Hi
A dice has 3 sides A B C
It is thrown twice and lands as A A
then what is the probability that next to land will be B or C?
That is A A B or A A C
The dice has no memory of what was thrown before but surely the probability can't be 2/3.
It is rare to see say B B B B B B B B to land consecutively so surely there is less probility that B will land again than an A or C? The A and the C have their '1/3 probability of landing destiny' to fullfill and the B has landed more than its fair share of occassions so surely A or C are more likely to land next?
What are the probabilty calculations?
Thankyou
You were right when you said that the die 'has no memory' . It's what's known in probability as independence.
All earlier rolls have no effect what so ever on the next roll, so the probability of getting, say, B OR C is ALWAYS 2/3 . Why would it change? Nothing has physically happened to the die, so long as it's a 'fair' die.
Also, all strings (of equal length) of rolls are equally likely. For example BBBBBBBB is just as likely as ABABCBCC.
I get this, but surely if you were sitting in a casino and the dice rolled BBBBBBB then surely you would feel confident that A or C may land next.
In an infinite time period you would see BBBBBBBB land as often as ABABCBCC but in the 1 hour (time limit) you may be sitting in the casino you won't see BBBBBBBB that often and you will expect the letters to be assorted.
I know what you say makes sense mathematically ..... but I'm sure that I'm correct in some deep parallel universe sort of way ..... and there are plenty of casino dwellers sitting there on the slot machines that know I'm correct.
If AAAAAAAAA land so many times in a row .... all the gamblers I've seen will put down a B or C on their bet.
Thanks Pomp ... (deep down know you're telling the truth but I can't admit it to myself)
That's because the human mind doesn't comprehend random chance - it tries to make order out of something. It's a major reason why casinos rake in the money
BBBBBBBB is as likely as ABABCACA or any other sequence. A simple tree diagram will show this
Admittedly if you were not picky about the order they came in but the actual number it would be different. For the above sequence you're more likely to get 4 As, 2Bs and 2Cs than 8 B's
What about my probability destiny theory.
If AAAAAAAAA does fall in this sequence then in a 1 hour time frame B and C have yet to fulfill their 1/3 probability destiny. So surely they will have an increased theorectical probability to be next.
Lets forget that the dice does not have a memory for a moment .... because that just spoils it.
How do you calculate the probability of B landing after A?
Is it 1/3+1/3 or 1/3x1/3 or something else?
Deep down I know you're all correct but I have a feeling that there is something in what I am saying?
Hi AceHero,
and Happy Christmas!
You stated "a dice has sides A, B, C".
Is this a standard dice of sides, with only of them marked,
or is this a -sided dice with two A sides, two B sides and two C sides?
From the discussion, I can "safely" assume it has sides.
Also, each side is equally likely if we are rolling an unbiased dice.
On "any" throw, the probability of an A is , B is , C is .
As the throws are independent, it does not matter that the first two throws
were A followed by A.
Previous throws do not alter the probability for subsequent throws.
Hence, on "any" throw, the probabilities remain the same.
A few of us did an experiment once...
The probability of achieving an unbroken series of heads by tossing a coin is very small, practically impossible according to our maths lecturer.
We threw heads in a row and if you work out that
probabilty, it's quite small.
Yet, the probability of any alternative sequence of is equal to it.
However, if you ask what is the probability of heads and tails, you are summing together the probability of all the different sequences that give
rise to heads and tails.
The moral of the story is..
The probability of a B or C landing "next" is the probability
of a B or C landing on "any" throw for this case.
This is .
The probability of B landing after A is equal to the probabilty of A landing after A,
which equals the probability of C landing after A, which equals the probability of B after any of A, B, C etc,
which is 1/3.
If A keeps occuring, B and C have "increased expectations of occuring in our minds".
According to quantum physics, this may bias the outcome !!
However, if there is never any bias, the probabilities never change.
This of course is saying that the outcome of the next throw is completely random.
There is a tiny probability that A will keep occuring for our lifetime.
However, I feel that what you are really asking here is....
Is not the probability of a sequence containing a B or C much greater than
AAAAAAAAAAAAA ?
and is not the probability of a sequence with a B or C even greater still than
AAAAAAAAAAAAAAAAAAAA ?
Absolutely!