Capt Ayhab

maghshoosh

where did you disappear to?

IRANdokht

Jaleh,

I understood your reasoning and said so. I was only picking on your notation, but I see you just meant "expectation value of the random number N," so the notation makes sense too. On these pages, I usually try to avoid mathematical notations as much as possible, just b/c many aren't used to it. I'd usually suppress arguments of functions as much as possible to make it look simpler, and would rather write out a sum than use the summation notation, etc. Even though these notations are not complicated, b/c most readers may not be used to them, they'll think it's just math majors talking shop & leave the scene. And IDokht & you would be stuck here w/ me.

Maghshoosh jan,

I had defined N to be the random number of flips in the game. The idea is to use recursion on N after 2 flips. After two flips, the game ends with prob. 1/2, i.e. N=2, or it returns to the initial case after the first flip. In the latter case, N= 1+N', where N' has the same probability distribution as N (this is the critical step in the recursion). I skipped the step E(N') = E(N) for the expected values in my derivation, but I think it should be clearer now.

 By the way, you are correct that this idea would not work if the game could not continue forever. e.g. if the game stopped after at most 10 flips.

Jaleh, I knew you wouldn't pass this up

Your explanation is sound and intuitive. Only your notation is a bit absurd. What's N supposed to signify in E(N)? Surely, the formula you wrote is not valid for any finite truncation of this coin toss game. So "N" would have to be your unique notation for infinity. Another solution, not as elegant as yours, would be to recognize that the terms in explicitly writing out the infinite series for this expectation value are the same, up to a constant and a multiplicative factor, as those in p_m(infinity)=1 that we wrote in previous posts.

yeh medale meydooni ham beh khodetoon bedid keh zaneh meydoonid.

Hello again to our Field Prize winner

This thread did fall off the cliff, but in Moshaereh thread Irandokht told me that you're palying coins in this Vegas again. So here I am hoping to win some, as long as you let me enter the game for less than $3. Here's neat way of figuring it out without complex calculations, hope you like it.

Suppose N is the number of times we need to flip the coins before we get one Head and one Tail. Suppose the first flip comes H (by symmetry, the same argument follows for T). With prob. 1/2, the second flip is T and the game ends after 2 flips; with the remaining prob. 1/2 , we flip another H, so we are back to the original situation after flip 1, but we have prolonged the game by one flip. So we can write for the expected number of flips E(N):

E(N) = (1/2)*2 +1/2*(E(N)+1)

which can solved to give E(N) = 3.  

If this recursive reasoning is not convincing to you, you can also solve this the long way, i.e. by writing out P(N=k) as 1/2^(k-1)and calculating the infinite series for E(N), but that is not as much fun! 

IDokht khanom,

Have we been on this page for a couple of weeks already? I'm sure you've reserved your option of taking risks on things that matter to you, not mindless coin tossing. Thanks for your invitation to the other pages. Perhaps, if the alignment of the constellations give it a green light.

Ciao

maghshoosh aziz

akheh chand dafeh begam... I am not a gambler so the most I'd pay is the $2 :o))  your question says how much I am willing to pay and I won't pay more than a guaranteed win. Others may not be as much a chicken as I am but $10? you sound like a daring type! wanna go to Vegas?  :o)

There is an expiration date associated with the "most discussed" items which is about 2 or 3 weeks I guess, after that, even if there are stubborn souls like us hanging around, the blog will be off that list anyway.

I know you can mosha-ereh and post nice music videos, why don't you try the current most discussed blog? you'll find JalehO and the Capt'n there too.

IRANdokht

Will you play this for a few bucks?

IDokht, I was only being moderately skeptical. I don't think the sleight of hand tricks need video editing, just trying to cover all logical possibilities. There are card tricks that are determined by mathematical rules, but magic tricks are mostly artistry and so could have different approaches.

So what happened to the ranking of this thread on the "most discussed" list? It was the highest & suddenly dood shod raft hava. Did we go so far up the cliff that we fell off the other side? In that case, I doubt if this coin toss problem will bring it back.

You're offered a game in which you flip a fair coin as many times as necessary to get at least one head & one tail, but as soon as you get there, the game stops & you win the dollar equivalent of the number of tosses you've made. So if your 1st toss is heads and so is your 2nd one, but the 3rd one is tails, then the game stops & you win $3, and so on. Obviously, the minimum gain is $2 & the sky is the limit as far as your possible loot. How much are you willing to pay to enter this game? In other words, what is the average win in this game?

The only solution I know that doesn't require calculus-level math actually uses the formula for the coin toss probability problem we discussed below; the probab I called p_m(k). So the solution may be somewhat esoteric for most, but I was just wondering what people's intuition would tell them. Would $10 be a fair price for this game? More, less?

Dear Maghshoosh, I'm right here

I have mastered the disappearing act myself, but not this time :o) I was just staying quiet in the presence of you guys because I couldn't compete with your answers nor add to the discussion.

video editing? c'mon you sound like such a skeptic. Does everything have to be 2 - 2 ta 4 ta?  I love magic tricks, I always have and sometimes they're nothing short of MAGIC!

;-)

there are plenty of card tricks online in the same line as your little mind reading game, have you seen those?

IRANdokht

IDokhtar khanom,

Was wondering if you'd switched "bases" & abandoned this thread. As far as your magic trick, I dunno, could be many things. Since it's a video, it could be camera trick or editing. Even if not, there are usually several sleights of hand that could accomplish a given trick. The guy probably has all 3 cards in hand, but is skillful at manipulating them w/o being noticed. He shows all 3 cards simultaneously only at the end, so is probably showing the same card 2ice when he is supposedly "shuffling" and flipping. From your fascination, I see that magic tricks bring out your inner child.

that's an easy one maghshoosh

and I will tell you how it's done, if you tell me how THIS one works:

//www.youtube.com/watch?v=tScm-eZInBE

IRANdokht

Guess the number.

I got this in an email the other day and the sender was impressed w/ it, but I thought it was relatively transparent. What do you guys think?
//www.quizyourprofile.com/guessyournumber.swf

Jaleh,

Thanks for the accolades, ama cheshmetoon "brilliant" mibineh. I guess being "vasvas in the head," as you put it before, can be useful if it induces critical thinking.

BTW, out of curiosity, since we did the case where one set has more coins than the other, the only other case is if both sets have the same number of coins, n, i.e. k=0. But the answer should be fairly clear. The probab of getting the same number of heads after tossing both sets would be p_s(0)=[2n,n]/2^(2n) so that the probab that one set gets more heads (which would be the same as getting less heads) is p_m(0)=(1-p_s(0))/2, which is n-dependent. As n becomes infinite, you can show that p_m(0) becomes 1/2.

Now that I've seen the beauty in this "beast," I'd never contemplate "killing" it, though it may be nearing retirement.

Anony.mathmojar

Shazdeh, don't get your panties in a knot . take a look at this:

---------------------------------

by capt_ayhab on Wed Mar 04, 2009 07:22 AM PST

Here is the original coin problem from Jaleho:[ Agha
Majid has one more coin than Captain. Both throw all of their coins
simultaneously and observe the number that come up "shir".  what is the probability that agha Majid gets more "shir" than Captain? Assume all the coins are fair. ]

Solution from Jaleho herself:[ Then I said that you can try K having 3 coins and M having 1 coin, still K has two more coins. Now for these four coins we have 2*2*2*2 or 16 possible outcomes.
Out of these, there is one case that K can have less heads (TTT and H);
4 case that they have equal heads (TTT and T) and another 3 cases of (_
_ _ and H) where M's first or second or third coin is a H. So, you have
1/16 for less head, 4/16 for equal heads, and remaining 11/16 for more
heads. ]

Here is my FIRST solution:[whether having one coin or 10 coins,
probability does not change, as a result. there are SAME probability of
SHIR in both of our tosses. this is the statistical calculation.

However i think your problem had a little bit of trick that being
the one who has 2 extra coins get MORE heads[so to speak] with SAME
PROBABILITY.]

-------------------------------------

Jaleho's math problem is NOT a probability problem, it is rather a COUNT [Combinatorics]
problem. Although she has asked [what is the probability of,,,,] but in
her solution and the solutions that has come after she has COUNTED [Combinatorics] COMBINATION of the number of accurance and not the PROBALITY of occurance.

Probability is defined as: Probability, or chance, is a way of
expressing knowledge or belief that an event will occur or has
occurred.

Combinatorics on the other hand is: "Counting" the objects satisfying certain criteria (Enumerative Combinatorics), deciding when the criteria can be met, and constructing and analyzing objects meeting the criteria.

Case closed ... ;-))

--------------------------------------------------------- 

any more questions I am at your service.

Cheers

-YT

A very good Sunday morning to you all!

Maghshoosh jan, that was brilliant! Thank you very much.

The Iranian.com Math Riddle thread awards you the Field Medal:

 **********************************************************

To Maghshoosh Ali Maghshoosh for resurrecting some dead neurons in everyone's brain. With gratitude.

**********************************************************

The answer

//mathforum.org/dr.math/faq/faq.missing.dolla...

Ali Mostofi

//www.alimostofi.com

wrong capt. what he gave you

wrong capt.

what he gave you is conditional probability and not as easy as you think.
review your books unless you majored in libr=eral arts.

Thinking of our coin toss problem

Since there was at least 1 request for the proof of the solution to the generalized coin toss puzzle, here's my proof. It's relatively straightforward, although lengthy, w/ no convoluted math involved. The problem consisted of tossing 2 sets of unbiased coins, S1 & S2, S1 containing n coins and S2 containing n+k coins (k>0) and calculating the probability, call it p_m(k), that more heads will be obtained from S2 than from S1. Since we'll need it shortly, lets denote by p_s(k) the probab that equal number of heads will be obtained from S1 as from S2.

For k=1 the answer is known to be p_m(1)=1/2, independent of n. Lets do k=2 and you'll see the pattern. Since the coin tosses are independent, we can think of tossing the coins in any order. So lets think of it in 2 steps. In the 1st step throw all the coins in S1 and n+1 coins of S2, & in the 2nd step throw the one remaining coin of S2. So there are 2 ways of ending up w/ more heads from S2 at the end. In the 1st instance, you get more heads from S2 in step 1 (in which case the result of step 2 ain't gonna matter). We already know that has p_m(1)=50% chance of happening. And in the 2nd instance, you get equal number of heads from S1 & S2 in step 1, in which case there is 50% likelihood you'll get a head in step 2 so that S2 will come on top. So the overall probab of ending up w/ more heads from S2 at the end may be written as p_m(2)=p_m(1)+p_s(1)/2.

In a moment I'll prove the result for p_s(k), but the pattern of argument should be clear now. You simply replicate the above argument for k=3 and then k=4 and so on. So, for general k, b/c you've already obtained the result for k-1, you think of the problem as happening in 2 steps. In step 1 all the n coins of S1 and n+k-1 coins of S2 are tossed, and in step 2 the one remaining coin of S2 is tossed. Then the overall probab of getting more heads from S2 than from S1 at the end is p_m(k)=p_m(k-1)+p_s(k-1)/2. Since we already solved for p_m(k-1) & I'm gonna derive p_s(k-1), you'll end up w/ the formula I quoted in a previous post by incrementing the value of k from 2 to whatever.

Here's one way to think of p_s(k), the probab of getting equal number of heads from S1 as from S2. If you get i heads from S1, it means you got n-i tails from S1, which can happen the same number of ways as getting n-i heads from S1. But getting n-i heads from S1 and i heads from S2 is like getting (n-i)+i=n heads from the combination of the 2 sets of coins. So p_s(k) has to be the probability of getting n heads after tossing n+(n+k)=2n+k coins. Remember that the number of ways of getting n heads after tossing 2n+k coins is given by the binomial coefficient [2n+k,n]=(2n+k)!/n!/(n+k)! as mentioned in a previous post. The overall number of outcomes when tossing 2n+k coins is 2^(2n+k). So that p_s(k)=[2n+k,n]/2^(2n+k), which was also quoted in a previous post.

So now all the pieces for writing the formula for p_m(k) are in place, and you should be able to replicate my earlier formula. As a check, suppose k=infinity, i.e. S2 has infinitely more coins than S1. Then, no matter how many coins in S1, as long as its a finite number, you'd expect the chances of S2 coming up w/ more heads should be 100%. Think of it this way: after the coins of S1 are tossed, the coins of S2 are tossed and if S2 doesn't produce more heads, you keep tossing more & more coins for S2 (because it's allowed infinitely many) until you get enough heads to beat S1. Indeed, using a math identity that I won't quote here, you can show that the formula I produced earlier does give p_m(infinity)=1.

So what do you think? Will you buy that for a dollar?

Thanks IRANdokht jan!

You are both baazogh, kind and brilliant my dear! A true user of both hemispheres. (But wait is this even a math riddle?) 

he flunked... ;-)

It is such a kam lotfi. lol

A is reserved for ones who earn it, and there are plenty who do earn it.

As to [Kh-da] he flunked my course couple of times ;-) and it was an open book exam. That is why he made me so ugly.

-YT

Baby Cat

I could give the answer but what is the fun in that?

-YT

riddle

They paid ..... 3 x $9 = $27

Bellhop kept ............ $2

Total ................... $29

Initial pay.............. $30
Missing $ ................. $1 Where did $1 disappear?? less $3 paid to them makes the final payment $27
so let them go to sleep since they are so tired and have to wake up early. no $1 is missing.

Monda jan

Of course I'd rather it attacks the alligator!!

:-P

IRANdokht

Please focus on this one now!

Would you rather a crocodile attack you or an alligator? why?

-YT

capt

Could this be the right time to know your own answer for this enigma? ;) I can't wait to see your answer.

maghshoosh jaan, pak maghshoosham kardi ha!

First of all, who told you to take K=1 and K=2 case and generalize it to any K to begin with?!

Not me!! It was probably your genocidal tendencies (to kill the beast as you'd say) which I was so affraid of when you killed that poor soul Anonymous on, and I was so nervous about a generalized n-case of that! Now you say:

"For the original k=2 for example, it's just
p_m=1/2+[2n+1,n]/2^(2n+2)
I'm sure you'll take that as closed form."

To be fair and square that is a neat form. Just tell me how to get it, margins permitting. Thanks again.

Payam

Payam jan

read the problem again, you need to get back to original 30.

-YT

3*9=25+2

dude what are you talking about maaan?!?!
only $27 gets passed around and not $30

$27 = $25(rent) + $2( Tip)

why in the world you are adding the $2 agian

Jaleh,

Dash Kapitan dareh maharo kook mikoneh to watch us squirm. As far as the Field prize, if you have any to dispense, it's probably stolen; you should give 'em back. But as far as your condition, you're misinterpreting the rules. Initially your condition was just for the k=2 case. I just realized that the derivation easily generalizes to higher k. 2ndly the form I wrote it as is as closed-form as it's gonna get. Remember that now we're doing general k, which is why there are k terms. For the original k=2 for example, it's just
p_m=1/2+[2n+1,n]/2^(2n+2)
I'm sure you'll take that as closed form.

Your objection is as if someone wrote a formula that had an exponential in it, and you object that the exp is really an infinite sum and that's not "closed form". There's no finite sum representation of the exp, and so exp (which is just a notation for the infinite sum) is considered "closed form." (My expression was not even an infinite sum.) For general k, there's no "briefer" expression than what I wrote. There may be alternate ways of rewritting the sum, but won't be more "closed form."

"Closed form" is not a precise term & is a matter of convention. Plenty of functions, such as exp, Bessel, log, etc., only have inifinite-sum or integral representations, but are considered "closed form," b/c there's no simpler form.

Now we're talking!

Good to see the conversation leaving the bowels of the Bermuda triangle of arithmetic and ending up with the Shams ensemble. I saw them at Wilshire Ebell Theatre a few months ago... they are fantastic!

After all the (non)responses I have seen from the Captn', I think he's one of those professors who would never give a full 100% of the grade to anyone. You know, we had one of those who used to say A maleh khodast and he would never give anyone an A no matter how hard they tried and how correct they answered...

:-(

Azarin I can't believe the shir ya khat affected you to that extent, what a sensitive soul you are! why did you ever become an engineer hun?   

IRANdokht