Capt Ayhab

Jaleho ;-)

Valiant effort but no dear

I'll check back later

-YT

Dear Captain,

First, I agree with this part of your last post:

"Case closed ... ;-)) "

Now on Half of 12 = 7 , I am going to use a captainian logic:

(Here, I am really interfering with Foad Khosmood's expertise!)

1. get the "/" sign from your keyboard and insert it between 1 and 2 in 12. You'll get 1/2. Keep that 1/2 safely on the left of what follows!

2. Take the said "/" out to get 12 back. So far, you have 1/2 and12 neatly.

3. insert the said "/" right in the middle heart of number 12, (slightly tilted) so it falls on the "/"-looking part of number 2. That is, cut the number 2 sharply with the slach to separate the curvy part of 2 on top! Now you are left with an inverted 7 on the right hand side of number 2!

4. Turn that upside-down 7 and stick it to the left of 1/2 of 12 is 7 ;-)

Ms. Azarin

Not too shabby for a NEW KID in the BLOG hey?????????

But really, I appreciate your kind words, and all the comments every one has made. Many consider Math being a boring subject, but people like it because it challenges the mind, specially when it comes to Iranians.

-YT

Now

Now how about some elbow grease on my problem:

Half of 12 = 7

Decimal[base 10] is assumed.

-YT

Lets see.........

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 ... ;-))

-YT

Dear Irandokht,

Actually, I always knew that Iranians love poetry more than sex (The proof: if you read most of the classics you can see that the whole story is lines and lines of poems and then the sex part is just one sentence: "And they lived happily ever after!")  ...but loving mosha-ereh more than sex? Oh my, oh my,...it is a new discovery for me :-)

Azarin jan

Going by that same logic, we also proved that Iranians love mosha-ereh 10 times more than sex :o))

IRANdokht

Isn't it surprising?

Dear Captain,

I just noticed that your great blog is the number one in the list of the most discussed (khiaanat kardam is behind you!) ...and I couldn't believe my eyes! It means that we Iranians love Mathematics more than Sex!

Wow...Congratulations dear Captain! You have done the impossible!

Azarin

maghshoosh jan

do comprehensive and vigorous math and infinite series of powers of binomial coefficients harmonize with ghahr o eshveh? come on hala ghahr nakon LOL

ta to bashi incognito javab nadi :o) especially when the "esghteshashat" give you away anyway...

so what's the symbol manipulation there? 

IRANdokht

PS: no disappointment at all 

Sorry to disappoint you ...

Thanks for the enthusiasm ID khanom: "I also thought we have a new fella to complete me for sure and this time with a smile too! YAY!" What am I to make of that? If I'd known it'd dash your hopes so much, I'd let the "anonymous" post stay that way. Can't believe the one who completed me says, "with a smile too"! Have I been too akhmoo otherwise? Don't you realize that comprehensive & rigorous math proofs don't mix w/ fake Persian pleasantries? Proofs by induction and infinite series of powers of binomial coefficients don't exactly harmonize w/ ghorboon sadagheh.

Jaleh, what's w/ a sex blog that's "on top," as you put it? Maybe I should start posting proofs there to induce eghteshashe ravani. A friend of mine who was looking at it, told me of a solution to the "half of 12", and if that's what Capt had in mind, it's not really mathematical, but symbol manipulation. BTW, I sent the site an email to correct the name on that post which gave you guys false hope. Will they do it?

LOL Jaleho

It sure is good to be on top! and I also thought we have a new fella to complete me for sure and this time with a smile too!  YAY! but the eghteshash gave him away anyway :o)

So did Capt'n Ayhab forget about our half a 12 in decimal or is he currently not allowed near electronics?

We have another blog to bump up Jaleh and I am looking for more poems to gear up. 

See Q?  we do arts and poetry too ;-)

IRANdokht

I should have guessed!

Anonymous on's post was so por-eghteshash, I shoulda guessed! But, I was hoping we might have found another fella to take care of you! Ok, if I found my other conjectures (hehe) should I post them here or in the new blog? You know Captain has produced another Math Puzzles II blog. But, I love the fact that this one finally is getting more comments than the sex blog one!

How rude of me not to intorduce myself ...

Hey Jaleh, I just noticed that I forgot to identify myself in the proof of your 3rd conjecture (what were the other 2?). I got drawn into checking the expressions, so I forgot to check the "name" part.

Thanks Anonymous on,

O wow, you generalized "Jaleho's third conjecture," I'll set aside a Field Prize for you, IFFF you write me the proof :-)

That's right Q, these guys are impressive. The good thing is that this "eghteshash" seems to be a bit contagious as Irandokht proved it by her half of 12!

O Captain My Captain

When I said "assume the coins are fair," I basically said what you are saying about the probability of getting a head (or tail) in a single toss is simply 1/2.

And, when I said that with one extra coin, K gets either more heads or more tails , but not both, and by symmetry probability of more heads is thus 1/2, I explicitly used the property that those events are mutually exclusive.

So, we did use your stuff! Thanks.

Iranian M&Ms

I'm truely in awe of all these hard puzzles being solved by Iranian friends left and right!

I have always said the Iranian M&Ms are "Math" and "Medicine". These seem to be the fields of choice for all the young kids, in no small part due to their parents "old world" input.

I see nothing wrong with this obsession and I wish all of you good luck. I was into math puzzles myself in my day!

But we need more lawyers, historians, artists, philosophers and sociologists too. Don't forget the other half!

Not that the Captain would buy any of this ...

Since we're not the sort of criminals who leave loose ends behind, I thought I'll just post the full solution to Jaleh's extension of her own puzzle, which was left unsolved. This is my quick calculation & I haven't rechecked it. Problem: Suppose person A is given n coins and person B n+2 coins. Each toss all their independent, identical and unbiased coins. What are the odds that B will get more heads than A?

I won't post the proof, since its long & involves math calculations. Just to recall some notation ... For a non-negative integer n, factorial of n is given by
n!=nx(n-1)x(n-2)x...x1, where we take 0!=1 (it's not 0), as you can see at //en.wikipedia.org/wiki/Factorial
Also, n^m means "n to the power of m."

The full answer I believe is (1/8)(5+(2n)!/(n!)^2/2^(2n))
We can check this w/ the simplest cases Jaleh calculated. Suppose one person is given no coins and the other 2, then n=0 and the formula gives (1/8)(5+1)=3/4. Suppose the players are given 1 & 3 coins respectively, then n=1 and we get
(1/8)(5+2/4)=11/16. So, double check.

So ID, even though this may not complete you completely, at least it adds a smile beneath the eyes!

I wasn't maghshoosh until I heard Capt.

Help!!! What is Capt. talking about?? Simple scenario of 2 coin tosses: HH, TT, HT & TH. How can HH be 50% of this? Is each of these outcomes 50% of total, so that we have 200% total probab??? Is this the Bermuda triangle of arithmetic?

maghshoosh

I have to tell you this LOL

[heads is (1/2)^n (where "^" means "to the power of"...]

Probability is a fraction[less than 1 greater than zero] so when to take a fraction to a power, you are really REDUCING the number and not increasing it.

I just thought you shoul know that LOL

.5^2 = .25 = 1/4

funny how numbers deceive us na?

-YT

Heads up!

What do you mean "in case of coin toss, you are not allowed to multiply the outcomes"?? Of course, you are. This doesn't have to do w/ Bayesian stats. Because the coin tosses are independent and identical, then the probability of getting 2 heads in tossing 2 coins is (1/2)^2: since they're independent you just multiply the probab of each. How can the probab of getting 2 heads in tossing 2 coins be 50%? Follow your own suggestion of enumerating possible and favorable outcomes: HH, TT, HT, TH. How can HH be 50% of this?? The probab of getting a head for each coin is 50%, but not for getting heads for all.

ID

LOL

You are mixing up number of coins with probability concept. Read my comments particularly about 2 sets of playing cards.

And trust me, I deal with this thought pattern a lot. Don't take me wrong please, but grasping CONCEPT of probability, although an easy discipline, is rather difficult at the beginning. Because people tend to mix the COUNT with the PROBABILITY.

Good starting point: //en.wikipedia.org/wiki/Probability

-YT

Again

No matter how many coins there are PROBABILITY remains the same for both players which is 50%.

player with 2 extra coin get 2 additional head, with SAME probability of getting shirs = 50%

this might help you guys. Probability[P] ranges between 0 and 1. Yani equal or greater than ZERO - equal  or less than ONE. So compute your theories see what numbers you get.

I shall grade you all in the AM, or later on this evening if [khanoom e khuneh] allows me to the computer room ;-)

RULE: in case of coin toss, you are not allowed to multiply the outcomes.

-YT

LOL Captn

What's more:

50% of 10 or 50% of 1010?

I'd say the probability of the one with 1010 coins having more shirs is about 100% don't you think?   ;-)

chera aziat mikoni akheh? LOL

IRANdokht

Coin toss IS a simple

Coin toss IS a simple probability by definition, for which I have given all the definitions.

By definition SIMPLE probability can NOT be added or multiplied,
otherwise will be conditional probability[Bayes' Theorem], which is NOT
applied to simple probability[coin toss, rolling die, spin wheel,
drawing cards, etc]

Conditional probability[as what you guys are proposing for this
problem] is used for complex forecasting and business models and is NOT
applicable to coin toss, which is a SIMPLE probability by definition.

Even if Agha majid had 1000 coins still probability would be 50% . Yani same chance for both.

-YT

capt jan

with 1001  coins you get more shirs than you get with one coin.

it's the probability of getting more shirs so if I toss 10 coins, I might get 5 shirs, but you would get 50 shirs with 100 coins so you get 45 more shirs than I do even though we both got shir 50% of the time.

IRANdokht

Captain,

I don't understand this statement, "IF both the players land all heads at the first toss, which is .5 probability ..." If a player has n coins, the probab of him landing all heads is (1/2)^n (where "^" means "to the power of"), so that its different if one has extra coins. Also, see ID's response as to what the original puzzle was. What are you saying should be the correct solutions to the 1-coin-deficit and the 2-coin-deficit problems as posed by Jaleh?

IRANdokht

coin toss IS a simple probability, for which I have given all the definitions.

Even if Agha majid had 1000 coins still probability would be 50% . Yani same chance for both.

-YT

capt'n jan

The question was not the simple probability. It was the probability of more heads. 

Here's the original problem posed by 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.

aslan nevermind,

I read your comment again and I think you and Jaleho are saying the same thing (as maghshoosh said already) and I am more confused now  thanksalot!  :o)

IRANdokht

maghshoosh khan jan

Probability does not change no matter how many coins there are. let me give you another example:

Assume you have 1 set of[52] playing cards. Probability of drawing lets say king of heart is 1/52.

Now lets assume you have 2 sets of[104] playing cards. Probability of drawing same king of heart is 2/104 = 1/52.

As you see probability does not change since it is independent from other cards. same is applied to coin toss, and rolling a die. In Jaleho's question  both players have the identical Probability of having head or tails. The player with more coins GETS more head or tail BUT with SAME probability.

Here is how IF both the players land all heads at the first toss, which is .5 probability, player  with 2 extra coins gets 2 extra head[so to speak lol] with SAME .5 probability.

In short, since it is a simple probability problem[mutually exclusive and collectively exhaustive] no matter how many coins are involved, probability does not change.

It is called SIMPLE probability because  we conduct a ONE stage or ONE object experiment. We choose an event[head of tails], then the probability of that event is found by counting the number of times the event is true (favorable) and dividing by the total number of  possible  and equally likely outcomes. Keep in mind please that probability is calculated by observing the occurrence of event when it is preformed infinite number of times.

-YT

oops Maghshoosh jan

you're right about that 7...  

I shouldn't try to solve problems when I am at work and completely distracted. I am embarrassed now ;-)

as for the Captn' saying it's in decimal anyway, I don't think I have the brain power to solve that, so lets see how YOU complete ME

:o)) 

IRANdokht