds was a pr…

I'm curious. When those people see the simulation, do they then go back to the analysis and uncover their mistaken reasoning? Or do they just continue to reject the analysis but begrudgingly accept the outcome of the simulation? The analysis of the Monty Hall problem is so very simple I find it very odd to staunchly reject it but then be persuaded by the simulation.

Re: Fair coins tend to land on the same side they started

#226

Post by joshuamorton »

evouga wrote: post #216
tacitusarc wrote: post #186

Earlier quoted context omitted.

For sufficiently analytical folks that works, but for lay people it tends to still be confusing. The best way I’ve heard it explained to help people get it through intuition is by changing the number of doors and goats. Say there are 100 doors, and they all have goats except one, which has a car. You pick door 1. Monty then proceeds to open doors 2 through 48, skips door 49, and then opens the remaining doors. After…

The situation is now counterintuitive in the other direction: if Monty Hall had opened those 48 doors at random and they just happened to not contain the car, then there is no advantage to switching, though many people would insist otherwise.

But the doors weren't opened at random. You know he won't open the car, because that's part of the rules of the game.

Let's demonstrate with a slightly different construction: You're no longer playing with monty, but with a demon. This demon wants you to lose, but also picked a very bad game for themselves. You pick a door, then the demon opens all-but-one of the remaining doors. Then, you can pick any door, open or closed, and you get what's in it.

If the demon opens doors at random, nearly all the time (with 100 doors) you'll see the car and be able to pick it directly. In this situation, switching between the closed doors doesn't really matter, but you'll usually know exactly which door to pick, because you can see the car.

So instead, the demon only opens doors that don't have a vehicle behind them. You only ever see goats. At this point, he's not opening doors at random. If he were, you'd see the car 98% of the time, but you never do. At this point, since he's using additional information, it is in your best interest to switch.

Re: Fair coins tend to land on the same side they started

#227

Post by thih9 »

noobermin wrote: post #9
fbartos wrote: post #2

About a year ago, we embarked on a quest to answer one of the most intriguing questions: If you flip a fair coin and catch it in hand, what's the probability it lands on the same side it started? Today, we are finally ready to share the results. Thanks to my friends, collaborators, and even strangers from the internet, we collected flippin 350,757 coin flips. We ran several "Coin Tossing Marathons" (e.g., https://you…

>If you bet a dollar on the outcome of a coin toss 1000 times, knowing the starting position of the coin toss would earn you 19$ on average. This is more than the casino advantage for 6-deck blackjack against an optimal player (5$) but less than that for single-zero roulette (27$). This sounds like the plot of a western where a man travels from town to town and gleans a little cash from the local waterhole a little e…

> in order to get just the $19, assuming you played a modest 20 times a day, it'd take 10 weeks (not including weekends)

What if one play session consisted of 10 coin tosses (each an independent $1 bet)? I guess 20 games like this per day would still be doable. Would that mean $19 per week?

Next we up the bet to $10 per throw.

Re: Fair coins tend to land on the same side they started

#228

Post by mucle6 »

jbandela1 wrote: post #16

Von Neumann described a very elegant way to get fair results from a biased coin. 1. Flip the coin twice 2. If you get the same result both times, goto 1 3. Now that you have different results for your pair of flips, use the first element of the pair of flips as your result. https://en.wikipedia.org/wiki/Fair_coin#Fair_results_from_a_...

This is very interesting, but it assumes a coin is biased the same way every flip.

If a coin is more likely to land on the side it starts on, then the bias can change between flips. To fix this, we just need to make sure the coin starts the same side up before every flip.

Re: Fair coins tend to land on the same side they started

#229

Post by zh3 »

dotancohen wrote: post #206
pmontra wrote: post #149

Earlier quoted context omitted.

That's how it went when I was solving problems at the Statistics course at university. I modeled the problem perfectly, got the wrong result. Changed assumptions, got the wrong result. Checked the solution, its reasoning didn't make much sense anyway. Run a simulation, got an approximate result close to the correct solution.

This sounds like the classic "tweak the model until the results fit with our preexisting conclusion". Very common across all industries unfortunately.

Also known in a derogatory fashion [0] as "adding epicycles" (after the Ptolemaic view of the heavens).

[0] https://en.wikipedia.org/wiki/Deferent_and_epicycle#Bad_scie...

Re: Fair coins tend to land on the same side they started

#230

Post by mucle6 »

a_c wrote: post #56

I always tell people that result of coin flip is highly start state dependent. Imagine a sequence of H(ead), T(ail), H, T, H, T, ... if the sequence starts with H first, in no way can the number of T exceed that of H, but the number of H might be 1 greater that that of T. I never tested my self, but I hypothesize that the propability will be more skewed if the number of revolutions is less, i.e. having a shorter Head…

I would have never considered this. Such an interesting way to think about the problem
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