UK 19:27 Friday, 23 October 2026
Breaking news and expert analysis on fast-moving AIM and LSE listed shares
Disclosure: I have no positions in any stocks mentioned, and no plans to initiate any positions within the next 72 hours. I wrote this article myself, and it expresses my own opinions. I am not receiving compensation for it (other than from ShareProphets). I have no business relationship with any company whose stock is mentioned in this article.
Sorry for the delay with this podcast. First up I am still not smoking, second I am delayed as one of my cats seems very ill indeed. Then bloody Andrew Monk of VSA has my mind in torment with a puzzle I cannot do and another where as a chemist - and a boring boatie from Oriel - he just cannot see that there are 2 correct answers. He should have done PPE. Finally a reader demands we show balance, i.e. parrot his unbalanced view. No can do compadre. Puzzle answers in comments section below please.
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Comments
Hanging is Far Too Good for Them
Following the pattern gives 96 but it looks mathematically invalid.
Mr Mensa
Add one to the second number and multiply…
1 + 4 -> 1 × 5 = 5
2 + 5 -> 2 × 6 = 12
3 + 6 -> 3 × 7 = 21
8 + 11 -> 8 × 12 = 96
Daniel Victor
It’s not 50-50 because if you happen to have picked from the bag with one white and one black marble,there’s only a 50% chance that you would have picked a white marble – but if you picked from the bag with two white marbles there’s a 100% chance.So there’s a two in three chance that the marble came from the bag with two whites,and a one in three that it came from the bag with one white and one black.So the chance of the other marble in the bag being black is only one in three.
Mr Mensa
Add one to the second number and multiply…
1 + 4 -> 1 × 5 = 5
2 + 5 -> 2 × 6 = 12
3 + 6 -> 3 × 7 = 21
8 + 11 -> 8 × 12 = 96
Oh… and your other one is Multiply and add the first number..
1 + 4 -> (1 × 4) + 1 = 5
2 + 5 -> (2 × 5) + 2 = 12
3 + 6 -> (3 × 6) + 3 = 21
8 + 11 -> (8 × 11) + 8 = 96
But.. they are mathematically equivalent in as much as adding one to the second number and multiplying by the first is the same as multiplying the numbers and then adding the first.
genuser063
For the marbles question, I think it goes like this…
1 in 3 chance of choosing the bag with 2 white marbles.
Taking one of the white marbles leaves 2 white and 3 black in total, so 2 in 5 chance there for another white marble.
Therefore my answer is (1/3) * (2/5) = 0.13 or 13%
Mr Mensa
Add one to the second number and multiply…
1 + 4 -> 1 × 5 = 5
2 + 5 -> 2 × 6 = 12
3 + 6 -> 3 × 7 = 21
8 + 11 -> 8 × 12 = 96
Oh… and your other one is Multiply and add the first number..
1 + 4 -> (1 × 4) + 1 = 5
2 + 5 -> (2 × 5) + 2 = 12
3 + 6 -> (3 × 6) + 3 = 21
8 + 11 -> (8 × 11) + 8 = 96
But.. they are mathematically equivalent in as much as adding one to the second number and multiplying by the first is the same as multiplying the numbers and then adding the first.
Oh.. and, tentatively, the bag thing is misleading or obfuscation. Just because you have a choice of bags with some combination of marbles in them in the first place and pick one does not affect the final result. You have no idea which bag you picked. You will still be finally selecting one marble from the total number of available marbles, Three White, Three Black. Having selected your White one you are left with Two White and Three Black… at this point my NCR/NPR falls over but the answer might be 2/3 or 33%
Simple Moving Avarice
The easiest way to solve the second one is to write down all the possible choices you have. There are only six marbles, so after you draw one you have these cases:
B -> B
B -> B
W -> W
W -> W
B -> W
W -> B
We know that you in fact drew a W marble so we can eliminate the B first cases. So there are two cases where you will have a W marble left and only one where you will have a B marble, giving (2/3,1/3) counter-intuitive though it is.
Daniel Victor
Sorry – I should have added – best wishes to Tara.I also admit that I was unable to solve the first problem.
Mr Mensa
More importantly. Best wishes from Cade to Tara.
bwakem
My 2 answers are 40 and 96.
DeepPurple
Classic Bayes Theorem of Conditional Probability.
The chance that you drew from the all white bag is 2/3, the chances that you drew from the all black bag is 0, the chances that you drew from the 1W1B bag is 1/3. Therefore that the next marble is white is =2/3*1 + 0 + 1/3*0 = 2/3
Think of the 3 bags having a 100 white, 100 black , and 1white/99 black. Then it becomes obvious why it is not 50%. The chances are you drew the white from bag 1 in the first place, Therefore higher prob of next one being white too.
Brian
Marble problem: I see a couple of people above have got correctly explained – it’s largely to do with the choice of the initial bag.
Of passing interest; this is a common problem in the card play part of Bridge, and is known as “the principle of restricted choice”. Outside bridge, it is probably better known as the Monty Hall problem https://en.wikipedia.org/wiki/Monty_Hall_problem.
DeepPurple
Classic Bayes Theorem and conditional probability
The chances that you drew the first white from Bag 1 is 2/3, from Bag 2 is 0, from Bag 3 is 1/3. Therefore the chances of drawing the next ball white is = 2/3*1 + 0 + 1/3*0 = 2/3
To see why it is not 50% then think of the bags having 100whites, 100blacks, and 1white/99blacks. Chances are you drew from Bag 1 so chances are higher that next will be white.
Nigel Somerville
There are 3 W marbles in all. In 2 cases out of the three that you can pull a W marble the second marble is W, since 2 of the 3 cases of pulling a W come from the same bag. So 2/3.
Alan Mills
40
1+4 +5
2+5(+previous total, 5) = 12
3+6 (+ prevous total 12) = 21
8+11 (plus21) = 40
manic seagull
I think the answer to your propability puzzle is 2/3rds. If bag A has 2 white and bag C 1 white & 1 black, the probability that the white ball taken is from bag A is twice the probability that it is from bag C, so probability that it was taken from bag A is 2/3rds. If it was taken from bag A the probability that the remaining ball is white is 1. If it was taken from bag C it is 0. Therefore the probability that the remaining ball is the same as the probability of the bag it was taken from. Intuitively the answer is 1/2, but this is wrong. We all make subconscious intuitive use of probabilities in making investment decisions. This is an interesting example of how intuitive reasoning can lead to erroneous answers.
One answer to the other problem is 40. I don’t know if that is the same answer you got, but here is my reasoning. In the generic formula for each line x + y = a there is something missing – what is it. We could express this as x + y + z = a (where z is the missing item). In the first line z is 0 (i.e. nothing missing – no previous line). In the second line z is 5 which is the answer to the previous line. In the third line z is 12 – the answer to line 2. Ergo the answer is 40. I can’t see another answer but I am sure there is one.
Do I win a prize?
Drunken Sailor
Andrew Monk is wrong. The probability of the next marble being white is 50%. SMA the way you work it out assumes there is a difference between marbles of the same colour, which there is not.
The overall probability from the start of picking the bag with 2 white marbles in is 1/3. But you have already picked 1 white marble and the question is will the next one be white? Thus there is zero probability that you have picked the bag with 2 black marbles and therefore you have either picked the bag with 2 white or 1 white and 1 black. Both have equal probability. The remaining marble in the bag that used to contain 2 whites is white (100% probability). The remaining marble in the bag that used to contain 1 white and 1 black is black (100% probability).
A company with a mine in Bongo Bongo land has a 25% chance its customers won’t pay and a 25% chance the corrupt government will just revoke its licence and steal the mine for itself. Baring other mining risks the company has a 50% chance of making money. The corrupt government is removed and replaced by colonial rule – the democratic choice of the people who remember how much better things were in the colonial days. The risk of the licence being revoked is now zero, but the customers are Chinese and still 25% likely not to pay. Because one possibility of not making money has now gone, the probably that the company will make money is now 75% – still a shit investment. If listed on the AIM Casino the Nomad will still be signing off on ramptastic RNSs saying how much dirt with a little bit of something valuable in is being dug out of the ground, being put through processing to concentrate the valuable bit and then shipping it off to China. It will only be when the accounts have been properly audited that you find out that trade receivables are by far the biggest part of current assets and the company has been issuing confetti not to explore more places in Bongo Bongo land, but just to cover what has not been paid by the Chinese, who when you do proper due diligence on them turn out to have been crooks all along who have pulled the same stunt dozens of times and therefore the probability they would not pay was always 99.9%. And of course to keep the lights on and to pay boards salaries plus crony capitalist fees – coke and hookers all round, where are the shareholders’ yachts?
Drunken Sailor
On the above. BBMs will not bother looking at the balance sheet and cash flow statements plus the detailed notes that go with them and will only read the ramptastic waffle that precedes them regarding EBITDA and adjusted EBITDA and great opportunities opening up in Bongo Bongo land now the government has changed. They will ignore the detailed analysis done by “the pizza boy” as you only need to look at what happened at Rivington to know that he does not know what he is talking about. Yet more confetti will then be issued.
Assuming Andrew Monk is a smart guy, despite having gone to the wrong university and being a boring boatie (the boring bit is redundant as they are all boring even at the right university). Was that puzzle an exercise in showing how easy it is to convince someone to come to the wrong answer by giving erroneous hints?
bwakem
Drunken Sailor you are wrong about the probability.
There are 3 options once you have picked the first white ball.
Option 1: You have the bag containing the black ball.
Option 2: You have white ball number 1 from the double white bag, in which case the other ball is white.
Option 3: You have white ball number 2 from the double white bag, in which case the other ball is white.
So 2/3.
I wrote a quick computer program to test it 1 million times and the results were:
Black ball first: 499224
Of the remainder the seconds ball was:
White: 332762
Black: 168014
wildrides
All the marbles in my house are down the back of the sofa ……… I wish I had kept my marbles in little bags .
Poor old Tara ……. sounds like liver probs what with all that weight loss .
David Russell
I am sorry about Tara and hope she gets well soon Tom. I lost my cat Molly last week after ill health and its killed me. We had her for 19 years. She would always be with me when I was at home and on the laptop on Share prophets Bearcast to day she is in my heart. Hope everything goes well with Tara I understand more than most what its like when they am ill.
bwakem
#!/usr/bin/perl
use strict;
my %bags = ( ‘bagA’ => [qw( W W )], ‘bagB’ => [qw( B B )], ‘bagC’ => [qw( W B )],
);
my ($white,$black,$black_first,$white_first) = (0,0,0,0);
foreach (1..1000000)
{ my $rand_bag = $bags{(keys %bags)[rand keys %bags]};
my $first_ball_index = int(rand(2));
my $first_ball = $rand_bag->[$first_ball_index];
if ($first_ball eq ‘B’) { $black_first++; next; } else { $white_first++; }
if ($rand_bag->[!$first_ball_index] eq ‘W’) { $white++; } else { $black++; }
}
print “Black first: $black_first\n”;
print “White first: $white_first\n”;
print “Second ball of those that were white:-\n”;
print “White: $white\n”;
print “Black: $black\n”;
$ perl marbles
Black first: 500173
White first: 499827
Second ball of those that were white:-
White: 333619
Black: 166208
Daniel Victor
The fact that you have picked a white marble means you are twice as likely to have picked from the bag with two white as the bag with one white and one black.That’s the ‘restricted choice’ bit.
wildrides
Its now turned into a “whos got the biggest swinging dick substitute compo” with MENSA written all the way through it like a stick of rock .
Drunken Sailor
The third answer to the 1 + 4 = 5 puzzle is: Who cares we are trading on a PE of 1 and very soon there is going to be so much cash that share buy backs will start, remortgage the house and get in at these crazy prices.
BWAKEM there is no difference between the 2 white balls, it does not matter which is picked first and which is picked second. Re program your computer with the right logic and you will see it is 50/50.
Mr Mensa
I’ll try the Marbles again. Three Bags. One with two black Marbles. One with Two White Marbles. One with a Black and a White marble. You pick a Bag and pull a White Marble. You now know you either have the Bag with two White Marbles or the Bag with a Black and White Marble but not which one. Given you have already pulled a White Marble then there is either a White Marble left in the Bag or a Black Marble left in the bag so despite the assertion by Andrew Monk that the answer is not 50:50 the answer is in fact 50:50.
Drunken Sailor
Mr MENSA,
There may still be a joke answer involving always getting black balled in the end.
However I think the psychology is interesting. If the question had been asked without the “hint” that it was not 50/50 everyone would have come up with 50/50 (well maybe not Wildes). Because people believed the hint they then came up with a whole variety of clever ways to arrive at the wrong answer. I was particularly impressed by BWAKEM’s computer program.
Is it any wonder so many people still believe in the global warming myth.
Hoggness
Add one to the second number and multiply…
1 + 4 -> 1 × 5 = 5
2 + 5 -> 2 × 6 = 12
3 + 6 -> 3 × 7 = 21
8 + 11 -> 8 × 12 = 96
or multiply 1st & 2nd number and add 1 …
1 * 4 + 1 = 5
2 * 5 + 2 = 12
3 * 6 + 3 = 21
8 * 11 + 8 = 96
Two answers but both give the same result
The 3 bag two marble answer is 66% or 2/3 in old money
phil
The answer is not 50%-50% nor is it 2/3rds
You are missing a key point. I think you are treating each pull from the bag as 2 separate transactions, in effect each time you reach into the bag you are playing a different game. You cannot calculate the prob in such a way, as you have introduced something into the game that you have all missed.
By reducing the number of balls into the game you have introduced entropy acting on a finite number of balls, ie the group of balls, as such you are causing the finite group to suffer the laws of the diminishing committee in that each pull out the bag is influenced by the outcome of the previous pull out of the bag by way of a linked common entropy.
The reason why some of you are choosing 1/2 and some 2/3 is because if you calculate the prob on the 1st pull, the answer is 1/2 however the 2nd pull from the bag is 2/3 if and only if the game doesn’t suffer entropy by reducing the number of balls.
Therefore prob’ acts not of the balls in bags but on the number of balls in the game, finite balls will each suffer 1/6th entropy regardless of what colour or number of balls in the game at any one time. Entropy works of the whole game from outside of the game, by way of predetermine .
If it were 50/50 when you remove a white ball leaving 2 white balls and 3 black your chance of pulling another white ball is not 50/50, it is 2/5th.
If your answer of 2/3rd is right then you can pull 2 white out of anyone of 3 bags on 2 pulls with 66% chance of success yet only 33% of pulling a single black ball, clearly that is wrong.
Had you put the 1st white ball back into the bag then 2/3rds would have been correct.
The link by common entropy is real, tangible and the outcome is the prob’s of 2 white balls is 20%,
.
Mr Mensa
Mr NEWT
As you can see I failed to spot the other possible ‘number’ answer so Mr Monk is on his back legs.. In respect of the marbles it comes down to how much information you have. I can’t decipher ‘Perl’, especially without comments, but it looks like Mr BWAKEM has fallen into the hole I originally fell into by assuming the Black Black Bag still played any part in the sums once it had already been excluded by the appearance of a White Ball. Looks like Mr Monk is 0 for 2. Perhaps he is more suited to Share Bullshitting than Chemistry.
DeepPurple
Phil
The question was “what is the probability that the next ball will be white from the same bag”.
You are clearly answering a different question. 20% is wrong. If that was the question posed by Andrew , then it is not the question that was posed by Tom.
To be honest it’s a great puzzler.
BRITTA
You’re all getting confused because this question deliberately plays on your assumptions. It’s a classic case that’s used a lot to illustrate these things.
(and yes, the answer is 2/3rd not 50%).
The first “spoof the market” trick is to have that bag that has only black balls in it. That makes no difference to the outcome, because you are told that a condition of the test is that the first ball is white. Ergo, you can tell from the start that it wasn’t drawn from Bag C (all black). This is used to test whether students can discount irrelevant information.
The second part is to test directionality and this is done by the trick of using two white balls. This also lays a little trap for the unwary.
The easiest way to think of this is to be a good philosopher and “re-frame the question” so as to avoid unnecessary mental assumptions. Here is how I would phrase it.
- You have two bags, A and B.
- one bag has 2W balls, the other 1B 1W
- the test is to work out which bag you have chosen, based on knowing only the colour of the first ball drawn. This re-framing makes it easier to follow.
There are only four potential outcomes when you draw that first ball, each with 25% probability, because there are 4 balls (3W 1B).
a. You choose Bag A and withdraw one of the white balls
b. You choose Bag A and withdraw the other white ball
c. You choose Bag B and withdraw the white ball
d. You choose Bag B and withdraw the black ball
Those are you only four options for the withdrawing of that first ball. There are no other options. And each will occur with a 25% frequency.
Now we are told that the first ball withdraw is white. We can therefore discount option d.
That leaves us with three options (a,b,c) each with equal probabilities.
In two thirds of the cases (a+b out of a+b+c) you have withdrawn a ball from Bag A, and the other ball is therefore white. Thus the probability of the next ball being white when the first one is white is 2/3rds, because when you choose Bag A the second ball is always white. (Hence the utility of re-framing the question)
The error people make is not realising that a. and b. are not the same, but are in fact two different options and therefore doubly weighted.
Simple Moving Avarice
Or for the first puzzle another answer is 131.
Compute x + y as (x3 + 37x – 2y)/6
1 + 4 => (1 + 37 – 8)/6 = 30/6 = 5
2 + 5 => (8 + 74 – 10)/6 = 72/6 = 12
3 + 6 => (27 + 111 – 12)/6 = 126/6 = 21
So 8 + 11 => (512 + 296 – 22)/6 = 786/6 = 131
In fact, since apparently we can invent any function we like such that it meets the first three equations, I think you can find an infinite number of different solutions just by using different polynomials. Polynomials of degree 2 all seem to give the answer 96 e.g. x2+4x or x-3y+y2
phil
Deeppurple
We differ
If the prob were 2/3 you have it that you will choose the right bag with 2 white balls in it 2 times in every 3 goes, therefore your chances of pulling a black ball is no more than 1 in every 3 goes. What you are saying is the prob of choosing another white ball in the right bag increases against a falling number of white balls. What you are also saying is if on your 1st choose you choose a black ball, you have the bag with 2 black balls.
I have it that the prob of finding the right bag with 2 white balls in it is 1 in every 5 goes and in 4 times out of 5 you will choose a bag that contains (at least) one black ball.
Prob acts of the number of balls in the game, the bags are irrelevant, the outcome of the choose is linked by entropy
Therefore prob 1st choose = ½, 2nd choose with 2 white 3 black = 2/5, linked by entropy ½ x 2/5 = 1/5 = 20%
Lies, dam lies and statistics, so I agree these riddles are devils, once you have it your way, you rarely change it. As long as you think you have the right call and the call is right for you, that is all that matters.
bwakem
Mr Mensa: “but it looks like Mr BWAKEM has fallen into the hole I originally fell into by assuming the Black Black Bag still played any part in the sums”
No my program drops the BB bag completely. I will explain the logic:
#!/usr/bin/perl
use strict;
# Assign the balls to the bags and the bags to a bags hash (a data structure)
my %bags = ( ‘bagA’ => [qw( W W )], ‘bagB’ => [qw( B B )], ‘bagC’ => [qw( W B )],
);
# Pre-declare these variables with a value of 0
my ($white,$black,$black_first,$white_first) = (0,0,0,0);
# Do the folowing 1 million times
foreach (1..1000000)
{ # Pick a random bag from the hash of bags
my $rand_bag = $bags{(keys %bags)[rand keys %bags]};
# Pick either the 0th or 1st ball in the bag
my $first_ball_index = int(rand(2));
# Assign the colour of the ball to the $first_ball variable
my $first_ball = $rand_bag->[$first_ball_index];
if ($first_ball eq ‘B’) { # If the first ball is black that’s the end of the loop, we start from the beginning again
$black_first++; next; } else { # The first ball is while
$white_first++; }
# At this point the first ball can only have been white # The ! sign negates the index so 0 becomes 1 and 1 becomes 0, so we are testing the ‘other’ ball from the same bag
if ($rand_bag->[!$first_ball_index] eq ‘W’) { $white++; } else { $black++; }
}
# This prints out the results
print “Black first: $black_first\n”;
print “White first: $white_first\n”;
print “Second ball of those that were white:-\n”;
print “White: $white\n”;
print “Black: $black\n”;
The logic is flawless IMO and the results are 2/3 white as I posted before.
Cynical Bear
@ Brian – Thank you.
Ever since I read this yesterday I knew that it reminded me of something but I couldn’t drag it from the edges of my memory but it was the bridge principle of restricted choice that I read about when I was studying the game last year.
Now I can sleep soundly tonight.
Drunken Sailor
Incorrect. You need to follow the logic not the total possibilities. There are only 2 bags you could have chosen. 1 ball has been removed and it is white. If you chose bag A with 2 white balls then there is a white ball left. If you chose bag B then there is a black ball left. There is no probability they you chose bag C with 2 black balls thus it is a straight A or B random choice so the probability is 50/50. Try it for yourself with only 2 bags one with 2 white and one with one white and one black that have already had a white ball removed ie one bag has a white ball left anf the other has a black ball left. There is no difference between the white balls thus the Bag A with one white ball and Bag A with the other white ball is an irrelevance.
Where Monk has gone wrong is equating it to Betrand’s box puzzle:
https://en.m.wikipedia.org/wiki/Bertrand%27s_box_paradox#sthash.lbJ7rXKo.dpuf
There are 2 separate drawers in this puzzle whereas in the bag there are not 2 separate drawers and it makes no difference which ball you pick.
Nigel Somerville
Drunken….but, but, but…
the bag with 1W 1B only qualifies for the test half of the times it is chosen whereas the bag with 2W qualifies every time it is chosen. So for every time the WB bag is selected and it comes up W, the 2W bag will have qualified twice.
The issue here is that there is a condition attached to getting to the question. In fact you are being asked what the probability is of having selected the WW bag, as opposed to the WB bag, when you drew a W, and looked at that way the WW bag comes up twice as frequently as the WB bag since half the time the WB will produce a B and therefore not qualify for the test.
Finally, if the correct answer here were indeed 50% then the total probabilities don’t add up to 1:
Prob of pulling a W to start with is 50%. The implication then of a second W being a 50% chance means that WW is 25% probability overall. The same maths works for BB.
Would you then argue that out of 3 bags there is a 50% probability of picking the WB bag? Surely that has to be 1/3 and thus one of the 50% figures has to be wrong.
Right…back under my stone!
phil
Bwakem
People are using the wrong starting point
I used an algorithmic approach not one reasoned. The game starts with 3 bags, 2 balls in each, a combination of 2 colours, the outcome of any choice is unknown..
This is the way I playing the game.
Start the game by choosing the 1st ball
You choose a black ball, No good, game over. Put the ball back, restart the game, choose another ball
A white, good now we can move on, choose the 2nd ball, black ball. no good, game over, put them all back, restart the game.. The new 1st chouce is white, your 2nd choice is white. You win the game
This is what you are doing
Choose a ball. Black ball, no good, put it back, have another go,
Choose another ball, a white ball, good now the game can start.
How many balls are in the bags when you started. 5, how many balls in the bags when the game started as defined by TW, when the game started. 6.
You are only counting the probability of the 2nd choice, which is indeed 40%
The game is played with choices, not one.
Prob 1st choice is white is 1/2
Leaving 5 balls, 2 white, 3 black, you have a 40% chance the next ball will be white
The prob of 2nd choice is 1/2 × 2/5 = 1/5 = 20%
The game as defined start with 6 balls in 3 bags, whereby the outcome of any choice is unknown. The game doesn’t start with the choosng of the 1st is white. The game starts with no choosing.
We would all like to play the lottery when we know the outcome of the 1st ball before the draw started, or a bet whereby we know the 1st winner of a 3 race combination bet.
It doesn’t mean this midway interpretations don’t have purpose. It is unlikley that a bookie will lay off that combination bet at the start but he sure would if he knew the 1st 2 results of a 3 horse combination bet had won
By using an algorithmic approach it is perfectly possible to calculate the prob before the start of then game, and indeed track the prob in real time as the game is played.
The algorithmic I produced can predict prob of any combination of n balls in m no of bags, be it 6 balls in 3 bags, or a million balls in 1000 bags,
By tying my model of prob using the common factor of entropy, my algorithmic approach can cope with n no of balls, in m no of bags, with p no of different colours, with all balls of different sizes and weights, and in a game played by q no of players, where none of the players will ever meet or contact each others, whether the outcome of previous choices were known or not would make no difference.
Now if anyone wants to try an reason all that lot, go ahead. This is why prob theory came into being, we can all play the game when we already know some of the outcome.
That ethic that this game suffers entropy, that each ball is discarded after being chosen, is the reason why the designer choose to include it, if you fail to include it, the prob that you calculate will always throw a false positive. Entropy acts of the game from outside of the game, not from within .
It is because TW is not including entropy (neither is anyone else) is why he keeps coming back to the false positive the solution is a 50/50
It is a key point for shareholders in the present state of the markets whereby the entropy of deflation is acting on the markets from outside of the markets, from the economy results in all balance sheet erosion of contributory value/assets and further leveraged erosion as a consequence an inverse proportional inflation of fixed cost liability, that all companies whole wide are struggling for the want of margin
For companies with no contributory value, just fixed costs liability their only method of support is cash injection in one form or another, and if they cannot raise the finance they will be become insolvent
GreatBritain plc is a high costs producer with colossal fixed costs and minimal contributory values, GreatBritian plc needs to reduce those fixed costs and pretty quickly.
It is impossible to reduce those fixed costs from inside the EU, if you fail to vote OKI Britain will end up exactly being another Greece, not if but when.
To give the prob of when Britain ends up like Greece I would calculate the prob factor of this happening to Britain by simply using the same algorithm to play this simple game
Daniel Victor
N.B. For anyone reading a lot of this comments thread and wondering why the same answers are repeated,it is because the comments only become visible after a while [perfectly sensible when readers are asked to solve a problem].So to begin with,we can’t see each others’ answers.
DeepPurple
Phil
We may differ but you are wrong. The key phrase you are missing is “GIVEN that a white ball has been chosen”. This makes it a conditional probability. The starting point is when a white ball comes out a bag at random.
Look up conditional probability. Even better actually do the trick yourself with 3 bags. Physically do it a number of times. Count the number of times you pick a white ball, then count the number of times you pick a white second ball. The ratio will be around 2/3
So you can actually prove it yourself by trying. Let me know when you’ve got it.
Mr Mensa
Mr BWAKEM:-
Ooooo-Errrrrrr
http://i.imgur.com/rQQS7Lc.png
http://i.imgur.com/AU4jFiF.png
http://i.imgur.com/LYK8DYk.png
Looks like I’ll have to eat some humble pie and delete my subscription to Mensa. :-(
Mr Mensa
So in mumbly words. You are twice as likely to end up with the White:White Bag and you are to end up with the White:Black Bag and this is because the requirement to discard the White:Black Bag if you pull the Black Ball from it first favours selection of the White:White Bag…. or something like that. At least that would be how I make sense of it.
Drunken Sailor
Nigel / Mr MENSA that at least makes more sense that being able to draw the second white ball first.
When I am bored out of my skull I will try it a few hundred times to get a statistically significant sample.
phil
DeepPurple
This really is my last go and attempting to convince anyone,
1)
According to Tom’s instructions the game starts by taking a white ball out of a random bag.
If the game starts with a white ball, then you didn’t choose a random bag, you chose a bag that has a white ball in it, if it had a black in it you never started the game. That is not a bag chosen at random, it has to have a white ball.
After the 1st choice you are left with 5 balls, so when you started the game you started with 5 balls not 6. So why did Tom even bother to describe the game has 3 bags with 6 balls, 2 in each bag, when the game really started with 3 bags, 5 balls when one of the bags only had one ball. If you are not going to calculate the prob of the 1st choice, why bother with an opening white ball at all. Just pick the ball from the bag with one ball and calculate the prob it will be white, which is what you have all done.
Somebody posted the 2nd ball is chosen out of the predetermined bag with the 1st ball. Foul, you can’t do that, if you are not going to calculate the prob of the 1st choice, you cannot use any outcome (and I mean any) of what happened on the 1st choice, it has to be disregarded. Again making the rules up
Clearly we are playing 2 different games. But even if I played your game 2/3rds is still the wrong answer. You have that you can choose 2 white ball from 5 balls with a cos of 66.2/3rd% yet only 33.1/3% chance of choosing a black ball from 3 out of five. Clearly wrong. The only combination to give you 2/3rd knowing you have 2 white balls is 2 white balls and 1 black, so you have managed to disappear 2 black balls, What have you done with them. The correct answer to your game is 2/5th = 40%
I’m perfectly happy to be proved wrong but no-one has proved that yet, I spent 6 weeks of my life quite happy being a Quenron, so I can screw up believe me.
We will all beg to differ,