Capt Ayhab

Azarin,

It's funny that you were playing shir-ya-khat w/ a shir who drew khat o neshoon for you. His question was a trick one; he wanted you to say it's 100% shir, so that he'd let his khorshid shine on you.

Shams gave a US tour a few months ago. It was memorable.

Maghshoosh, then let's write

the probability of the (n+k)-coin guy getting more Heads than the n-coin guy as p_m in a less maghshoosh form:

p_m= 1/2+1/(2^2n)SUM (j=0 to k-2){[2n+1+j),n]/2^(2+j)}

where Sum(...) is the "nakh-looking-thingie" :-)

For K=1, we quickly get our infamous 1/2 that dash Kapitan seems to be obsessed with, even when he agreed on "case closed ;-))"

For K=2, then n=0,1,2 gives the 3/4,11/16 and 21/32 that it is supposed to give. We will punish Dash Kapitan by having him actually write a chart of heads and tails, enumerate, and come up with the correct probability for k=2 and n=3 and compare with the formula! Good, needed, and fair punishment for him.

As for your Field prize, I'll still keep it here for your future. Hint: the reason is in the "nakh-looking" thingie I wrote p_m. I wouldn't want to sound like Dash Kapitan now, but remember the condition for your prize was: 

"I'll still keep your Field Prize handy here in case you found a closed form for that probability, or in case you found a book with a larger margin." Meany, I know.

Maghshoosh Jaan,

I don't know about any killed beast, but you've defintely killed me! Last night, I fell asleep after going through your links! Unfortunately, I had this nightmare of a lion tossing coins with me... and I woke up in horror when I got more heads and the lion was asking me "you have 15 seconds to give me the formula!"

Azarin

BTW, I found the drum link with Shams ensemble great!

50% of us is wrong.

Captain, you've guided this ship deep into the bowels of the Bermuda triangle of arithmetic, where results such as 1=2 become legit and every probability is 50%. Of course, the binomial coefficients are calculating the combinatorics, but those are then used to calculate probabilities per your recipe of favorable_outcomes/total_outcomes. Evidence? The combinatorics are divided by 2^n type numbers, and the probabilities are less than 1.

You still haven't answered me as to how in tossing 2 unbiased independent coins, where the possible outcomes are HH, HT, TH, TT, can the HH outcome have 50% probability?

maghshoosh

Nope, not yet

Probability is still the same, as in 50%. what you are calculating is COMBINATION of having H or T. 

Proof? You are using [!] factorial. 

-YT

Just to complete the k-differential coin toss ...

In my previous post we wrote the probability for the guy w/ n+k coins to get more heads than the guy w/ just n coins. Call that probability p_m. The probability for both of these guys getting the same number of heads (which, of course, will have to be less than n+1) is given by:
p_s=[2n+k,n]/2^(2n+k)
where, as explained in the last post, [n,k] is the binomial coefficient, so that [2n+k,n]=(2n+k)!/n!/(n+k)!
(The expression for p_s might give you a hint as to how p_m was derived.) Then the probability that the guy w/ n+k coins gets less heads than the guy w/ n coins, is 1-p_s-p_m.

Have we killed the beast yet?

????

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

-YT

P/S You guys all rock.............

k-differential coin toss problem & solution, or ...

... bezanim beh simeh akhar, or lets drive off the cliff w/ Thelma & Louise.

2 persons are given 2 sets of identical & unbiased coins, one is given n coins and the other n+k coins (k>0), all of which they toss. What is the probability that the one with more coins will end up w/ more heads than the other one?

First some notation: Recall the factorial & its properties, ala //en.wikipedia.org/wiki/Factorial and that 0!=1. Let [n,k] denote the binomial coefficient for the 2 non-negative integers n & k ala //en.wikipedia.org/wiki/Binomial_coefficient so that
[n,k]=n!/k!/(n-k)! if k<(n+1), and [n,k]=0 if k>n. The binomial coefficient is relevant b/c [n,k] is the number of ways you can get k heads if you toss n coins. Also, let a^b mean "a to the power of b".

Then the desired probability is given by 1/2 if k=1, and by
[drum role: //www.youtube.com/watch/v/Jws_tM-20do ]
1/2+2^(-2n){[2n+1,n]/(2^2)+[2n+2,n]/(2^3)+...+[2n+k-1,n]/(2^k)}
for k>1.

Anybody care to check?

maghshooh, where you?

waitin'!

It's not over till Hayedeh sings; math-symbol help needed

Azarin khanom, Lotf darid, but we all have our confusions. I'm not sure what made you maghshoosh here, but it could be that some of these puzzles, including the ones I posed, actually have ambiguities and may have different interpretations, but we pretend otherwise. How does the fact that you don't understand everything set you apart from the rest of us?

Jaleh, not so fast. I think I have the general solution for the k-coin deficit problem. But I'd like to know how to input math symbols on these pages. I noticed Foad put superscript on numbers. Does anyone know how to input math symbols here?

I second that

Jaleh jan, you just said it all! I think you guys are "totally awesome" and I had a lot of fun hanging out with you and yes I was also shocked to find out that seeing a number in another base could actually be that sexy ;-)

It was nice of everyone to stop by and participate, it helped not feeling too geeky for a change...

Maghshoosh better not stay away or else!

Thanks Ostad Capt'n you made me reach some brain cells that I had forgotten I had, and then maghshoosh made me lose track of them again. :o)

IRANdokht

I loved this blog too!

It was a pleasure to get to know some other sides of tasteful friends that I have known in other blogs. It was also a great pleasure to get to know our hero, Maghshoosh Aziz. His creating eghteshash was most welcomed. Thanks to him, we'll never look at any number without thinking how sexy they might look in another base!

The games were all fun, and I kept on thinking about wonders following Flying Solo's dainty question. Thanks to all of you.

Of course I hope that even if the blog gets slow, it might continue, or just like Moshaereh we might bring things like it here and there. So, Maghshoosh jan, if this blog disappeared, you don't please.

I feel so maghshoosh now..:-)

Dear Maghshoosh,

I'm still confused, almost maghshoosh (but not in  a good way like you!)... But it's ok! A long time ago, I accepted that I cannot understand everything and my kids keep reminding me that "Mom, you don't understand!"...(sigh, sigh)

Thanks to all of you, my intelligent IC friends, for this on-going fun blog! Now, I have to focus on writing another depressing page..:-)

Azarin

cap'n

ssshhhh... majid is busy...lol.  poor baby has been working his butt off moving boxes.   but i'm heading to LB after work with a cerce.  we'll have a picnic on our (hard)wood dining room floor (no rugs or big furniture yet...:-(

dang...  i actually knew that one.  too slow...lol

If I've told you once, I've told you a 1000 times:

888+88+8+8+8
Id, Can we find an answer in another base?!

Id aziz,

The answer I had in mind was Jaleh's. They're all muddied, so that each kid sees his 2 bros as such. So at first you may think that the dad's statement, which is true, doesn't seem to add any extra information, but each son doesn't know a priori that the other ones are also seeing 2 muddied faces & so dad's statement helps to resolve it.

?

Where is Agha Majid when we all need him?????

now lets see if we can rivive it by some cpr

Q: How can you add eight 8's to get the number 1,000? (only use addition)

-YT

maghshoosh jan

Is that the final verdict? so we don't have anyone with a muddy face and the Dad ain't lying either? 

Now I feel all maghshoosh-ed all over again LOL

IRANdokht

It's not dead if it lives in our minds.

Azarin & Captain, If the dad were lying, so that all the sons were clean, and if the sons believed their dad, then each one would immediately assume that he himself is the muddied one & would leave. So the fact that they all paused is supposed to rule this out. If the sons don't trust the father, then they can't make a definitive decision. So Jaleh's solution is the most consistent.

Nafase Akhereshe

Seems like this blog has taken its last breath. Khoda biamorzatesh.

It was fun guys, Thanks

-YT

Good morning Sobh! and muddy guy

Little Lion has done a more general version of your puzzle couple of pages ago. You'd enjoy going back a couple of pages and check that one also.

And maghshoosh jan, why do you think Irandokht and I were happy to see that Anonymous guy, ha? Because we needed "somebody to put you back into your place," my entire household loooooves queen :-) Ok, ready? Assuming that all of these guys are maghshoosh:

We can not have 2 clean guys, becasue then the muddy guy who saw two clean guys would have immediately gotten up to go wash, and we wouldn't have any pause in the beginning.

We also can not have 1 clean guy, becasue if we did, call him Mr. clean Z; and X and Y muddy. Then Z would argue: If I were clean and X saw Y muddy and Z clean, then X would have argued that Y who saw Z clean, should have seen X as muddy, otherwise he'd have gotten up and go wash immediately. Since Y didn't rush to wash, so he should have seen me (X) muddy, and in that case X would have rushed to wash immediately. So, Z ain't clean either.

Now maghshoosh jan, I got all maghshoosh just writing that! But, between us, this puzzles are more tautology than real puzzles ;-)

muddy face and ASHOORA

Neither one had mud on their face, each saw the others with clean face assumed himself being the muddy one, then all with out communication washed up their faces.

dorosteh?

-YT

IRANdokht

[ cuz roman numerals ain't decimal your hint was not accurate after all]

How so? Roman numerals ARE decimal base only difference is instead of numbers they use letters. That is the only difference.

You are so aziyating ;-)

-YT

Is it possible that the father has lied?

If any of the brothers has mud on his face, the other two would see it and so they would assume that father is talking to that brother, so they wouldn't move.

But if any brother sees that the two other brothers have clean faces then he would assume that he is the one with muddy face, so he would go to wash his face!

What if the father has lied to push the boys go to the washroom to clean their faces (and their hands too!!) Because the only case that all 3 of them act the exact same way is if all 3 of them have clean faces.

Azarin

"You got mud on your face, you big disgrace"

3 brothers who have been working in their back yard go inside. Their dad looks at them & says, "at least one of you has mud on his face." The brothers look at each other. There's a pause for a while when no one moves. Then all 3 simultaneously go to the wash room to clean their faces. There was no communication amongst them, besides what the dad uttered, they cannot see their own faces and there aren't any mirrors or the like. How did they realize that they all had mud on their faces?

Assume they're all perfectly logical, and think and act at the same speed.

صبح عزیز

اینها همه نشان از یک عمر خرخونی و دوستانی خرخونتر از خودمه. اکثر این معما ها و ریاضیات مشکوک و مغشوش را قبلأ شنیدم

در هر صورت مرسی از لطفتون

To Irandokht

Did you know the answer or did you resolve it by yourself? In any case good job.

1=-1

I'll assume familiarity w/ the exponential and the logarithm, which are inverse functions of each other. Also, you should recall the imaginary number "i", which satisfies i^2=-1 (i squared is -1). Let "p" designate the infamous number "pi"=3.1415... As you would then know,
exp(-2ip)=exp(2ip)
Take logarithm of both sides and you get -2ip=2ip. But i and p are not zero, so it's perfectly legal to eliminate them from both sides to get 1=-1.

resistance is futile sobh

When you have one guard who tells the truth and one who lies, you ask one of them: "which door would the other one say leads to freedom" and whichever door he says you pick the other door, because one telling the truth and one telling a lie, he will always point you to the wrong door.

 IRANdokht

PS: Jaleh jan shayeh-ast.. bavar nakon  ;-)

ایراندخت، خیلی‌ زرنگی ها

خودمونیم!!