Thread: Calculate the odds

Calculate the odds
You have three coins you flip the first one it comes up heads, you flip the second it comes up heads.
Question #1
what are the odds that coin 3 will come up tails?
a. 1 in 2
b. 1 in 3
Lets say it does come up tails and all three coins are covered by nut shells and moved around like in a shell game. The object is to chose the shell containing the face up tail if you pick correctly you will win a billion dollars!
You have shell a, b, and c. To chose from.
You chose (c), however before you can look under shell (c) you are shown what’s under shell (a) and its heads.


You are told you can change your mind and pick either b or keep your original chose of c.

Question # 2

What are the odds that coin c. will be tails?
a. 1 in 2
b. 1 in 3

Question #3

What are the odds that coin b. will be tails?
a. 1 in 2
b. 1 in 3

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the answer to all 3 questions is:

a. there is a 1 in 2 chance.

unless you have thrown in some sick twist that i missed.

there is also a 1 in 2 chance of me guessing the answer to this question however i am not one for such guesses.

Incorrect: no sick twist just mathematical odds

If I don't win the billion dollars can I at least keep the coins?

Originally Posted by Ophiolite

If I don't win the billion dollars can I at least keep the coins?

LOL sure

ok where was i wrong?

it all looks straight forward to me.

1 in 2, for all, using mathematical principles, and leaving out my damnable bad luck.

It does look straight forward but is actually counter intuitive.
Its about how we perceive information. I set this up to show that there are levels to which we associate linear thinking with non-linear thinking.
How information paradigms are constructed by information as opposed to patterns that are random.

Question #1
Some assume that two consecutive out comes would increase the odds of a opposite outcome. (believe me I know people that think this way) Some would assume this is a pattern or a trend will increase the odds of repeating itself. Of coarse most of us realize that each flip of the coin has no connection to one another. So the answer is 1in 2
But in real life people make this mistake all the time not only in games of chance but in how we perceive the world.

Question #2 # 3
What are the odds that coin b. will come up tails?
1 in 2
What are the odds that coin c. will come up tails?

1 in 3
We should change from c to b increasing our odds of becoming a billionaire from 1 in three to 1 in two.

Our first choice of c is from the paradigm of 1 to 3 chance model. When we are then shown a is heads it changes the model for c but not b.
the new information can only be applied by giving up the first choice that was not relying on new information but only chance.
The reason it is counter intuitive is that we do not recognize a. being heads as new information as it relates to our current position. The one being nothing more than a choice based on an assumption.

WikipediaBayesian statisticians claim that methods of Bayesian inference are a formalisation of the scientific method involving collecting evidence that points towards or away from a given hypothesis. There can never be certainty, but as evidence accumulates, the degree of belief in a hypothesis changes; with enough evidence it will often become very high (almost 1) or very low (near 0).

As an example, this reasoning might be
The sun has risen and set for billions of years. The sun has set tonight. With very high probability, the sun will rise tomorrow.
Bayesian statisticians believe that Bayesian inference is the most suitable logical basis for discriminating between conflicting hypotheses. It uses an estimate of the degree of belief in a hypothesis before the advent of some evidence to give a numerical value to the degree of belief in the hypothesis after the advent of the evidence. Because it relies on subjective degrees of belief, however, it is not able to provide a completely objective account of induction. See scientific method.

Bayes' theorem also provides a method for adjusting degrees of belief in the light of new information

Wait, if [a] is heads, then of the two remaining coins, one is heads and one is tails. So, the chance of either coin being tails, is 1 in 2.

Now we have the crux of then problem, this is the obvious answer but it is incorrect.
Almost everyone makes this same assumption including my self.

remember our original pick of c was in a 1 in 3 paradigm this will not change.
If we chose b however it is a new informational paraphgm of 1 in 2.

Originally Posted by Metatron

Now we have the crux of then problem, this is the obvious answer but it is incorrect.
Almost everyone makes this same assumption including my self.

remember our original pick of c was in a 1 in 3 paradigm this will not change.
If we chose b however it is a new informational paraphgm of 1 in 2.

Wrong. When we originally picked [c], we had a 1 in 3 chance of being right; when we stuck with [c] after knowing that [a] was heads, we had a 1 in 2 chance of being right. This was a new choice.

Originally Posted by j

Originally Posted by Metatron

Now we have the crux of then problem, this is the obvious answer but it is incorrect.
Almost everyone makes this same assumption including my self.

remember our original pick of c was in a 1 in 3 paradigm this will not change.
If we chose b however it is a new informational paraphgm of 1 in 2.

Wrong. When we originally picked [c], we had a 1 in 3 chance of being right; when we stuck with [c] after knowing that [a] was heads, we had a 1 in 2 chance of being right. This was a new choice.

No, think you went from 3 to 2 choices when a is shown it changed the odds to 1 in 2 but only for b, c remains 1 in 3.

Originally Posted by Metatron

No, think you went from 3 to 2 choices when a is shown it changed the odds to 1 in 2 but only for b, c remains 1 in 3.

so you have a 1 in 3 chance of guessing right when you origionally pick c. however when a is shown the number of choices you can make becomes 1 in 2, heads or tails, b or c.
since tails only lies under one of these there is a 1 in 2 chance of guessing right.

a tree diagram can be drawn to illustrate this.

Originally Posted by Metatron

Originally Posted by j

Originally Posted by Metatron

Now we have the crux of then problem, this is the obvious answer but it is incorrect.
Almost everyone makes this same assumption including my self.

remember our original pick of c was in a 1 in 3 paradigm this will not change.
If we chose b however it is a new informational paraphgm of 1 in 2.

Wrong. When we originally picked [c], we had a 1 in 3 chance of being right; when we stuck with [c] after knowing that [a] was heads, we had a 1 in 2 chance of being right. This was a new choice.

No, think you went from 3 to 2 choices when a is shown it changed the odds to 1 in 2 but only for b, c remains 1 in 3.

I have to agree with with wallaby and j on this one. Probabilities depend on information not choices. The fact that you chose c before a is revealed is irrelevant, however the fact that you have more information after a is revealed does change the probabilities.

This is an application of Baye's theorem. After a is revealed the probability of c being tails is now what the probability that c was tails given that a was heads which by Baye's theorem is the probability that c was tails divided by the probability that a was heads.
Formally,
P'(c=t) = P(c=t|a=h) = P(c=t)/P(a=h) = (1/3)/(2/3) = 1/2.
where P is the probability function before a is revealed and P' is the probability function after a is revealed.

Probably. :wink:

Metatron wrote;

remember our original pick of c was in a 1 in 3 paradigm this will not change.
If we chose b however it is a new informational paraphgm of 1 in 2

b is actually 2 in 3, oop's! my mistake

no it's not, its 1 in 2.

This sounds a lot like the "Monty Hall" problem posed to Marilyn Vos Savant (see Wikipedia) years ago. The easiest way to tackle such a problem is to simply make a table.

I like your twist on the problem though.

w

Originally Posted by william

This sounds a lot like the "Monty Hall" problem posed to Marilyn Vos Savant (see Wikipedia) years ago. The easiest way to tackle such a problem is to simply make a table.

I like your twist on the problem though.

w

This is a zombie thread, but what the heck.

This would only qualify as a Monty Hall problem if the person who flipped the shell with a "heads" coin under it knew in advance that it had a head coin and deliberately set out to reveal a "heads" shell from the two shells that the player didn't pick. In that case switching increases your odds of winning to 2/3.

If the person running the game were to randomly select a shell, look under it, and it turns out to happen to be a "heads" shell then there is no advantage to switching - you have a 1/2 chance of winning either way. Your odds increased from 1/3 to 1/2 when the shell luckily didn't turn out to contain the "tails" coin.

The original poster didn't make it clear which situation was supposed to apply.

Let's get to the bare essentials.
After the three coins are tossed, we know we have HHT.
Before a is revealed success is 1/3.
After a is revealed as H, we know the other two are HT.
We are now selecting from a different set.
There are only two choices.
Success is 1/2.

Originally Posted by phyti

Let's get to the bare essentials.
After the three coins are tossed, we know we have HHT.
Before a is revealed success is 1/3.
After a is revealed as H, we know the other two are HT.
We are now selecting from a different set.
There are only two choices.
Success is 1/2.

Only if you assume that the coin which was revealed was selected at rondom. If we know that the host set out to reveal a "heads" coin from the two remaining shells, then it's a Monty Hall problem and you can increase your chances of winning from 1/3 to 2/3 if you switch. Like I said, the original poster didn't make it clear which he meant.

If a revealed T then we know the others are HH.
Success is now 0/2. The probability depends on what you learn.
After the revealing of a, we have more information than before.
It doesn't matter what his intentions are. You are still selecting from
a two outcome set. If you want to be mislead by his misdirection
that's your choice. By definition, probability of success is one choice
divided by the possible choices. After a is revealed, the possible choices
go from 3 to 2. That's as simple as I can explain it.

Originally Posted by phyti

If a revealed T then we know the others are HH.
Success is now 0/2. The probability depends on what you learn.
After the revealing of a, we have more information than before.
It doesn't matter what his intentions are. You are still selecting from
a two outcome set. If you want to be mislead by his misdirection
that's your choice. By definition, probability of success is one choice
divided by the possible choices. After a is revealed, the possible choices
go from 3 to 2. That's as simple as I can explain it.

I suggest you google "monty hall problem" for a through explanation of why you are wrong. It's been done to death here, so I doubt anyone is very interested in explaining it again. To put it simply, you are making the common mistake of assuming that just because there are two choices the odds must be 50/50. You only determine probability by counting the number of choices if you don't have any extra information - since in this case you do have extra information, the odds are more complicated.

I checked the Monty Hall sites.
Without all the facts, anyone can be wrong.
Given an honest game however, which we thought it was,
i.e. if his choice of revealing a shell at random, the rest of us
would have been correct. I prefer real problems over flimflam. :-D