# Interesting probability problem

• Alkatran
In summary, the conversation revolves around trying to solve a problem involving the payout from a probability game. The game involves flipping a coin and winning either $2 or$0, depending on the outcome. There is also an event with a certain probability that forces the player to play again. The objective is to determine the exact payout for this game. The conversation includes a calculation for the payout if the player always wins, but it is wrong and negative for all values of p from 0 to 1. The correct equation for the payout is given as the sum of payoffs for an n-round game, which is also discussed in the conversation. The conversation also mentions mistakes in the derivation of the initial sum and clarifies that the game in question only

#### Alkatran

Homework Helper
I'm trying to solve a problem involving the payout from a probability game.

Here it is:

The game goes something like this:

pay a dollar
Flip a coin
If it's heads, you win 2$, otherwise you win 0$
If (some event which has chance P)
you are forced to play again

You must stop if your initial dollar is lost. (Essentially, you either move up, down, stop, or are forced to stop if you hit 0).

At first glance this looks like a fair game, and maybe it is, but what is the EXACT payout?

All I've tried to calculate so far is the payout if you always win (but stop) (not including cost of playing).

$$payout = \sum_{n=1}^\infty (\frac{1}{2})^n*p^{n-1}*(n+1)$$
$$= \sum_{n=0}^\infty (\frac{1}{2})^{n+1}*p^n*(n+2)$$
$$= \frac{1}{2} \sum_{n=0}^\infty (\frac{1}{2})^n*p^n*(n+2)$$
$$= \frac{1}{p} \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2)$$
$$= \frac{1}{p} \frac{d}{dp}{\int \sum_{n=0}^\infty (\frac{p}{2})^{n+1}*(n+2) dp}$$
$$= \frac{1}{2*p} \frac{d}{dp} \sum_{n=0}^\infty (\frac{p}{2})^{n+2}$$
$$= \frac{1}{2*p} \frac{d}{dp} ( \frac{1}{1 - p/2} - 1 - \frac{p}{2} )$$
$$= \frac{1}{2*p} \frac{1}{(1 - p/2)^2} * \frac{-1}{2} - \frac{1}{2}$$
$$= \frac{1}{2*p} *( \frac{-1}{(2 - p)^2} - \frac{1}{2} )$$
$$= \frac{-1}{2*p} * ( \frac{1}{(p - 2)^2} - \frac{1}{2} )$$

I know this is wrong because it's negative for all p from 0 to 1 (and the game should be at least fair in this case)

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Payoff for an n-round game:

$$P_1=0/2+2/2-1=0$$
$$P_2=0/2+0/4+4/4-1=0$$
$$P_3=0/2+0/4+2/8+6/8-1=0$$
$$P_3=0/2+0/4+0/16+4/16+4/16+8/16-1=0$$
. . .

You can then sum these

$$\sum_{i=1}^\infty(1-p)^{n-1}pP_n$$

CRGreathouse said:
Payoff for an n-round game:

$$P_1=0/2+2/2-1=0$$
$$P_2=0/2+0/4+4/4-1=0$$
$$P_3=0/2+0/4+2/8+6/8-1=0$$
$$P_3=0/2+0/4+0/16+4/16+4/16+8/16-1=0$$
. . .

You can then sum these

$$\sum_{i=1}^\infty(1-p)^{n-1}pP_n$$

I think you have the wrong game in mind. You can only win or lose a dollar in this game, not double or nothing. I already tried the double-or-nothing and found you broke even.

I'm not differentiating with respect to p: I'm integrating AND differentiating with respect to p, effectively doing nothing. We did this in calculus to solve series with factors of n+x in them.

I know this is wrong because it's negative for all p from 0 to 1 (and the game should be at least fair in this case)

It's not negative for all p from 0 to 1. (For example, p = 1/2)

Did you notice your initial sum only converges for -1/2 <= p < 1/2? Frankly, I'd be surprised if your answer wasn't negative for p = 1.

(It disturbs me that it's positive for p=1/2 and undefined for p = 0, though)

I'm not yet convinced your initial sum is right though...

I see two mistakes in your derivation:

(1) You got the derivative of 1/(1-p/2) wrong.
(2) When you factored out the minus sign, you only pulled it from one of the terms.

(I'm not promising there aren't more!)

Alkatran said:
I think you have the wrong game in mind. You can only win or lose a dollar in this game, not double or nothing. I already tried the double-or-nothing and found you broke even.

I'm not calculating for double or nothing -- look at my numbers again. The possible payoffs for the game I'm calculating are -1, 1, 3, 5, ...

Can you give an example of play? You said "Essentially, you either move up, down, stop, or are forced to stop if you hit 0", but this was a little confusing.

Oh, I know why I wasn't convinced your sum was right -- I didn't notice you were only modelling the case of always winning!

## 1. What is a probability problem?

A probability problem is a mathematical question that involves determining the likelihood or chance of a certain event occurring. It requires the use of mathematical concepts and formulas to calculate the probability of a particular outcome.

## 2. What makes a probability problem interesting?

A probability problem is considered interesting when it involves a situation that is relatable or relevant to real-life scenarios. It may also be interesting if it challenges traditional assumptions about probability or involves a unique and creative approach to solving it.

## 3. How do you solve a probability problem?

To solve a probability problem, you must first clearly define the event or outcome you are interested in and identify all the possible outcomes. Then, use the appropriate formula or method to calculate the probability of the desired outcome. Finally, interpret the results in the context of the problem.

## 4. What are some common types of probability problems?

Some common types of probability problems include calculating the probability of a single event, calculating the probability of multiple events occurring together, and using probability to make predictions or decisions.

## 5. Why is understanding probability important?

Understanding probability is important because it allows us to make informed decisions and predictions based on the likelihood of certain events occurring. It is also essential in fields such as statistics, finance, and science, where probability is used to analyze data and make accurate conclusions.