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melamori03 [73]
3 years ago
7

100-99+98-97+96-95+...+8-7+6-5+4-3+2-1

Mathematics
2 answers:
lina2011 [118]3 years ago
8 0
100-99+98-97+96-95+...+8-7+6-5+4-3+2-1\\\\=(100-99)+(98-97)+(96-95)+...+(4-3)+(2-1)\\\\=\underbrace{1+1+1+...+1}_{50}=\fbox{50}

sergij07 [2.7K]3 years ago
7 0
x=100-99+98-97+96-95+...+8-7+6-5+4-3+2-1=\\\\=1+1+1+...+1+1+1+1=1\cdot n=n\\\\and\ \ \ 2=100-2\cdot(n-1)\ \ \  \Rightarrow\ \ \ 2n=100\ /:2\ \ \ \Rightarrow\ \ \ n=50\\\\x=50
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Answer:

Expected value = 40/26 = 1.54 approximately

The player expects to win on average about $1.54 per game.

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============================================================

Further Explanation:

Let's label the three scenarios like so

  • scenario A: selecting a black card
  • scenario B: selecting a red card that is less than 5
  • scenario C: selecting anything that doesn't fit with the previous scenarios

The probability of scenario A happening is 1/2 because half the cards are black. Or you can notice that there are 26 black cards (13 spade + 13 club) out of 52 total, so 26/52 = 1/2. The net pay off for scenario A is 2-1 = 1 dollar because we have to account for the price to play the game.

-----------------

Now onto scenario B.

The cards that are less than five are: {A, 2, 3, 4}. I'm considering aces to be smaller than 2. There are 2 sets of these values to account for the two red suits (hearts and diamonds), meaning there are 4*2 = 8 such cards out of 52 total. Then note that 8/52 = 2/13. The probability of winning $10 is 2/13. Though the net pay off here is 10-1 = 9 dollars to account for the cost to play the game.

So far the fractions we found for scenarios A and B were: 1/2 and 2/13

Let's get each fraction to the same denominator

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  • 2/13 = 4/26

Then add them up

13/26 + 4/26 = 17/26

Next, subtract the value from 1

1 - (17/26) = 26/26 - 17/26 = 9/26

The fraction 9/26 represents the chances of getting anything other than scenario A or scenario B. The net pay off here is -1 to indicate you lose one dollar.

-----------------------------------

Here's a table to organize everything so far

\begin{array}{|c|c|c|}\cline{1-3}\text{Scenario} & \text{Probability} & \text{Net Payoff}\\ \cline{1-3}\text{A} & 1/2 & 1\\ \cline{1-3}\text{B} & 2/13 & 9\\ \cline{1-3}\text{C} & 9/26 & -1\\ \cline{1-3}\end{array}

What we do from here is multiply each probability with the corresponding net payoff. I'll write the results in the fourth column as shown below

\begin{array}{|c|c|c|c|}\cline{1-4}\text{Scenario} & \text{Probability} & \text{Net Payoff} & \text{Probability * Payoff}\\ \cline{1-4}\text{A} & 1/2 & 1 & 1/2\\ \cline{1-4}\text{B} & 2/13 & 9 & 18/13\\ \cline{1-4}\text{C} & 9/26 & -1 & -9/26\\ \cline{1-4}\end{array}

Then we add up the results of that fourth column to compute the expected value.

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Having an expected value of 0 would indicate a mathematically fair game, as no side gains money nor do they lose money on average.

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