Challenge Questions

If you toss first, you have a fifty percent chance of losing on that toss. Your friend has a zero percent chance of losing on that toss and a fifty percent chance he will have to toss the next time. If you don't lose on the first toss, your friend has a fifty percent chance of losing on the second toss (25% overall) and you have a zero percent chance. If a head is not tossed on either of the first two tries, you get your second toss with another fifty percent chance of losing (12.5% overall). If you don't toss a head, your friend's chance of losing on the fourth toss is fifty percent (6.25% overall). You can see that sums of the probabilities are just as Physicist? indicated. He gets my GA.

Thanks,

Jim

Nice answer. I thought I'd take physicist? up on his generous challenge of the weighted coin.

Let p_H be the probably of tossing heads with the coin.
Assuming that you toss first as your friend suggested your probability of paying is:
p_first = p_H + p_H(1-p_H)² + p_H(1-p_H)⁴ + ...
Using the limit rules for geometric series:
p_first = p_H / (1-(1-p_H)²)
Assuming that you convince your your friend to toss first your probability of paying is:
p_second = p_H(1-p_H) + p_H(1-p_H)³ + ...

p_second = p_H(1-p_H) / (1-(1-p_H)²)

Knowing the odds, what is the ethical choice in this situation?

We did a simulation in Excel using 10,000 trials and you are correct.

The probability of the coin coming up heads is 50%, no matter how many times the coin has been tossed.

The odds of paying is equal to your friends'.

There is a 50% chance the friend wins without ever having to toss the coin.

That's gotta count for something right?

"The odds of paying is equal to your friends' ..."

That would be true if you changed the rules to read:

'Each person shall toss the coin once. If one gets a head (while the other one gets a tale) s/he will pay. If no-one is a clear winner on the first round, each shall toss again. This will repeat until there is one and only one head tossed in a pair of tosses.'

What the OP is asking is who is likely to be the first to come up heads. Of course, on each toss, the chance of heads is 50%. But if the first toss is tails, and the second toss is tails, then the chances of heads coming up increases. If the first toss is yours, and you get tails, and the second toss is tails, then the third toss, which will be yours, is more likely to come up heads.

Not much math needed on this, just some common sense. A fair coin is not likely to come up the same 3 times in a row. I know it could happen, and someone will probably post a story about it happening to them, or they had seen it happen. But still, the chances are the coin is going to come up heads in the first 3 tosses, and if you are making 2 of those tosses, you have a 2/3 chance of getting heads.

In short, listen to Fyz, and be prepared to pay if you ever eat out with him.

"... But if the first toss is tails, and the second toss is tails, then the chances of heads coming up increases. ..."

I thought that the chances will always be fifty-fifty each time, as past history should not influence a particular toss. =TeeSquare=

the bet is not symmetrical. it can be won on toss one, but not lost. It is as fiz says. It resembles those goofs that play coin toss and when they lose, say "double or nothing", you know then you have a piker on your hands

Correct the odds remain the same at 50/50, but the probability changes.

In one of Indian movies, same trik was shown. But one of the friends has coin with both sides tail!

Except that the challenge says the same coin every time. I suppose you would continue throwing coins until the police arrived?

0.5 probability

My probability it's better, because in every set of turn's (mine and my friend), always I toss the coin first, and my friend only will do it after, if it's not heads, that's represents a 50% of a free lunch winning probability for my friend just for seen me tossing, then if he had to toss the coin, get's another 50% chance to pass me the turn to toss the coin. And the cycle repeats. that's represents 25% of probability for my friend to pay the lunch in every set of turn's, 50% (from his toss) of the 50% (of my toss passing him the turn).

The above explain a probabitlity totally different to the classical case of always toss a coin a fixed number of turn's

You have to pay up-front in McDonald's.

OK, what if it's a seen like Russian roulette ; Half the chambers in the gun are loaded. I'd want to go second.....

On any amount of tossing the coin the probabilities are 50% to any one of the players. No matter what. Totally random chances. Would just try to avoid being the first.

On any amount of tossing the coin the probabilities are 50% to any one of the players. No matter what. Totally random chances. Would just try to avoid being the first.

my probability of paying the bill will be more than my friend's.

3/4 times probability of me paying the bill.

I make probability of you paying = 2/3, probability of your friend paying = 1/3.

Codey

the probability of me paying the bill is 50% if I start the tossing and that of my friend tossing second is 25% therefore I have a greater probability of paying the bill

My friend's probability is better since I may end up paying the bill. Remember that I am the one tossing first.

Be careful!! Some Canadian quarters are "Two headed".

There are 4 (really 3) possible scenarios:

Tosser #1 Tosser #2

H T

H H

T T

T T

The first and second scenario really don't require tosser #2 to even toss as he's already won. The third scenario is a "push" or do over and the 4th scenario tosser #2 losses. This shows that 2 out 4 possible outcomes tosser #1 loses while only 1 out 4 possible outcomes can tosser #2 lose. The remaining 1 out of 4 just starts another round so Tosser #1 loses 2/3 of the time while Tosser #2 loses 1/3 of the time.

Shawn

What a bunch of tossers!

Sorry, the 4th scenario should be

T H

Thanks to John for pointing that out.

Shawn

The answer is 50% no matter how many times the coin is flipped, it will eventually come out that way.

Each time you the first person starts the toss, he has a 50% chance as does the other person.

Yes, but it's the 1^st to throw heads that pays for the meal, not the average fraction of heads over many tosses. The probability of the 1^st tosser being 1^st to throw heads is > 0.5, as several posters have pointed out.

Cheers...........Codey

I would guess that the first to toss would have a better chance of paying.since the odds are 50/50 of tossing a head on the first turn and the second to toss odds are 0 of tossing a head because he has not tossed the coin yet. a fairer way would be for them to toss a coin at the same time then the odds would be the same for both.

oilcan13

I actually spent 5 minutes reading all the comments very amusing,I would like to Thank everyone for making me smile.

If your name is Rosencrantz or Guildenstern and you haven't already encountered a group of perverted travelling performers and you toss first you will be buying the meal.

It makes no difference who tosses the coin, even if a third person tosses the coin, the result will be same.50%

Accidentally flip the coin into the shell of your friends lobster, then do the honourable thing and forfeit your unfair toss. Repeat if required (preferably into their wine/coffee/liqueur etc).

How about you say you need to use the restroom before the tossing begins. Then you 'forget' to tell your friend you intend to use the restroom next door, or one that's on the other side of the street.

hee he he.......sidle up to the manger first, and whisper that your companion is a dangerous felon..... you're just going around the corner to make sure back-up is in place.......

"What is the probability that you will end up paying the bill?"

0% with a slight change to the agreement - My strategy is simple.

Heads I win, tales you lose.

I never understood statistics and probability, but the answer (#1 by Fyz, also elaborated by Shawn33 and others) appeals to me intuitively **. The first person to toss obviously (?) has a fifty-fifty chance of winning/losing. If the outcome does indeed lead to a second toss, the next tosser has already reaped his 50 % benefit, and still has a further chance of a favourable result, so it must be overall in his favour. Whether that works out to 67% or 75% or something else is for the pundits to wrangle over -- but I would stay out of the sizable '50%' group present in this thread.

I wonder if the result changes in any way if at each throw the 'winning' outcome of heads or tails is declared by the person making the toss at his discretion. (Hope the Shes and Hers don't object to being left out in this post -- just can't handle the linguistic convolutions).

** I know that subjects like probability (and quantum theory, relativity, non-Euclidean geometry, and myriad others) do not lend themselves to intuition -- at least not my basic mechanical Newtonian sort. Nevertheless I often allow my intuition to overrule the logical arguments of experts, when I feel (intuitively) that their logic could be faulty but cannot prove it. Maybe it's plain and simple bias! I've never studied 'logic' either, but I suspect it is based on some axioms to start with, and perhaps there are alternative 'non-Euclidean' versions, or at least scope for semantic sleight-of-hand.

=TeeSquare=

First tosser's probability of paying = SUM (1/(2^(2n-1)) for n=1 to infinity (i.e., 1/2 + 1/8 + 1/32.......) or 2/3

Second tosser's probability of paying = SUM (1/(2^2n)) for n=1 to infinity (i.e., 1/4 + 1/16 + 1/64....) or 1/3

any how ,if i would go for dinner with a friend ,i think both of us may have same chances.

me: i would love to pay bill for my friend but it may be possible that he may not let me pay ,so the idea of tossing the coin again may increase my chances of paying the bill .

friend : same as me.

As I recall from Mr. Lariton's physics class back in the darkest ages during WWII, the odds are 50% either way on the first toss of coming up heads and 50% tails, 50% heads and 50% tails on the second toss, 50% heads and 50% tails on the third and so on.

There is no way of predicting which comes up when. It's always 50% heads, 50% tails, no matter who tosses when, nor how many times the coin is tossed.

Or did something happen to change the laws of chance since I took the course?

Ken Leigh

Imagine that something bizarre happens, and they keep tossing forever with no sign of a head; Next, consider the odds when each person has had one go. The first person has a 50% chance of getting a head. The odds of the second person being in the position of tossing a head are 25% (ie, 1/2 * 1/2, the lead player would have to have got a tail for the game to reach this stage). So, the first tossers chance of losing (ie getting a head) is twice the other persons. In this imaginary infinite game, the sequence of 1 toss each is repeated, with each pair of tosses carrying risk in the ratio of 2:1. Consecutive pairs of tosses carry the same relative risk. In other words, when the head eventually appears, the first tosser has taken 2/3rd of the risk.

That's just my fractal explanation, which will hopefully horrify maths purists.

Yes, the coin should be assumed at 50% on each throw. But as Kris indicated (almost taking the worlds right out of my mouth), it's the consequence of each throw that you need to consider. The first throw results in a 50% probability of paying immediately; but even if the first person to toss doesn't pay immediately, there is still a 50% chance the first person will throw again - with a 50% chance of paying on that occasion (total = 1/8th additional probability of paying on a second opportunity to throw). And so on, giving a 2/3 total probability of the first tosser paying. Hope that helps.

"indicated"....."almost"....

Why, Fyz, I pulled, rolled, boiled, and canned your tongue ! My numerically starved explanation was worthy of "most marvelous", even if do I say so myself.

The first throw results in a 50% probability of paying immediately

I beg to differ; It they've lost, they might 'put it on the slate', or pay by credit card.

Since we're nearly done here, this is a good, and almost topical, oldie that I chanced upon this morning ;

Three diners on finishing their meal are presented with a bill for £30, which they agree to split between them. They each gave the waiter £10, not knowing that the waiter had rechecked the bill and found that it was only £25. The waiter realised that the £5 change would not divide equally between the three men, and they were unaware of the mistake, so he returned to the table, apologised, and gave each diner £1, keeping the other £2 for himself.
The diners have all paid £9 each for their meal - totalling £27.
The waiter has kept £2 in his pocket - totalling £29.
If the men originally paid £30, where has the missing pound gone?

He he. 30-3 or 27-2? (Distraction is also the method used for most magic tricks!)

Hello Kris

The 3 diners question - there is no missing £. The men have paid £27 (original £30 minus £3 returned). £27 has been received, £25 by the restaurant, £2 by the waiter.

There is no reason (except to create an apparent paradox) to add the £2 in the waiter's pocket to the £27 paid by the diners, and then ask why it doesn't = £30. The £2 and the £27 belong on opposite sides of the equation.

That's my take on it, anyway.

Cheers.........Codey

Yep, it's a verbal deception/theft depending on how you look at it. If I were the waiter, I'd have been honest, but returned the change as a £2 coin and a £3 coin. The diners would take the hint and give me a nice tip of £3.

" ... and a £3 coin" ????? - Don't see too many of them around!

The solution gives

P_A = (1/2) + (1/2)³ + (1/2)⁵ ....

P_B = (1/2)² + (1/4)⁴ + (1/2)⁶ ....

Why bother* with the elaboration when it's obvious P_B = 1/2 P_A and P_A + P_B = 1

_{* I probably put my neck on the block with that !}

I worked it out by summing an infinite series as in published answer (as a check I also did it for Tosser B to confirm P_A + P_B = 1).

But I like shawn33's solution in #22 as it is more elegant and avoids the math. In fact I'd be inclined to give him a belated GA!

As a related problem, how about if, to make it fairer, the tossers alternate after every inconclusive round? What are the probabilities then? I've got an answer but any offers? (if everybody out there isn't tired of it by now )

Cheers......Codey

All (correct) answerimals are equal...

My personal preference is for the answer as given by Kris above - but it's hard to explain for those who don't find it "obvious", which is why my original summed the expression. Shawn33 largely overcomes the major explanation problem, but requires careful reading.

But perhaps some animswearls are more two-legged than others?