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Five bankers are sharing 12 golden ingots.
They decide to proceed that way :
The elder one will suggest an ingots
allotment.
The rest will vote for or against it.
If the majority accepts, the sharing is ratified.
If not, the elder will be dismissed.
So, the sharing would be done between the remaining banker with the same
rules.
Knowing that they are set from left to right in a diminishing
order of their ages, how would be the allotment ?
THIS should be the UNDENIABLE Allotment!
Source
What were rules of allotment process?
The eldest should allot ingot like 9, 0, 1, 0, 2 among 5 bankers.
Let's name bankers as Banker 5, Banker 4 ...... Banker 1 according to decreasing order of their ages.
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CASE 1 :
Suppose there are only 2 bankers left then the youngest will always deny whatever elder offers so that he can take away all 12 ingots on his turn.
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CASE 2 :
If case is reduced to 3 bankers then the eldest knows that the youngest is not going to agree with him in any case. With that, the eldest will be dismissed and case reduced to CASE 1 where youngest can take away all.
So, the eldest here proposes allotment as 11, 1, 0. The Banker 2 has no option than to accept this proposal otherwise he won't get anything if case is reduced to 2 bankers as in CASE 1 above.
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CASE 3 :
With 4 bankers, the eldest would propose allotment 9, 0, 2, 1.
In the case, Banker 3 will never accept any proposal as after banker 4 is dismissed he would be getting 11 ingots as in CASE 2 above.
The Banker 2 will happily agree with the eldest as he would be getting 1 more ingot than the CASE 2.
And Banker 1 knows he will be getting nothing if the case is reduced to 3 bankers as in CASE 2. So, he too will agree with the eldest in the case.
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CASE 4 :
With 5 bankers, the eldest i.e. Banker 5 should propose allotment 9,0,1,0,2.
Obviously, any how Banker 4 is going to deny any proposal as he wants the distribution among 4 bankers where he will be getting 9 ingots as in CASE 3 above.
And if Banker 5 is eliminated and the case is reduced to CASE 3 where 4 bankers are left then Banker 3 knows he won't be getting anything. So, better he should be happily agree this proposition where he is getting at least 1 ingot.
Finally, offering Banker 5 one more extra ingot than the case where 4 bankers will be left, makes him in favor of this proposition.
Notice that the Banker 5 has to give 3 ingots at least to banker 2 to get his vote as he will be getting 2 ingots in case of 4 bankers as in CASE 3. Whereas, in the same case Banker 3 is not getting anything & would be happily agree if getting 1 ingot at least in this case.
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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!
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%.
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!
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.
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?
