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Wizard
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Wizard
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April 27th, 2022 at 1:17:20 PM permalink
This thread is a split-off from this post in the thread THE TOP FACE WILL BE MORE PROBABLE, that alleges the roll of a dice will favor the side that started on the top.

What is the probability that if a coin is flipped beginning with heads face up it will land on heads face up? You may assume the mean number of complete 360-degree flips is 3 and the actual number is distributed according to the Poisson distribution.

For extra credit, what is the probability of a mean number of flips of m?

Please note I'm not asking this in the Easy Math Puzzles thread because it's not easy. I continue to encourage the forum to make hard problems their own thread.

The question for the poll is what do you think the general answer is?
It's not whether you win or lose; it's whether or not you had a good bet.
Dieter
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April 27th, 2022 at 3:40:47 PM permalink
This seems like a math puzzle, which usually implies fair (ideal) coin, fair toss, and fair catch. Is that correct?
May the cards fall in your favor.
ThatDonGuy
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April 27th, 2022 at 5:43:44 PM permalink
Is there a way to solve this without knowing the standard deviation in terms of the number of tosses?


I am assuming that the answer is not 1/2 because, with the Poisson distribution, the number of flips eventually becomes negative, so if, say, the SD = 1, you are looking only at the portion of the graph with x >= -3.

AlanMendelson
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April 27th, 2022 at 6:32:06 PM permalink
US government minted coins are not fair coins. The heads or obverse is heavier.
Gandler
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Dobrij
April 27th, 2022 at 7:15:05 PM permalink
I think it would depend on the coin.

I remember reading in some article years ago that cent coins for examples (Lincoln.01USD) have such an elaborate face that it leads to more tails landing in tosses because the head is heavier (though this was also in defense of not ending the coin so there could be bias), but it would not surprise me that even if this offshoot comment was made in jest, the faces of the coin could influence the landing position of a coin tossed into a glass bowl say.

Also, looking at the U.S. mint the portraits and graphics are at different location on different coins (the cent is perfectly in the center for examples), I have no idea if this has a massive impact, but I would imagine it has some effect:
https://www.usmint.gov/learn/coin-and-medal-programs/coin-specifications

Actually, I googled it out of curiosity, and this Smithsonian article seems to confirm two things (based on summarizing other studies), that a coin is slightly more likely to land on the heads up side that it was tossed from, and that a coin spin is (as opposed to a toss) far more likely to land on the lighter side up (which can mean the "heads" being face down). This even says that for spins cents are close to 80% tails up which seems to go along with what I recall reading. It seems like the basic premise is, the heavier the side, the more likely it to end up down (though having the side up that you want up on the flip can improve this further).

https://www.smithsonianmag.com/science-nature/gamblers-take-note-the-odds-in-a-coin-flip-arent-quite-5050-145465423/

So probably the best advice is to find the most unbalanced coin and toss it with the light side up for the best advantage (if you are betting that side, or even better for a coin spin, though the face up side would not matter of course for a spin as it starts on its side).
Wizard
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Dieter
April 27th, 2022 at 9:15:22 PM permalink
Quote: Dieter

This seems like a math puzzle, which usually implies fair (ideal) coin, fair toss, and fair catch. Is that correct?
link to original post



Certainly a fair coin. I already stated what to assume on the toss.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
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April 28th, 2022 at 8:50:19 AM permalink
Quote: ThatDonGuy

Is there a way to solve this without knowing the standard deviation in terms of the number of tosses?


I am assuming that the answer is not 1/2 because, with the Poisson distribution, the number of flips eventually becomes negative, so if, say, the SD = 1, you are looking only at the portion of the graph with x >= -3.


link to original post


Oh, Poisson distribution...


First of all, I am assuming the coin does at least one half-flip - otherwise it is not a "valid toss."

We are given that the expected number of half-flips is 6
The coin lands on its starting side if the number of half-flips is even
P(the coin makes 2 half-flips in the next toss) = 6^2 / (2! e^6) = 6^2 / (2 e^6) = P(1) x 6 / 2
P(the coin makes 4 half-flips in the next toss) = 6^4 / (4! e^6) = 6^4 / (24 e^6) = P(2) x 6^2 / (3 x 4)
P(the coin makes 6 half-flips in the next toss) = 6^6 / (6! e^6) = 6^6 / (720 e^6) = P(2) x 6^2 / (5 x 6)

P(the coin makes an even number of half-flips) = 18 / e^6 x (1 + 6^2 / (3 x 4) + 6^4 / (3 x 4 x 5 x 6) + 6^6 / (3 x 4 x 5 x 6 x 7 x 8) + ...)
= 18 / e^6 x (1 + 6^2 x 2 / 4! + 6^4 x 2 / 6! + 6^6 x 2 / 8! + ...)
= 0.49752432

If you do allow for a zero-flip toss, add 1 / e^6 to this; the result is 0.500003



P(the coin makes an even number of half-flips) = (2m)^2 / (2 e^(2m)) x (1 + (2m)^2 x 2 / 4! + (2m)^4 x 2 / 6! + (2m)^6 x 2 / 8! + ...)
= 2 m^2 / e^(2m) x (1 + (2m)^2 x 2 / 4! + (2m)^4 x 2 / 6! + (2m)^6 x 2 / 8! + ...)

Wizard
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April 28th, 2022 at 9:58:58 AM permalink
Quote: ThatDonGuy

Oh, Poisson distribution...


First of all, I am assuming the coin does at least one half-flip - otherwise it is not a "valid toss."

We are given that the expected number of half-flips is 6
The coin lands on its starting side if the number of half-flips is even
P(the coin makes 2 half-flips in the next toss) = 6^2 / (2! e^6) = 6^2 / (2 e^6) = P(1) x 6 / 2
P(the coin makes 4 half-flips in the next toss) = 6^4 / (4! e^6) = 6^4 / (24 e^6) = P(2) x 6^2 / (3 x 4)
P(the coin makes 6 half-flips in the next toss) = 6^6 / (6! e^6) = 6^6 / (720 e^6) = P(2) x 6^2 / (5 x 6)

P(the coin makes an even number of half-flips) = 18 / e^6 x (1 + 6^2 / (3 x 4) + 6^4 / (3 x 4 x 5 x 6) + 6^6 / (3 x 4 x 5 x 6 x 7 x 8) + ...)
= 18 / e^6 x (1 + 6^2 x 2 / 4! + 6^4 x 2 / 6! + 6^6 x 2 / 8! + ...)
= 0.49752432

If you do allow for a zero-flip toss, add 1 / e^6 to this; the result is 0.500003



P(the coin makes an even number of half-flips) = (2m)^2 / (2 e^(2m)) x (1 + (2m)^2 x 2 / 4! + (2m)^4 x 2 / 6! + (2m)^6 x 2 / 8! + ...)
= 2 m^2 / e^(2m) x (1 + (2m)^2 x 2 / 4! + (2m)^4 x 2 / 6! + (2m)^6 x 2 / 8! + ...)


link to original post



I agree! There is a formula for the probability the coin lands on heads, given Poisson and the mean, that doesn't involve an infinite sum. I show the formula and how to get there here.
It's not whether you win or lose; it's whether or not you had a good bet.
gordonm888
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April 28th, 2022 at 10:56:50 AM permalink
Here's another scientific journal article, maybe the best, on the dynamics of coin tossing for those of you who are completists. It's from 2008 and about 30 pages in "Physics Reports" the most prestigious journal in the world for physics publications. This is physics, not math. The article has some beautiful figures.

Dynamics of coin tossing is predictable
J. Strzałko, J. Grabski, A. Stefański, P. Perlikowski, T. Kapitaniak∗
Division of Dynamics, Technical University of Łódź, Stefanowskiego 1/15, 90-924 Łódź, Poland
Physics Reports 469 (2008) 5992

Link for Downloading the PDF

It has four conclusions that I feel are notable:
1. If you know the initial conditions (position, orientation, velocity vector and angular momentum vector) precisely enough then a coin toss is predictable. However, small uncertainties in those initial conditions will essentially make the coin toss unpredictable.

2. Unless a coin is dropped from tens of meters, the atmospheric drag is unimportant. If it is dropped from tens of meters, however, it becomes like a leaf falling.

3. Any coin that is not infinitely thin will have a finite chance of landing on its edge.

4. If the surface below the coin flip is hard, i.e., partly elastic, a coin is capable of flipping as it bounces and thus any partially elastic collisions with the surface will further randomize the outcome.
******************************
There is an important difference between these physics analyses and what the Wizard is referring to. The Wizard refers to coin flips, and the articles refer to coin tosses. The term "coin tosses" includes the possibility that the initial angular momentum is so small that the coin never flips in the air and essentially just drops onto the surface below. Of course, in physics, it may then elastically bounce, causing the coin to flip additional times. Wizard refers to a toss in which the coin must by definition flip at least once during its drop and (perhaps) lands onto a soft (inelastic) surface and does not further bounce.
Last edited by: gordonm888 on Apr 28, 2022
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Dobrij
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April 28th, 2022 at 2:27:21 PM permalink
I knew one person who can to drop a coin so that it falls on the side he needs. I didn't learn it myself, but it's all about throwing technique. So I think to answer the question you need to clarify:

1. Are there any conditions for how (technics) a coin will be tossed?
2. From what height will the coin fall?
3. Where can a coin fall, on a hand, on a hard floor, or on the ground?

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