creating a 2:1 shot with a fair coin

[2ndGoat]

so here's a probability curiosity I've been pondering. I imagine there may be a very trivial way to do this, but I can't figure it:

You've got a *finite* number of fair coins (Pr(H) = Pr(T) = .5). Any number you like, but not infinite. How do you define a test, with any *finite* number of coin flips using any combination of the coins, that have a Pr(Success) of exactly 1/3?

Basically, I'm saying you and your buddy have a handful of pennies, and you want to create an even-money game based entirely on coin flips where one player lays the other 2:1 odds.

Obviously, to create an even-money game, one player could place two coins facing heads up, one tails, and have the other try to pick tails. But that wasn't a random coin flip. No using infinite series either, obviously one can create a situation with a Probability arbitrarily close to 1/3 by adding more coin flips and picking the binomial expiriment with the probability closest to 1/3. We're talking finite numbers here.

Is this possible? I can't get away from probabilities that are sums of powers of two, I'm starting to believe it can't be done.

Goat

[BruceZ]

It can be done with conditional probabilities. Flip two coins at the same time. It is important that you cannot tell the coins apart. The possible outcomes are HH,TT,HT,TH. Show your friend one of the coins. You have to select the coin to show at random to be fair. The probability that the other coin is the same will be 1/3. Notice that if the coins can be distinguished, the probability will be 1/2. An alternative to having indistinguishable coins is to tell your friend the value of one of the coins, but don't show him either coin until after he guesses. A way you can select one of the coins at random to show is to flip a 3rd coin. If it comes up heads, show or tell him about the coin that is a head, if it comes up tails, show or tell him about the coin that is a tail. If both coins were the same, it doesn't matter which one you show or tell him about.

[irchans]

"...You have to select the coin to show at random to be fair. The probability that the other coin is the same will be 1/3..."

I really don't understand your post. Could you explain why the probability is 1/3? Why is selecting a coin at random important?

[BruceZ]

Say I tell you one of the coins is heads. Now the only possible outcomes were HH,HT,TH, so the probability that the other one is heads is 1/3. This is the same as the classic problem where I tell you I have two children and one is a boy; the probability that the other one is a boy then becomes 1/3.

The reason I have to choose randomly is because if I always told you say "heads" when it was HT or TH, and you pick up on this, then whenever I tell you "tails" you would know it was TT. Or if one of the coins had some distinguishing feature, and I always chose that coin to tell you about, and you knew this, then the probability becomes 1/2 since when I say "heads", now the only outcomes are HH,HT, since you would know one particular coin was heads.

[brad]

its the monty hall switching thing

[BruceZ]

In the monty hall problem, monty *must* show you the wrong door when one of the doors you didn't pick is wrong and the other is right, so your chances improve from 1/3 to 2/3 by switching. That is what I am trying to avoid here by making a random selection of which coin to show.

[brad]

yeah i just meant the method

[Ray Zee]

easy, one player starts with two coins and the other has one. they flip until one person has all three coins. the person with two coins is a two to one favorite or 2/3 the other persons chances is 1/3.

[BruceZ]

There's a slight problem with that Ray. He said a finite number of flips. Your coinflip freeze out can go on forever. Now you're saying, that's ridiculous, of course it will always end. But for fair coins with exactly a 50% chance of heads, it can be proven that the average value of the length of the game is infinite! What is true is that it becomes very unlikely that the game will last more than N flips when N becomes very large, but this probability never goes to zero for any N, and it turns out that N gets large faster than the probability approaches zero, so the expected value of the length of the game is infinite.

Now you'll say you're still not swayed because I sense you're a practical kind of guy, and that is what you say about such things, and you will say you'll play this freeze out, and you will have every confidence that the game will end every time, and most likely it will end very quickly. You would be right to be as confident as you please, as long as you're not exactly 100% confident. You can be 99.99999% certain that it will end within a certain period of time, but never 100% certain that it will end within any period of time no matter how long. It will almost always end in a short time, but on average it will take forever.

[Ray Zee]

of course you are correct. but its between two people so they can just quit if it starts taking too long. but that wasnt the question as you say. but again since its between two people it is a finite number of flips, unless you know how to add infinity to your lifespan.

[Ed Miller]

but again since its between two people it is a finite number of flips, unless you know how to add infinity to your lifespan.

You don't need to live forever if you can learn to flip really freaking fast. /forums/images/icons/smile.gif

[BruceZ]

I said one thing that isn't right. The average length of this game isn't infinite. The average length of the game is 2 flips. The average length of the game would be infinite if one opponent started with an infinite number of coins (so he could never go broke). That is the classical result I was thinking of, and I didn't bother to do the math for this case. I was under the influence of powerful opiates (legal ones). For this case, the probability is 1/2 that it will end in 1 flip, 1/4 that it will end in 2 flips (guy with 2 coins loses twice), 1/8 that it will end in 3 flips (LWW), etc. So the average length of the game is 1(1/2) + 2(1/2)^2 + 3(1/2)^3 + ... = 2.

It's still true that there is a possibility of the game not ending for any finite number of flips, which was the main point.

If they agree in advance to quit after a certain number of flips, it will still be 2:1 if they quit after an even number of flips, but now there will be the possibility of a tie, so it's really 2:1:t where t is some fractional number of ties. If they quit after an odd number of flips, then the guy that starts with more coins will be better than 2:1 since they are quitting when the other guy is ahead, and he would have won most of these if the game had continued. If they quit after an odd number, then it is back where it started, so it would still be 2:1 if they continued.

[BruceZ]

If they quit after an odd number, then it is back where it started, so it would still be 2:1 if they continued.

Make that an even number. So you want to quit after an even number for 2:1.

[BruceZ]

This will illustrate the difference between a probability of 1 and a dead certainty that someone was asking about earlier. If we can flip as fast as we please, we can flip an infinite number of times in 1 minute. We'll just take 30 seconds for the 1st flip, 15 seconds for the second flip, 7.5 seconds for the 3rd flip, etc. After 1 minute we will have flipped an infinite number of times, or else the game will have ended at some point. It is still possible that the game never ended, and the probability of that it never ended is 0. That is, for each flip we can assign a winner and a loser so that the game doesn't end. There is only 1 way to do this, that is the winners and losers must alternate. Since there are an infinite number of ways the flips can occur, the probability that the game never ends is 0. That is, the probability that the game does end is 1. But it is not a dead certainty that it will end, because there is 1 way that it will not end.