Coin Toss Challenge

[Little Fishy]

Assume that by flipping any coin either a heads or a tails will always show. is there any way that by flipping a coin or multiple coins you can create an event with a probablilty of 1/3?

Note the odd man out method is not a true solution to this question.

[gaming_mouse]

[ QUOTE ]
Assume that by flipping any coin either a heads or a tails will always show. is there any way that by flipping a coin or multiple coins you can create an event with a probablilty of 1/3?

Note the odd man out method is not a true solution to this question.

[/ QUOTE ]

Flip the coin twice.

HH - Person 1 wins
HT - Person 2 wins
TH - Person 3 wins
TT - Begin again.

A winner will eventually be selected with probability 1. Also, in practice, it is unlikely that you'll ever have to do it more than 4 or 5 times to get a winner.

[BruceZ]

[ QUOTE ]
Assume that by flipping any coin either a heads or a tails will always show. is there any way that by flipping a coin or multiple coins you can create an event with a probability of 1/3?

Note the odd man out method is not a true solution to this question.

[/ QUOTE ]

Sure, here are 2 ways:

1. Play a coin flip freeze out where player A starts with 1 coin and player 2 starts with 2 coins. Player A will win 1/3 of the time.

2. Flip 2 coins, randomly choose 1, and announce the result as "selected coin is heads" or "selected coin is tails". The other party cannot see the coins, so he cannot distinguish which coin was selected. The probability that the other matches is 1/3.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Assume that by flipping any coin either a heads or a tails will always show. is there any way that by flipping a coin or multiple coins you can create an event with a probability of 1/3?

Note the odd man out method is not a true solution to this question.

[/ QUOTE ]

Sure, here are 2 ways:

1. Play a coin flip freeze out where player A starts with 1 coin and player 2 starts with 2 coins. Player A will win 1/3 of the time.

2. Flip 2 coins, randomly choose 1, and announce the result as "selected coin is heads" or "selected coin is tails". The other party cannot see the coins, so he cannot distinguish which coin was selected. The probability that the other matches is 1/3.

[/ QUOTE ]

Number 2 is 1/2 the way I have it worded, and I can't guarantee 1/3 every time without multiple flips. What I'm trying for is the old "I tell you I have 2 kids and at least one is a boy", so then the probability that I have 2 boys is 1/3. We would have to flip until we get at least 1 head, and then announce that one is a head, so the probability that the other is a head would then be 1/3.

The problem with 1 is that the game could theoretically last forever, though with probability 1 it will end after a finite number of flips.

[jason1990]

With a fixed, finite number of flips, it is not possible to create an event of probability 1/3, since any event will have probability k/2^n for some k. But as others have pointed out, there are procedures that work and that end in a finite number of flips, but this number is necessarily not almost surely bounded (there is always some positive probability that the number of flips it takes will be greater than N, no matter which N you choose).

[probman]

There is a nice answer to this type of question. Let's say you have a fair coin and want to design a game where player A wins with probability p and player B with probability 1-p. Here p can be any number between 0 and 1, even something irrational like pi. Just write the number p down in it's binary expansion. This may be infinite, but with high probability you will only need the first few digits. Then start flipping the coin. The first time the coin deviates from the binary expansion of p the game is over. If the deviation was at a place where the binary expansion of p was a 1, then player A wins. If the deviation occurs at a place where the binary expansion of p was a 0, then player B wins. The expected number of flips needed is just two, so the game will not take long with high probability. The probability of player A wining is exactly p.

[Siegmund]

As others have mentioned, there are various ways to do this, all of them requiring a random number of flips.

The shortest procedure is this one:

Flip a coin until heads comes up for the first time.
This will take an odd number of tosses 2/3 of the time, and an even number of tosses 1/3 of the time.

Expected number of tosses, 2.

[gaming_mouse]

[ QUOTE ]
As others have mentioned, there are various ways to do this, all of them requiring a random number of flips.

The shortest procedure is this one:

Flip a coin until heads comes up for the first time.
This will take an odd number of tosses 2/3 of the time, and an even number of tosses 1/3 of the time.

Expected number of tosses, 2.

[/ QUOTE ]

Clever. I like this solution.

[probman]

[ QUOTE ]
As others have mentioned, there are various ways to do this, all of them requiring a random number of flips.

The shortest procedure is this one:

Flip a coin until heads comes up for the first time.
This will take an odd number of tosses 2/3 of the time, and an even number of tosses 1/3 of the time.

Expected number of tosses, 2.

[/ QUOTE ]

The solution is interesting, but it is no faster on average than the one previously given. Since the coin is fair, the waiting time for the first head is statistically identical to the waiting time for the coin flips to deviate from a predetermined series.

[Siegmund]

Yes, it's just a special case of the binary series method you proposed (the case of 00000.... or 11111....). What I meant is that it was faster than solutions of the "flip two coins, and if they are both heads, flip them again" type.