# Find the expected value of a coin flipping game

• docnet
In summary: EV = P(3) * 0 + P(4) * 4 + P(5) * 5 + P(6) * 6EV = .23445 * (4 + 5 + 6) = 3.46065In summary, the game involves tossing six coins and getting a certain amount of money based on the number of heads obtained. If you get 3 or fewer heads, you play again and get the same amount of money. However, if you get 4 or more heads, you get the number of heads as dollars. The expected value of the game is approximately 3.46.
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Homework Statement
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Relevant Equations
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summary: you toss six coins. if you have 3 heads, you don't get any money. if you have 4 or more heads, you get the number of heads amount of dollars. if you have 2 or less heads, you toss six coins again and get the number of heads amount of dollars.

the number of ways to get 0 head ##\frac{6!}{(6)!1!}=1##
the number of ways to get 1 head ##\frac{6!}{(6-1)!1!}=6##
the number of ways to get 2 heads ##\frac{6!}{(6-2)!2!}=15##
the number of ways to get 3 heads ##\frac{6!}{(6-3)!3!}=20##
the number of ways to get 4 heads ##\frac{6!}{(6-4)!4!}=15##
the number of ways to get 5 heads ##\frac{6!}{(6-2)!2!}=6##
the number of ways to get 6 heads ##\frac{6!}{(6-2)!2!}=1##

$$P(X<3)=\frac{22}{2^6}$$
$$P(X=3)=\frac{20}{2^6}$$
$$P(X>3)=\frac{22}{2^6}$$

$$E(X<3)=E(X)=3$$
$$E(X=3)=0$$
$$E(X>3)=\frac{15}{21}\cdot 4+\frac{6}{21}\cdot 5+\frac{1}{21}\cdot 6=\frac{32}{7}$$

the expected value of the game is
$$P(X<3)E(X<3)+P(X=3)E(X=3)+P(X>3)E(X>3)$$
$$=\frac{22}{2^6}\cdot 3 +\frac{20}{2^6}\cdot 0 + \frac{22}{2^6} \frac{32}{7}=\frac{583}{224}\approx 2.6$$

If I was given this question, the first thing I would say is "what happens if you get exactly 3 heads?" The question doesn't say. You seem to be assuming this means you get nothing, but is that correct? I'd say the game is not fully specified.
PS in calculating E(X>3), the probabilities should be 15/22 etc., not 15/21 etc.

Master1022, docnet and FactChecker
mjc123 said:
If I was given this question, the first thing I would say is "what happens if you get exactly 3 heads?" The question doesn't say. You seem to be assuming this means you get nothing, but is that correct? I'd say the game is not fully specified.
PS in calculating E(X>3), the probabilities should be 15/22 etc., not 15/21 etc.
oops yes rhat was a mistake. It turns out my professor changed the prompt to "equal to greater than 3 heads".$$P(X\geq 3)=\frac{42}{2^6}$$
$$E(X\geq 3)=\frac{20}{22}\cdot 3+\frac{15}{22}\cdot 4+\frac{6}{22}\cdot 5+\frac{1}{22}\cdot 6=\frac{78}{11}$$

the expected value of the game is
$$\frac{22}{2^6}\cdot 3 +\frac{42}{2^6}\frac{78}{11}= \frac{2001}{352}\approx5.68$$

docnet said:
oops yes rhat was a mistake. It turns out my professor changed the prompt to "equal to greater than 3 heads".$$P(X\geq 3)=\frac{42}{2^6}$$
$$E(X\geq 3)=\frac{20}{22}\cdot 3+\frac{15}{22}\cdot 4+\frac{6}{22}\cdot 5+\frac{1}{22}\cdot 6=\frac{78}{11}$$

the expected value of the game is
$$\frac{22}{2^6}\cdot 3 +\frac{42}{2^6}\frac{78}{11}= \frac{2001}{352}\approx5.68$$
Can you write a computer simulation to check that?

In any case, that number must be too high. It means you're getting 5-6 heads almost every time.

docnet
PeroK said:
Can you write a computer simulation to check that?

In any case, that number must be too high. It means you're getting 5-6 heads almost every time.
ahhhhh! I forgot to divide everything in ##E(X\geq 3)## by 42 instead of 22 so ##E(x\geq 3)## is smaller.

$$E(X\geq 3)=\frac{20}{42}\cdot 3+\frac{15}{42}\cdot 4+\frac{6}{42}\cdot 5+\frac{1}{42}\cdot 6=\frac{26}{7}$$

the expected value of the game is
$$\frac{22}{2^6}\cdot 3 +\frac{42}{2^6}\frac{26}{7}\approx 3.46$$

And I have never tried writing a computer simulation to check my answers. I wouldn't know where to start, tbh.

PeroK
docnet said:
ahhhhh! I forgot to divide everything in ##E(X\geq 3)## by 42 instead of 22 so ##E(x\geq 3)## is smaller.

$$E(X\geq 3)=\frac{20}{42}\cdot 3+\frac{15}{42}\cdot 4+\frac{6}{42}\cdot 5+\frac{1}{42}\cdot 6=\frac{26}{7}$$

the expected value of the game is
$$\frac{22}{2^6}\cdot 3 +\frac{42}{2^6}\frac{26}{7}\approx 3.46$$

And I have never tried writing a computer simulation to check my answers. I wouldn't know where to start, tbh.

I think it's easier to observe that if you get fewer than 3 heads the first time, you play a game whose expected outcome is $3 and if you get at least 3 heads the first time you get an amount equal to the number of heads. So the expectation should be $$\frac{22}{2^6} \cdot 3 + \frac{1}{2^6}\left( 20 \cdot 3 + 15\cdot 4 + 6 \cdot 5 + 6 \right)$$ which is your result. However, rather than do that calculation I would say that the expected number of heads from flipping six coins is 3 and this game has a strictly higher expectation. So at$3 I know my expected outcome is positive.

pasmith said:
So the expectation should be $$\frac{22}{2^6} \cdot 3 + \frac{1}{2^6}\left( 20 \cdot 3 + 15\cdot 4 + 6 \cdot 5 + 6 \right).$$
That is the calculation that leads to ##3.46##.

another approach, binomial with ... n=6 ... k=0, 1, 2 ... p=0.5

P = [n! / (k! * (n-k)!)] * [(p**k) * (p**(n-k))]

2nd portion in square brackets will always be .01563 when p=0.5 and n=6

P(0) = [6! / (0! * 6!)] * .0156 = 1 * .0156 = .01563
P(1) = [6! / (1! * 5!)] * .0156 = 6 * .0156 = .09378
P(2) = [6! / (2! * 4!)] * .0156 = 15 * .0156 = .23445

P(3+) = 1 - 01563 - .09378 - .24445 = .65614

Last edited:

## 1. What is the expected value of a coin flipping game?

The expected value of a coin flipping game is the average amount that a player can expect to win or lose over the long run. It is calculated by multiplying the probability of each outcome by its associated payout and then summing up all the values.

## 2. How do you calculate the expected value of a coin flipping game?

To calculate the expected value of a coin flipping game, you need to multiply the probability of each outcome (heads or tails) by its associated payout (usually 1 or -1). Then, add up all the values to get the final expected value.

## 3. What is the significance of the expected value in a coin flipping game?

The expected value in a coin flipping game helps us understand the long-term outcome of the game. It tells us the average amount that a player can expect to win or lose over time, and helps us make informed decisions about whether or not to play the game.

## 4. Is the expected value always the same in a coin flipping game?

No, the expected value can vary in a coin flipping game depending on the rules and payouts. For example, if the payout for a correct guess is 2 instead of 1, the expected value will change accordingly.

## 5. How can the expected value be used to make decisions in a coin flipping game?

The expected value can be used to make decisions in a coin flipping game by comparing it to the cost of playing the game. If the expected value is positive, it means the player can expect to win money over time and it may be worth playing. If the expected value is negative, it means the player can expect to lose money over time and it may be best to avoid the game.

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