Author Topic: Coin flippin quiz (math question)  (Read 3458 times)

kav

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Coin flippin quiz (math question)
« on: April 24, 2015, 10:45:03 AM »
This was posted on Wizzard of Vegas. It is an interesting puzzle

You and your friend are both in prison.  You are both informed that you will shortly be separated.  After separation, you will then each flip a fair coin.  You will then be asked to submit a prediction of the other person's flip.  If you are both correct then you will both be set free. If one or both are wrong, then you will both be put to death.

You are allowed some time to devise a strategy.  What strategy should you devise? 

What is the maximum probability of success?


 

scepticus

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Re: Coin flippin quiz (math question)
« Reply #1 on: April 24, 2015, 02:35:20 PM »
This was posted on Wizzard of Vegas. It is an interesting puzzle

You and your friend are both in prison.  You are both informed that you will shortly be separated.  After separation, you will then each flip a fair coin.  You will then be asked to submit a prediction of the other person's flip.  If you are both correct then you will both be set free. If one or both are wrong, then you will both be put to death.

You are allowed some time to devise a strategy.  What strategy should you devise? 

What is the maximum probability of success?


They should both agree to bet the same outcome - either head or tails.Probability is 50 %. not being a maths geek I could be wrong .As I am banned from W.O.V. I  didn't look it up !
« Last Edit: April 25, 2015, 12:45:17 AM by kav »
 

scepticus

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Re: Coin flippin quiz (math question)
« Reply #2 on: April 25, 2015, 12:42:37 AM »
was I right kav ? or wrong - again !
 

kav

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Re: Coin flippin quiz (math question)
« Reply #3 on: April 25, 2015, 12:46:09 AM »
They should both agree to bet the same outcome - either head or tails.Probability is 50 %. not being a maths geek I could be wrong .As I am banned from W.O.V. I  didn't look it up !

The answer 50% is correct but I don't understand your reasoning. Can you elaborate?
 

scepticus

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Re: Coin flippin quiz (math question)
« Reply #4 on: April 25, 2015, 01:57:00 AM »
SH*T ! I never logged in before typing my answer so have had to do it over again  !
There are 4 possibilities - HH -HT -TH -TT.
Intuitively we think that a 1 in 4 chance is 25 %. But look at it this way.
HH and TT  both give the second toss the SAME as the first toss  no matter which one we choose - while TH and HT don't.
So we now have 2 versus 2 which is 50 %.
I have no qualifications whatsoever in maths so don't ask me to explain it further !
 

kav

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Re: Coin flippin quiz (math question)
« Reply #5 on: April 25, 2015, 07:48:46 AM »
I better like this explanation
Quote
If 1 person were to do this and call the coin before they flipped it:
P(guessing 1 flip correctly) = 1/2 = .5 or 50%
P(guessing 2 flips correctly) = P(guessing 1 flip correctly) * P(guessing 1 flip correctly) = .5 * .5 = .25 or 25%

There's a lot more options here than I think people realize. The outcome of the coin flip is binary 00, 01, 10, 11... but that's not factoring your choices in to each outcome.

0000 (RIGHT)
0001
0010
0011
0100
0101
0110 (RIGHT)
0111
1000
1001 (RIGHT)
1010
1011
1100
1101
1110
1111 (RIGHT)

4/16 = 2/8 = 1/4 = .25 or 25%

Flip-Guess, Flip-Guess, Correctness
T - Heads, T - Heads, WRONG
T - Heads, T - Tails, WRONG
T - Tails, T - Heads, WRONG
T - Tails, T - Tails, RIGHT

T - Heads, H - Heads, WRONG
T - Heads, H - Tails, RIGHT
T - Tails, H - Heads, WRONG
T - Tails, H - Tails, WRONG

H - Heads, T - Heads, WRONG
H - Heads, T - Tails, WRONG
H - Tails, T - Heads, RIGHT
H - Tails, T - Tails, WRONG

H - Heads, H - Heads, RIGHT
H - Heads, H - Tails, WRONG
H - Tails, H - Heads, WRONG
H - Tails, H - Tails, WRONG

Now after I flip, let's say Tails, then I can eliminate the 2 groupings of 4 where my flip was Heads. I'm down to 8 possibilities. If I tell my partner I'm always going to guess the opposite of what I flip, then that means I'll be guessing Heads in this scenario. This puts me down to 4 possibilities, 3 of which are wrong and 1 which is right. Because even if my partner does flip Heads, he may guess my coin incorrectly. The 4 resulting options where I flip tails and guess Heads are:
T - Heads, T - Heads, WRONG
T - Heads, T - Tails, WRONG
T - Heads, H - Heads, WRONG
T - Heads, H - Tails, RIGHT

If my partner ALSO agrees to always guess opposite, then that eliminates 2 more of the choices leaving only 2 choices (one being right) for a 50/50:
T - Heads, T - Heads, WRONG
T - Heads, H - Tails, RIGHT

Therefore the answer would be 50% if you both either A) agree to always guess the same as you flipped, or B) agree to always guess the opposite of what you flipped.
 

scepticus

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Re: Coin flippin quiz (math question)
« Reply #6 on: April 25, 2015, 08:22:06 AM »
It will be interesting to find out the WOV answer.Mine was derived from the 3 card trick.
 

kav

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Re: Coin flippin quiz (math question)
« Reply #7 on: April 25, 2015, 09:29:16 AM »
It will be interesting to find out the WOV answer.Mine was derived from the 3 card trick.
the answer is 50%
 

scepticus

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Re: Coin flippin quiz (math question)
« Reply #8 on: April 25, 2015, 09:42:57 AM »
Yes . But how did they work it out ? Sometimes the workings are more important than the answer .The "HOW TO". As we learned in our schooldays.
 

dobbelsteen

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Re: Coin flippin quiz (math question)
« Reply #9 on: April 25, 2015, 01:56:41 PM »
The throw of the friend has always a 50% chance that the result is the same The chance of the friend is undepended on the result of the first throw . See the triangel of Blaise Pascal.

I have a  roulette puzzle .
The number of corners of the France table layout is 45. What is the chance of  a hit of a corner?
 

scepticus

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Re: Coin flippin quiz (math question)
« Reply #10 on: April 25, 2015, 04:01:10 PM »
What it really boils down to is that the maths seem to indicate that it is ONLY the probabilities at the start that matters. What happens in the interval from start to finish is irrelevant. Food for thought . Nevertheless that is still the expectation not the inevitable result.Uncertainty still rules the roost !
 

BlueAngel

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Re: Coin flippin quiz (math question)
« Reply #11 on: April 26, 2015, 05:58:51 AM »
The throw of the friend has always a 50% chance that the result is the same The chance of the friend is undepended on the result of the first throw . See the triangel of Blaise Pascal.

I have a  roulette puzzle .
The number of corners of the France table layout is 45. What is the chance of  a hit of a corner?

I like your thinking! 8)
The answer to your question is 4 to 37 or 10.7 %

As about the quiz of WOV,if the two imprisoned persons agree to predict the same and their release depends only from knowing what the other person have predicted,then they are free men.
If the point is to guess right about the coin toss and not to guess what the other person predicted,each of them would have 50 % chance to be released.

It's quite simple,it's about binomial distribution,according to this theory,within a large amount of eight-set results,the combinations of equal distribution (4 heads and 4 tails) within each set of 8 results,would be significantly higher than the ones with the uneven distribution (regardless the sequence)
If you are interested in the subject,just google "Bernoulli trial"
 

dobbelsteen

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Re: Coin flippin quiz (math question)
« Reply #12 on: April 26, 2015, 08:50:47 PM »
BA you are right. In this case the number of corners is not relevant. Every number has a hit chance of  2,7%
A corner has 4 numbers, so the hit chanceis 10.8%