Flipping The Unusual Coins

You have three coins. One always comes up heads, one always comes up tails, and one is just a regular coin (has equal change of heads or tails). If you pick one of the coins randomly and flip it twice and get heads twice, what is the chance of flipping heads again?

Chances of flipping head again are - .......% Click to know!

Chance of Flipping Head Again

What was the problem?

For a coin to always show head on flip we assume both it's sides are heads and the coin which is showing tail always we assume both of it's sides are tails.

There is no way that you have selected tail only coin since there are 2 heads in first 2 flips.

So it could be either head only coin say D coin or regular fair coin say F.

Let H1 and H2 be the sides of head coin and H, T are side of fair coin.

If it's head only coin D, then possible scenarios on 2 flips are -

DH1 DH1
DH1 DH2
DH2 DH1
DH2 DH2

And if it's fair coin F then possible scenarios on 2 flips are -

FH FH
FH FT
FT FH
FT FT

There are total five combinations (all 4 of head coin + first one of fair coin) where there are 2 consecutive heads on 2 flips.

So, the chances that you have picked a head coin is (4/5) and that you picked fair coin is (1/5).

For head coin, the probability of getting head again is 1 and that for fair coin is (1/2).

Since you holding either head coin or fair coin,

Probability (Head on third flip) = 
Probability (You picked Head coin) x Probability (Head on head coin) + Probability (You picked fair coin) x Probability (Head on fair coin) 


Probability (Head on third flip) = (4/5) x 1 + (1/5) x (1/2)

Probability (Head on third flip) = 9/10.

Hence, the chance of flipping head again on third flip is 90%.

Find Working Batteries for Flashlight

You have a flashlight that takes 2 working batteries. You have 8 batteries but only 4 of them work.

What is the fewest number of pairs you need to test to guarantee you can get the flashlight on?

You need to test at least THESE pairs! 

Selection of Working Batteries for Flashlight

What was the task given?

Divide batteries into 3 groups - 2 of them having 3 batteries each and 1 with 2 batteries. 

Then the working pair of batteries has to be in 1 of these groups and now it's easier test to each group. That is 4 working pairs might be distribute as - 
(2,1,1) or (1,2,1) or (1,1,2).

If A, B and C are name of these groups then possible combinations for testing group A are - 

A1-A2, A1-A3, A2-A3

Similarly, for B group testing pairs are - 

B1-B2, B1-B3, B2-B3

And finally, if we don't find any working pair in above testings then the C1-C2 pair of group C has to be working pair.

You may find in the working pair in testing those 6 pairs from group A or B or can conclude that C1-C2 is the working pair. 

The Watermelon Paradox

There is a 100 pound watermelon laying out in the sun. 99 percent of the watermelon's weight is water. After laying out for a few hours 98 percent of the watermelon's weight is water.

How much water evaporated?

The calculation is TRICKY one! 

Tricky Watermelon Water Weight Calculation

What is the question?

Initially, since 99% of watermelon is water, the weight of water must be 99 pounds and weight of other stuffs should be 1 pound.

After evaporation, 98% of watermelon is water indicates that the there is 2% of other stuff present at the point of time. 

That is, if we assume the weight of other stuff unchanged during evaporation, now 1 pound itself is equal to the 2% of total weight of watermelon.

In other words, the process of evaporation increased percent weight of other stuffs in total weight of watermelon from 1% to 2%.

If W is the total weight of watermelon, then weight of other stuffs (which is 1 pound)

W x (2/100) = 1 pound

W = 50 Pounds.

Out of these 50 pound, 1 pound (2% of total weight) is other stuffs and 49 pounds is water (98% of total weight).

That means, out of 99 pounds of water, 99 - 49 = 50 pounds of water is evaporated in the process. 

गोष्ट विषमासूर नावाच्या राक्षसाची !

विषमासूर नावाच्या राक्षसाच्या समोर 1000 माणसे एका रांगेत उभी आहेत. त्या रांगेमधील विषम क्रमांकावर [ उदा. 1,3,5,7,9 ] उभ्या असलेल्या सर्व लोकांना विषमासूर खावून टाकतो. 

त्या नंतर उरलेल्या माणसांची रांग शिल्लक राहते. आता त्या रांगेतील विषम क्रमांकाच्या माणसांना विषमासूर खावून टाकतो. 

असे करत करत रांग लहान होत जाते आणि शेवटी एक माणूस शिल्लक राहतो. तर तो शिल्लक राहिलेला माणूस मूळ 1000 लोकांच्या रांगेत कितव्या क्रमांकावर उभा असेल ?

या क्रमांकावरील व्यक्ती शेवटी उरेल. पहा कोणता आहे तो क्रमांक! 

विषमासुराच्या गोष्टीतील शेवटचा माणूस !

काय बरं होती हि गोष्ट ?

समजा,  त्या रांगेत फक्त १०च माणसे होती तर प्रत्येक फेरीनंतर वाचलेली माणसे या क्रमांकांवर असतील -

(१)  २  ४  ६  ८  १०
(२)  ४  ८
(३)  ८

तर ८ व्या क्रमांकावरील व्यक्ती शेवटी उरेल.

आता त्या रांगेत ५० माणसे होती असे समजूयात.

(१)  २  ४  ६  ८  १०  १२  १४  १६  १८  २०  २२  २४  २६  २८  ३०  ३२  ३४  ३६  ३८  ४०  ४२  ४४  ४६  ४८  ५०
(२)  ४  ८  १२  १६  २०  २४  २८  ३२  ३६  ४०  ४४  ४८
(३)  ८  १६  २४  ३२  ४०  ४८
(४)  १६  ३२  ४८
(५)  ३२

म्हणजेच, ३२ क्रमांकावरील व्यक्ती शेवटी उरेल.

ही रांग आता १०० माणसांची होती असे समजूयात.

(१)  २  ४  ६  ८  १०  १२  १४  १६  १८  २०  २२  २४  २६  २८  ३०  ३२  ३४  ३६  ३८  ४०  ४२  ४४  ४६  ४८  ५०
      ५२  ५४  ५६  ५८  ६०  ६२  ६४  ६६  ६८  ७०  ७२  ७४  ७६  ७८  ८०  ८२  ८४  ८६  ८८  ९०  ९२  ९४  ९६
      ९८  १००

(२)  ४  ८  १२  १६  २०  २४  २८  ३२  ३६  ४०  ४४  ४८  ५२  ५६  ६०  ६४  ६८  ७२  ७६  ८०  ८४  ८८  ९२  ९६  १००

(३)  ८  १६  २४  ३२  ४०  ४८  ५६  ६४  ७२  ८०  ८८  ९६

(४)  १६  ३२  ४८  ६४  ८०  ९६

(५)  ३२  ६४  ९६

(६)  ६४

या वेळी ६४ क्रमांकाचा व्यक्ती शेवटी राहील.

वरील तिन्ही उदाहरणे बारकाईने पाहिल्यास असे लक्षात येते कि शेवटी उडणाऱ्या व्यक्तीचा क्रमांक हा त्या एकूण क्रमांकांपैकी सर्वात मोठा २ चा घात (Biggest Power of 2) आहे. उदा. १ ते १० साठी ८, १ ते ५० साठी ३२ आणि १ ते १०० साठी ६४.

याचे कारण, अंतिमपूर्व फेरीमध्ये या क्रमांकाच्या निम्मा क्रमांकाचे प्रथम स्थानावर असणे होय जेणेकरून हा क्रमांक सम स्थानांवर उभा असेल. उदा. १ ते १० साठी ४,८ तसेच १ ते ५० साठी १६,३२ आणि १ ते १०० साठी ३२,६४.

याच तर्कानुसार आपण निष्कर्ष काढू शकतो कि, १००० माणसे रांगेत उभे असतील तर ५१२ क्रमांकाचा व्यक्ती शेवटी उरेल कारण ५१२ हाच २ चा सर्वात मोठा घात (Biggest Power of 2) आहे.

आणि समजा १०२५ माणसे रांगेत असती तर १०२४ क्रमांकाचा व्यक्ती शेवटी उरला असता.

अंतिम फेरीमध्ये मात्र हा व्यक्ती विषम स्थानावर (१) असेल आणि त्याचे काय करतो हे गोष्टींमध्ये नमूद केले नाही.

Plan The Best Chance of Winning!

You are playing a game of dodge ball with two other people, John and Tom. You're standing in a triangle and you all take turns throwing at one of the others of your choosing until there is only one person remaining. You have a 30 percent chance of hitting someone you aim at, John has a 50 percent chance, and Tom a 100 percent change (he never misses). If you hit somebody they are out and no longer get a turn.

If the order of throwing is you, John, then Tom; what should you do to have the best chance of winning?

This should be you plan to increase chances of winning! 

Planning The Best Chance of Winning!

What was the game?

You should miss the first shot for the purpose.

Remember, about one of your 3 shots (30% accuracy), John's 1 out of 2 shots (50% accuracy) and Tom's every shot is on target.

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

CASE 1 : If you target Tom and hit him then John has to hit you. Even if he fails to target you in first attempt he will be successful in his second attempt. 

And since, your first shot was on target your next 2 has to be off the target one of which will give John a second chance.


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

CASE 2 : If you target John then Tom will certainly hit you to be winner of the game.

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

CASE 3 : Better miss the first shot and then 1 of next 2 shots will be on the target.

Now, John has to target Tom otherwise assuming John as stronger player, Tom will eliminate him immediately. 

    CASE 3.1 : If John hits Tom and eliminates him then it's your turn

                     now and John's next attempt has to be off the target. 

                     So, even if you fail in first try after John's

                      unsuccessful try you can surely hit John in second try.

    CASE 3.2 : And if John misses Tom then Tom will throw John 

                     out of game in his first attempt. Now, it's your turn 

                     and you can target Tom with 50% accuracy.

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

This is the best plan to get a chance of winning this game. 

Angles on Christmas Tree in Puzzle!

Four angels sat on the Christmas tree amidst other ornaments. Two had blue halos and two - yellow. However, none of them could see above his head.

Angel A sat on the top branch and could see the angels B and C, who sat below him. Angel B, could see angel C who sat on the lower branch. And angel D stood at the base of the tree obscured from view by a thicket of branches, so no one could see him and he could not see anyone either.

Which one of them could be the first to guess the color of his halo and speak it out loud for all other angels to hear?

THESE angels have maximum chance of correct guess!


Similar Puzzle 

Logical Angels on Christmas Tree

How their logical skills are challenged?

Case 1 :

The angel A observes that the aureoles of B and C are of the same color. Then, A can be sure that the color of own aureoles must be other than that of B and C.

So, A will be the first to guess the correct color in case he observes 2 same colored aureoles B and C.

Case 2 :

The angle D notices that the color of aureoles of B and C are different. Now, A has to remain silent. The silence of A suggests B that the color of aureole of B must be different than the C.

So, in that case, B will be the first one to guess correct color of own aureole.

However, C has no chance to guess the color of own aureole unless, A or B reveals color of own aureole. And D has absolutely no chance of correct guessing the color of own aureole.

Will the Sheep Survive?

Hundred tigers and one sheep are put on a magic island that only has grass. Tigers can live on grass, but they want to eat sheep. If a Tiger bites the Sheep then it will become a sheep itself. If 2 tigers attack a sheep, only the first tiger to bite converts into a sheep. Tigers don’t mind being a sheep, but they have a risk of getting eaten by another tiger.

All tigers are intelligent and want to survive. 

Will the sheep survive?

Survival chances of the sheep are - Click Here! 

Survival Chances of the Sheep

Why sheep is in danger?

First let's see what happens when there are different number of tigers present on 
the island. Remember we are talking about survival of the sheep that is initially present n the island and not sheep converted from tiger.

1. Suppose there are only 1 tiger and 1 sheep on the island. Then, the tiger will eat 
    the sheep and won't have fear of being eaten up after transformation into sheep 
    as there is not tiger left. 

    The sheep will not survive.

2. Let's suppose there are 2 tigers and 1 sheep present. Each intelligent tiger can think - 

    ----------------------------------------------------------------------------------------------------------
   "If I eat a sheep then I will be converted into sheep and other tiger would eat me
    as it would result into '1 tiger and 1 sheep' scenario above in (1). 

    So, I should not take risk"
    ---------------------------------------------------------------------------------------------------------- 

    The sheep will be survived.


3. Now suppose there are 3 tigers and 1 sheep are on the island. Each tiger would think-  
     ----------------------------------------------------------------------------------------------------------
    "If I target the sheep and get converted into sheep itself then on the island there
     would be 2 tigers and 1 sheep as above case (2). 

     As per that, none of other 2 tiger would dare to attack me and I would be 
     survived as a sheep in the end. 

     So better I should attack the sheep and anyhow I will be survived in the end as a
     sheep"
     ----------------------------------------------------------------------------------------------------------
      The sheep will not survive.

4. Finally, suppose there are 4 tigers and 1 sheep. Now, each tiger can put logic like -       ----------------------------------------------------------------------------------------------------------
    "If I attack the only sheep and get myself converted to sheep then this case 
     will be reduced to '3 tigers and 1 sheep' as in case (3). 

     In that case, the sheep has 0 chance of survival in the end. 

     That means, my life will be in danger as in above case (3), if I attack 
     this sheep toget converted into sheep. Better, I shouldn't attack"
     ----------------------------------------------------------------------------------------------------------
     The sheep will be survived.

Conclusion : 

If observed carefully, it can be concluded that the sheep will be survived when there are EVEN number of tigers (Case 2 and Case 4) are present. And obviously, will be in danger when there are ODD number of tigers present on the island.

In the given situation, there are 100 tigers on the island which is EVEN number. That means, as per above conclusion the only sheep on the island will be survived.

 

The Last Bean in the Pot

A pot contains 75 white beans and 150 black ones. Next to the pot is a large pile of black beans. 

A somewhat demented cook removes the beans from the pot, one at a time, according to the following strange rule: 

He removes two beans from the pot at random. If at least one of the beans is black, he places it on the bean-pile and drops the other bean, no matter what color, back in the pot. If both beans are white, on the other hand, he discards both of them and removes one black bean from the pile and drops it in the pot.
 
At each turn of this procedure, the pot has one less bean in it. Eventually, just one bean is left in the pot. What color is it?

And the color of last bean is...... 

The Color of Last Bean in the Pot

Little story behind the title!

There are 75 WHITE beans in the pot i.e. they are odd in numbers. Since, they are always taken out in pair, in the end there will be 1 WHITE bean left out.

At some point, when there are 3 bean are left in the pot then there has to be 2 BLACK and 1 WHITE beans in the pot. They can't be 1 BLACK and 2 WHITE beans as for that 73 WHITE beans need to be taken out which is not possible since WHITE are always taken in pair.

So if cook pick 2 BLACK beans (or BLACK & WHITE) at this point then anyhow BLACK and WHITE will be left in the pot. Now, when he pick this pair of BLACK and WHITE then he puts BLACK on the pile and drop WHITE back to the pot as per his rule.

Eventually, WHITE bean will be left in the pot.

MPSC मध्ये विचारला गेलेला प्रश्न

सोडवा गणित

तुम्ही एका यात्रेत आहत,

त्या यात्रेत तुम्हाला काही प्राणी खरेदी करायचे आहेत.त्या प्राण्यांची किमंत खालील प्रमाणे....

*१० रुपयाला - १ हत्ती

* १ रुपयाला - १ घोडा

* १ रुपयाला - ८ उंट

आणि तुमच्याकडे १०० रुपयेच आहेत.प्राण्याची संख्या १०० च आली पाहिजे.

वरील सगळे प्राणी घेणे बंधनकारक आहे.

कसे येतील.???

प्रश्न कठीण वाटतोय का ? उत्तरासाठी येथे क्लिक / टॅप करा.

👉  वाचा -  गोष्ट विषमासूर नावाच्या राक्षसाची !  

MPSC प्रश्नाचे उत्तर : यात्रेतील प्राणी खरेदी

जाणून घ्या काय होता प्रश्न ?

यात्रेमध्ये कोणता प्राणी किती किंमती मध्ये मिळतो ते पाहुयात.

* १० रुपयाला - १ हत्ती

* १ रुपयाला - १ घोडा

* १ रुपयाला - ८ उंट

म्हणजेच १ उंट १/८ रुपयांना मिळतो.

समजा आपण ' क्ष ' हत्ती, ' य ' घोडे आणि ' ज्ञ ' उंट घेतले तर क्ष हत्तींची किंमत होईल १०क्ष, य घोड्यांची किंमत होईल १य आणि ज्ञ उंट (१/८)ज्ञ रुपयांना पडतील. 

आपल्याजवळ १०० रुपये आहेत म्हणजे - 

१०क्ष + य + (१/८)ज्ञ = १००   ......... (१)

आणि आपल्याला एकूण १०० प्राणी घ्यायचे आहेत म्हणजेच -

क्ष + य + ज्ञ = १००

य = १०० - क्ष - ज्ञ          ....... (२)

समीकरण (२) हे (१) मध्ये टाकल्यानंतर,

१०क्ष + (१०० - क्ष - ज्ञ) + (१/८)ज्ञ = १००

९क्ष + १०० - ज्ञ + (१/८)ज्ञ = १००

९क्ष - (७/८)ज्ञ = ०

९क्ष = (७/८)ज्ञ 

क्ष/ज्ञ = ७/७२. 

याचाच अर्थ क्ष = ७ आणि ज्ञ = ७२ व म्हणूनच य = १०० - क्ष - ज्ञ = १०० - ७ - ७२ = २१ असू शकतो.  

म्हणजेच आपण ७० रुपयांचे ७ हत्ती, २१ रुपयांचे २१ घोडे आणि ९ रुपयांचे ७२ उंट खरेदी करायला हवेत जेणेकरून १०० रुपयांमध्ये १०० प्राण्यांची खरेदी पूर्ण होईल.  

एकूण खर्च = ७० + २१ + ९ = १००. 

एकूण प्राणी = ७ + २१ + ७२ = १००.  

One More Alphamatic Problem?

In the following  puzzles, replace the same characters by the same numerals

so that the mathematical operations are correct.
               

Note - Each letter represents a unique digit and vice-versa.

ABCB - DEFC = GAFB

     :          +      -
  DH  x     AB =    IEI
---------------------------
 GGE + DEBB = DHDG

Here is the SOLUTION 

One More Alphamatic Solution!

Look at the problem first!

Rewriting the problem once again,

ABCB - DEFC = GAFB
   :         +       -
  DH x   AB    =    IEI
-------------------------
 GGE + DEBB = DHDG

We have 6 equations from above -

(1) A B C B - D E F C = G A F B  

(2) G G E + D E B B = D H D G  

(3) G A F B - I E I = D H D G 

(4) D E F C +  A B = D E B B 

(5) A B C B : D H  = G G E  

(6) D H x A B = I E I  

Steps :

1. From (1), we have B - C = B. That's possible only when C = 0.

2. If C = 0 then in (1), for tens' place subtraction i.e. C - F = F the carry need to 

    be taken from B. And that subtraction looks like 10 - F = F. Obviously, F = 5.

3. From (3), we see D in result seems to be carry and carry never exceeds 1 

    even if those numbers are 999 + 9999. So, D = 1. 

4. From (1), since C = 0, at hundreds' place (B - 1) - E = A and from (4),

    we have F + A = B (since first 2 digit of first numberremain same in result

    indicating no carry forwarded in addition of FC + AB = BB.

    So placing F = B - A in (B - 1) - E = A gives, F = E + 1. Since, F = 5, then E = 4. 

5. In (3), G at the thousands' place converted to D without actually subtraction 

    of digit from IEI. Since, G and D are different numbers some carry must have been

    taken from G.

    As D = 1 then G = 2.

6. From (1), A - D = G and D = 1 and G = 2 then A = 3 since if carry had been taken 

    from A then A = 4 which is impossible as we already have E = 4. 

7. From (2), E + B = G i.e. 4 + B = 2 only possible with B = 8.

8. With that, in (2), carry forwarded to  G + B = D making it 

    1 + G + B = 1 + 2 + 8 = 11 = 1D  i.e. carry 1 forwarded to G + E = H making it 

    1 + G + E = H = 1 + 2 + 4 = 7.

    Therefore, H = 7 and no carry forwarded as digit D in second number remains

    unchanged in result. 

9. Now (6) looks like - 17 x 38 = 646 = IEI = I4I. Hence, I = 6. 

To sum up,

A = 3, B = 8, C = 0, D = 1, E = 2, F = 5, G = 3, H = 7 and I = 6. 

Eventually, all above 6 equations after replacing digits in place of letters look - 

1. 3808 - 1450 = 2358  ✅

2. 224 + 1488 = 1712   ✅

3. 2358 - 646 = 1712   ✅

4. 1450 + 38 = 1488    ✅

5. 3808 : 17 = 224      ✅

6. 17 x 38 = 646         ✅  

Rewriting in the given format,

3808 - 1450 = 2358
      :         +       -
    17 x     38 =  646
-----------------------------
  224 + 1488 = 1712

Ultimate Test of 3 Logic Masters

Try this. The Grand Master takes a set of 8 stamps, 4 red and 4 green, known to the logicians, and loosely affixes two to the forehead of each logician so that each logician can see all the other stamps except those 2 in the Grand Master's pocket and the two on her own forehead. He asks them in turn if they know the colors of their own stamps:

A: "No."

B: "No."

C: "No."

A: "No."

B: "Yes."

What color stamps does B have? 

'THIS' could be his color combination! 

Earlier logicians had been part of "Spot On The Forehead!" and  "Spot on the Forehead" Sequel Contest

The Wisest Logic Master!

What was the challenge in front of him?

Let's denote red by R and green by G. Then, each can have combination of RR, RG or GG.

So, there are total 27 combinations are possible.

1.  RR RR GG
2.  RR GG RR
3.  GG RR RR

4.  GG GG RR
5.  RR GG GG
6.  GG RR GG

7.  RR RG GG
8.  GG RG RR
9.  RG RR GG
10.RG GG RR
11. RR GG RG
12. GG RR RG

13. RR RG RG
14. GG RG RG
15. RG RR RG
16. RG GG RG
17. RG RG RR
18. RG RG GG

19. RR RR RG
20. GG GG RG
21. RG RR RR
22. RG GG GG
23. RR RG RR
24. GG RG GG

25. RR RR RR
26 .GG GG GG

27. RG RG RG.

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1. Now, obviously (19) to (26) are invalid combinations as those have more than 4 red or green stamps.

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2. In first round, everybody said 'NO' thereby eliminating (1) to (6) combinations. That's because, for example, if C had seen all red (or all green) then he would have known color of own stamps as GG (or RR). Similarly, A and B must not have seen all red or all green.

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3. For (9 - RG RR GG), A would have responded correctly at second round as NO of B had eliminated GG and NO of C had eliminated RR for A in first round. Similarly, (10) is eliminated.

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4. For (11 -  RR GG RG ), C would would have responded correctly immediately after NO of A had eliminated GG and NO of B had eliminated RR for him in first round. With similar logic, (12) also get eliminated.

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5. Remember, B has guessed color of own stamps only in second round of questioning.

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6. For B in (13), his logic would be I can't have RR (total R>4) but GG [ No.(11) - RR GG RG] and RG can be possible. But (11) is eliminated by C's response in first round. That leaves, (13) in contention.

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7. Similarly, if it was (14 - GG RG RG) combination, then B's thought would be - I can't have GG (total G>4) but can have RR as in (12) - GG RR RG which is already eliminated by C's NO response at the end of first round. Hence, I must have RG. That means (14) also remains in contention.

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8. On similar note, (17), (18) remains in contention after A's NO at the start of second round.

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9. If it was (15), then A would have been responded with RG when C's NO in first round eliminates RR (as proved in 2 above) and GG (as proved in 4 above) both. Similarly, (16) is also eliminated after C's NO in first round as above.

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11. For (27), B's logic would be -

"If I had RR then A must had seen RR-RG and had logic - 

"Can't have RR (total R>4); if had GG then C would have answered with RG after I and B said NO in first round itself. Hence, I must tell RG in second round."

Similarly, A's response at the start of second round eliminates GG for me.

Hence, I must have RG combination."

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10. So only possible combinations left where only B can deduce color of own stamps are -

7.  RR RG GG
8.  GG RG RR

13. RR RG RG
14. GG RG RG

17. RG RG RR
18. RG RG GG

27. RG RG RG.

If observed carefully all above, we can conclude that B must have RG color combination of stamps after observing A's and C's stamps as above to correctly answer in the second round.


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