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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?

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.

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).

Quote:DieterThis seems like a math puzzle, which usually implies fair (ideal) coin, fair toss, and fair catch. Is that correct?

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Certainly a fair coin. I already stated what to assume on the toss.

Quote:ThatDonGuyIs 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.

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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! + ...)

Quote:ThatDonGuyOh, 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! + ...)

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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.

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) 59–92

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.

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?

Quote:DobrijI 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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Scarne (self)reportedly could flip a coin, catch it in his hand, flip that hand over the back of the other. Then know by feel through his hands if it were heads or tails, and be able to reverse the coin between his hands without the eye being able to tell he did anything.

Quote:unJonQuote:DobrijI 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?

link to original post

Scarne (self)reportedly could flip a coin, catch it in his hand, flip that hand over the back of the other. Then know by feel through his hands if it were heads or tails, and be able to reverse the coin between his hands without the eye being able to tell he did anything.

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It is harder to flip the coin between your hands unnoticed than to catch it after the right number of flips in the air. At least for some. Kudos if you've got the moves.

Quote:billryanI knew a bartender who would let you call the coin while it was in the air and he'd catch it in one hand and slam it down on the back of his wrist. 100% of the time, it would be the opposite of what you called. He must have had a way to feel the coin but as far as stupid bartender tricks go, I rated it behind playing the bottles.

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That is the issue with coin tosses when the tosser grabs it from the air and places it on his wrist or back of hand (which is common, not even for tricks, but regular tosses), this seems to be a common trick, there is a magician that specializes in this trick with silver dollars (my guess is he uses this specific type of coin because of the large face, and from using it for so many years he can know well what the sides are in his hand by feel). My guess would be neither the magician or your bartender could do this with a random foreign coin that they have never seen before (where they don't know the feel or images on either side).

The other issue with this is you do not let the coin naturally land (even if he is not cheating), so any slight advantage from the weight of the faces or the starting position of the coin, goes out the window. As opposed to tossing the coin into a bowl or letting it fall to the floor.

Except you high rollers... Google US coins.

LOL

2 young mathematicians have (according to the sub headline) astonished colleagues with a full proof of a famous conjecture re randomness using coins to demonstrate their point

this is above my pay grade - but I would guess there are some here that will appreciate it

.

https://www.quantamagazine.org/elegant-six-page-proof-reveals-the-emergence-of-random-structure-20220425/

.

Quote:AlanMendelsonPortraits on the obverse of US coins have fewer "bumps" or tactile ridges. Look at the change in your pockets.

Except you high rollers... Google US coins.

LOL

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Palladium Eagles still seem to be high relief, although it seems unlikely someone trying to hustle a coin flip would fish one from their pocket.

Quote:lilredrooster____________

2 young mathematicians have (according to the sub headline) astonished colleagues with a full proof of a famous conjecture re randomness using coins to demonstrate their point

this is above my pay grade - but I would guess there are some here that will appreciate it

.

https://www.quantamagazine.org/elegant-six-page-proof-reveals-the-emergence-of-random-structure-20220425/

.

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I read the article, as far as my English allows me. I understand that there we are talking about something like fractals, which are created on theoretical/abstract numbers. Not applicable in this matter, since we are dealing with elements of physics

I'll even say more, this mathematician girl Huy Tuan Pham is a real fool, she speaks with the importance of ordinary nonsense

Quote:WizardThis is rather obvious, but my grandfather could make you lose or win any flip. He would simply catch it in his palm and call the flip on the face up side, if he favored that side. Otherwise, he just flipped it over and placed the coin on his arm.

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My older brother always used to win at coin tosses. While the coin is in the air he calls, “Heads I win; tails you lose.”

Even after I figured it out, I didn’t say anything, so as not to get beaten up!