### Brain Teasers

# Heads Up!

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

One day, two friends named Al and Bob found a shiny quarter on the ground. To decide who would get it, Al would flip the coin. If it came up heads, then Al would keep it, but if it was tails, Bob would have to flip for it. If Bob got heads, Bob would keep the coin, but if he got tails, then Al would flip it and the entire process would repeat. Assuming that Al goes first and that the coin has an equal chance of landing on either heads or tail, what is the probability that Al would keep the coin?

### Hint

It isn't 1/2.### Answer

2/3If Al flipped heads, then he would win, and the probability of that happening would be 1/2. However, he could also win if he flipped tails, Bob flipped tails, and Al flipped heads afterwards. The probability of this happening would be:

(1/2)(1/2)(1/2)=1/8

If this process were to continue indefinitely, then the probability of Al winning would be:

(1/2)+(1/8)+(1/32)+...+(1/2)(1/4)^n

in which n equals an integer one greater than the end preceding it and one less than one integer after it. Using the formula to obtain a solution for the summation of a geometric sequence, it is possible to deduce that the summation equals:

(1/2)/((1-(1/4))

which is simplified to 2/3.

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## Comments

Very clever.

Nice job!!!

Wahoo now I can tell all the tutees a use for the summation of a geometric series

There is a simpler solution:

Let P be the probability of keeping the coin, then with prob. 1/2 Al gets heads ans keeps the coin, and with prob. 1/2 he has to give the coin to Bob. In the latter case, Bob is in the same situation as Al, so probability that he will keep the coin is P.

Using this, we can write

P = 1/2 + 1/2 * (1 - P).

Which gives P = 2/3.

Let P be the probability of keeping the coin, then with prob. 1/2 Al gets heads ans keeps the coin, and with prob. 1/2 he has to give the coin to Bob. In the latter case, Bob is in the same situation as Al, so probability that he will keep the coin is P.

Using this, we can write

P = 1/2 + 1/2 * (1 - P).

Which gives P = 2/3.

Neat problem. Even neater solution Ulan!

Fun!

Since the problem is recursive, I solved it via the following:

Let p=the chance of Al winning when he flips the coin first; q=chance of Al winning when Bob flips the coin.

We then have p=1/2+1/2*q and q=1/2*p. Solving for two equations and two unknows, we have p=2/3.

Fun teaser!

Let p=the chance of Al winning when he flips the coin first; q=chance of Al winning when Bob flips the coin.

We then have p=1/2+1/2*q and q=1/2*p. Solving for two equations and two unknows, we have p=2/3.

Fun teaser!

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