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So, this one is pretty easy...

How can you get a fair (equal probability) outcome using only an unfair coin (where unfair means that it will land head with probability p and tails 1-p where p != .5)?

Another similar one:

How can you get a fair outcome using only a loaded/weighted die (where loaded/weighted means that it will land one one number more often than any other)?

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So, this one is pretty easy...

How can you get a fair (equal probability) outcome using only an unfair coin (where unfair means that it will land head with probability p and tails 1-p where p != .5)?

Another similar one:

How can you get a fair outcome using only a loaded/weighted die (where loaded/weighted means that it will land one one number more often than any other)?

flip it enough times and base win/lose on the resultant percent.

For example, if p = 0.51 for H and 1 - p = 0.49 for T, you'd flip 100 times and if you got more than 51 H, then H is the winner. The drawback is that 51/49 is a tie, kind of like the coin landed on its edge.

If it was 60/40, you'd only have to flip 10 times.

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flip it enough times and base win/lose on the resultant percent.

For example, if p = 0.51 for H and 1 - p = 0.49 for T, you'd flip 100 times and if you got more than 51 H, then H is the winner. The drawback is that 51/49 is a tie, kind of like the coin landed on its edge.

If it was 60/40, you'd only have to flip 10 times.

But what if you don't know p...You just know that it doesn't equal 0.5?

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both flip the coin/ roll the die a given number of times, the person who gets the most number of heads or tails wins.

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both flip the coin/ roll the die a given number of times, the person who gets the most number of heads or tails wins.

How do you know how many times would be needed to ensure a fair outcome? What if I'm not flipping against someone...I just have a decision I need to make, and I assign heads to one option, tails to another...how can I use this unfair coin to get a fair outcome for my decision?

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Just to show what I mean...I wrote a simple program to demonstrate that it can be done:

So, given a "coin" that has 99% probability of heads, I was able to use this "coin" to get:

Flipped 5031 heads and 4969 tails...

Flipped 5091 heads and 4909 tails...

Flipped 4966 heads and 5034 tails...

Flipped 4963 heads and 5037 tails...

Flipped 4992 heads and 5008 tails...

...

That's obviously out of 10000 turns, and as you can see, that's pretty even (It can be shown that it truly is 50/50...)...The outcome is the same regardless of what probability I assign to heads...

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So, this one is pretty easy...

How can you get a fair (equal probability) outcome using only an unfair coin (where unfair means that it will land head with probability p and tails 1-p where p != .5)?

Another similar one:

How can you get a fair outcome using only a loaded/weighted die (where loaded/weighted means that it will land one one number more often than any other)?

you both call it at the same time and keep flipping until there is not a tie?

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Draw a line and flip the coin if the the coin falls to the left is heads and if it falls to the right is tails.

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Draw a line and flip the coin if the the coin falls to the left is heads and if it falls to the right is tails.

Hmm...definitely not what I was thinking, but I suppose if there was absolutely no bias and perfect accuracy, that could work...but I definitely wouldn't let someone else do the flipping for me in this case...What I am thinking of makes it completely fair no matter who is flipping the coin.

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Just to show what I mean...I wrote a simple program to demonstrate that it can be done:

So, given a "coin" that has 99% probability of heads, I was able to use this "coin" to get:

Flipped 5031 heads and 4969 tails...

Flipped 5091 heads and 4909 tails...

Flipped 4966 heads and 5034 tails...

Flipped 4963 heads and 5037 tails...

Flipped 4992 heads and 5008 tails...

...

That's obviously out of 10000 turns, and as you can see, that's pretty even (It can be shown that it truly is 50/50...)...The outcome is the same regardless of what probability I assign to heads...

I also just wrote a program to show the results of my solution the die problem as well:

I said that the die I was using was weighted so that 50% of the time, it would roll a 1, and then 10% of the time for everything else...Here are the results:

Rolled 1705 ones, 1610 twos, 1680 threes, 1636 fours, 1699 fives, 1670 sixes...

Rolled 1639 ones, 1668 twos, 1692 threes, 1600 fours, 1745 fives, 1656 sixes...

Rolled 1631 ones, 1640 twos, 1652 threes, 1657 fours, 1733 fives, 1687 sixes...

Rolled 1656 ones, 1665 twos, 1667 threes, 1637 fours, 1685 fives, 1690 sixes...

...

Which is again out of 10000 turns, and as you can see, it is again pretty even (again shown that it is equal probability)...again the outcome is the same regardless of what probabilities I assign to the numbers...

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So, this one is pretty easy...

How can you get a fair (equal probability) outcome using only an unfair coin (where unfair means that it will land head with probability p and tails 1-p where p != .5)?

Another similar one:

How can you get a fair outcome using only a loaded/weighted die (where loaded/weighted means that it will land one one number more often than any other)?

Coin: Flip the coin once for each outcome ("heads" and "tails"). If they flip the same, repeat. Stop when one flips heads (arbitrarily chosen as the target), and the other flips tails. Return the outcome that flipped heads.

Die: Roll the die for each outcome ("1", "2", ..., "6"). Remove all outcomes that do not have the highest number rolled (for that round). Repeat until there is only one outcome, and return it.

Essentially this turns the problem into a fair contest between the different outcomes. Of course, these won't necessarily complete in any given number of flips/rolls.

Edited by EventHorizon

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Alternate what you call "Heads" and flip multiple (even) times. And for the die, rotate the numbers. Should average out.

Hmm...interesting solution. Different than the solution I came up with...

Not sure I see exactly what you mean...basically If I was given an unfair coin, and I was asked to make a fair decision using only that coin...how does this method get me a completely fair decision?

I assume you mean flip the coin an even number of times (let's say 10 times...)...the first 5 times you flip you call side A "heads" and the next 5 times you flip you call side A "tails"...how does that translate to a single outcome of heads/tails? so for example, if I did that and the first 5 flips yielded 4 Side A "heads" and 1 Side B "tails" and the next 5 flips yielded 3 Side A "tails" and 2 Side B "heads"...that would mean it had 6 "heads" and 4 "tails"...would it be a "heads" decision since I had more "heads" outcomes than "tails"? What if I had flipped 5 and 5? Just trying to figure out your solution... I agree that in the long run, you can get a "50/50" split of heads/tails if you do this, but I'm just trying to figure out how it would help me make a fair decision...

My solution is along the same lines as this one in the fact that it may take more than 1 flip/roll to get a single outcome...But I think it may be slightly simpler to keep track of in practice...

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Coin: Flip the coin once for each outcome ("heads" and "tails"). If they flip the same, repeat. Stop when one flips heads (arbitrarily chosen as the target), and the other flips tails. Return the outcome that flipped heads.

Die: Roll the die for each outcome ("1", "2", ..., "6"). Remove all outcomes that do not have the highest number rolled (for that round). Repeat until there is only one outcome, and return it.

Essentially this turns the problem into a fair contest between the different outcomes. Of course, these won't necessarily complete in any given number of flips/rolls.

Just wondering if you could show an example of each of these...I think you're on the right track (maybe even have it the same as mine...but I'm just not following your description fully)

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Instead of calling heads or tails, you could call

Heads then tails OR Tails then heads

Then you flip the coin in groups of two. After every two flips the sequence starts over.

If the probability is only to determine one of two choices then you can pretend [1] is heads and [2,3,4,5,6] is tails. This gives you a 50 50 shot at heads or tails.

Edited by IDoNotExist

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Just wondering if you could show an example of each of these...I think you're on the right track (maybe even have it the same as mine...but I'm just not following your description fully)

Here's an example for the die (coin is similar).

roll once for outcome "1"...lets say the outcome is a 1.

roll once for outcome "2"...lets say the outcome is a 1.

roll once for outcome "3"...lets say the outcome is a 1.

roll once for outcome "4"...lets say the outcome is a 1.

roll once for outcome "5"...lets say the outcome is a 1.

roll once for outcome "6"...lets say the outcome is a 1.

Since they all rolled the highest rolled that round, none are removed.

roll once for outcome "1"...lets say the outcome is a 3.

roll once for outcome "2"...lets say the outcome is a 1.

roll once for outcome "3"...lets say the outcome is a 5.

roll once for outcome "4"...lets say the outcome is a 2.

roll once for outcome "5"...lets say the outcome is a 5.

roll once for outcome "6"...lets say the outcome is a 4.

"3" and "5" rolled a 5 (highest seen that round), so remove all the others.

roll once for outcome "3"...lets say the outcome is a 1.

roll once for outcome "5"...lets say the outcome is a 3.

Outcome "5" is the only outcome that rolled the highest, so return it.

The method is similar for the coin.

"H" flipped heads.

"T" flipped heads.

Need to try again.

"H" flipped heads.

"T" flipped heads.

....

"H" flipped heads.

"T" flipped heads.

yeah, this is getting old...

"H" flipped heads.

"T" flipped heads.

ugh...

"H" flipped heads.

"T" flipped tails.

Finally!

Since "H" flipped heads and "T" didn't, the outcome returned is "H".

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So, this one is pretty easy...

How can you get a fair (equal probability) outcome using only an unfair coin (where unfair means that it will land head with probability p and tails 1-p where p != .5)?

Another similar one:

How can you get a fair outcome using only a loaded/weighted die (where loaded/weighted means that it will land one one number more often than any other)?

Assign 1 or 0 to a "win" count based on the head/tail of each toss, but switch the "winning" case each turn.

So toss #1, if it lands heads then win =1, else win =0.

So toss #2, if it lands tails then win =1, else win =0.

So toss #3, if it lands heads then win =1, else win =0.

if there is an advantage in toss 1, it is met with an equal disadvantage in toss 2

tossing should happen in pairs

am i close?

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There is one flaw that throws everyone except burninator777.

What if the probability is of a heads is 1 and tails will never come up. You would end up with a tie every time.

Edited by zzembower

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Ok. Here is the solution.

Step 1: Toss coin twice.

Step 2: If coin is same outcome both times, repeat Step 1. If coin is different outcome, take the first outcome (or the second outcome, but make sure which out outcome you're going to take is predetermined before you toss the coin).

Step 3: Sucker punch that shady dude who tried to give you a bum deal in the face.

Step 4: Run away because you probably didn't hurt the shady dude at all since you're doing brain teasers. Seriously, RUN! He's going to kick your face in. OR

Step 4 (alternatively): Clutch your hand in pain as you hurt your hand more than his face because you've never thrown a real punch AND get your face kicked in.

Step 5: Throw your hands up and tell him, "Whoa whoa. I don't want any trouble. I'll tell you what, we can toss this here coin to determine whether or not you can kick my face in."

Step 6: Get your face kicked in.

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There is one flaw that throws everyone except burninator777.

What if the probability is of a heads is 1 and tails will never come up.

I think that Burninator's solution has its own problems. [spoiler='another way to do it

']If the probability of heads was 1 than I would spin the coin and time how long it spins. I would call even or odd and count the seconds until it falls.

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I think that Burninator's solution has its own problems. [spoiler='another way to do it

']If the probability of heads was 1 than I would spin the coin and time how long it spins. I would call even or odd and count the seconds until it falls.

Is anything ever for sure in this world? Can anything attain the probability of 1?

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If you figured out that the probability of heads was 1 then you could super glue the tails side to the back side of jelly bread which also always lands face down and according to the rules of probability it would never land.

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So, it seems like people have gotten the coin problem...basically my solution is:

Flip the coin in sets of 2.

HT = heads

TH = tails

HH = redo, TT = redo...

This will ensure an equal probability of getting heads or tails (yes, if the coin were so unfair as to not allow any tails, then you would be flipping for quite a while.. and I would punch the guy who made that coin). This strategy works for all other probabilities.

With the die, my solution is this:

Roll the die 3 times, until you get 3 results that are NOT the same.

The outcomes would then have a high value, medium value, and low value...so we have these equal possible outcomes:

HML, HLM, MHL, MLH, LHM, LMH...

Which is 6 possible...so just assign a value to each one and that's your roll:

HML = 1

HLM = 2

MHL = 3

MLH = 4

LHM = 5

LMH = 6

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So, it seems like people have gotten the coin problem...basically my solution is:

Flip the coin in sets of 2.

HT = heads

TH = tails

HH = redo, TT = redo...

This will ensure an equal probability of getting heads or tails (yes, if the coin were so unfair as to not allow any tails, then you would be flipping for quite a while.. and I would punch the guy who made that coin). This strategy works for all other probabilities.

With the die, my solution is this:

Roll the die 3 times, until you get 3 results that are NOT the same.

The outcomes would then have a high value, medium value, and low value...so we have these equal possible outcomes:

HML, HLM, MHL, MLH, LHM, LMH...

Which is 6 possible...so just assign a value to each one and that's your roll:

HML = 1

HLM = 2

MHL = 3

MLH = 4

LHM = 5

LMH = 6

I like my solution....it extends seamlessly to any number of sides for a die very easily. In fact, even with an unfair coin it could simulate a fair die of any number of sides (just let heads=1 and tails=0).

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If you figured out that the probability of heads was 1 then you could super glue the tails side to the back side of jelly bread which also always lands face down and according to the rules of probability it would never land.

That's like toasting and buttering a cat's back. Cats always land on their feet, toast always lands buttered side down. You drop em and they just float there until they've licked all the butter off! :lol:

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I like my solution....it extends seamlessly to any number of sides for a die very easily. In fact, even with an unfair coin it could simulate a fair die of any number of sides (just let heads=1 and tails=0).

Ahh...not only that, but you can simulate unfair coins/dice by adding in multiple copies of the same outcome. The probabilities for the unfair coins need to be rational numbers though.

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