Suppose I am playing a game.

I bet $2.

There is a 40% chance that I win my $2 back plus another $1
There is a 20% chance that I win $2 back (break even)
There is a 40% chance that I lose my $2 bet.

Using the box model and R, calculate the EV of the sample sum of playing this game 10 times.

Question

Suppose I am playing a game.

I bet $2.

There is a 40% chance that I win my $2 back plus another $1
There is a 20% chance that I win $2 back (break even)
There is a 40% chance that I lose my $2 bet.

Using the box model and R, calculate the EV of the sample sum of playing this game 10 times.

BlackTom AI · Accepted Answer

Each play's net gain values are +1 (win $2 back plus $1), 0 (break even), and −2 (lose the $2). Represent these as tickets in a box with probabilities 0.4, 0.2, and 0.4 respectively.

1) Relevant concepts
- Box model: represent each possible outcome as tickets; the expected value (EV) of one draw is the weighted average of ticket values.
- Linearity of expectation: the EV of a sum of independent draws equals the sum of their EVs (so n times the single-play EV).
- Independence: repeated plays are independent, so we can multiply the single-play EV by the number of plays.

2) Step-by-step reasoning
- Compute EV for one play (X):
  E(X) = P(+1)*(+1) + P(0)*(0) + P(−2)*(−2).
- Then compute EV for 10 plays S10 = X1 + X2 + ... + X10:
  E(S10) = 10 * E(X) by linearity.

3) Calculations (intermediate steps)
- E(X) = 0.4*(+1) + 0.2*(0) + 0.4*(−2)
       = 0.4 + 0 + (−0.8)
       = −0.4.
- E(S10) = 10 * (−0.4) = −4.

R implementation (box model):
- box <- c(rep(1,4), rep(0,2), rep(-2,4))  # 10 tickets reflecting 40%,20%,40%
- mean(box)            # gives −0.4
- 10 * mean(box)       # gives −4

4) Why this is correct
- The weighted average correctly accounts for probabilities. Multiplying by 10 is justified because expectation is linear and each play is independent, so total expected net loss equals 10 times the per-play expected loss.

5) Other possible mistakes (why alternatives would be wrong)
- Forgetting to multiply by 10 yields −0.4 (per-play EV), not the sample sum EV.
- Misinterpreting payouts (e.g., using +2 instead of +1) would give an incorrect EV.
- Multiplying probabilities incorrectly or swapping signs could produce wrong positive/negative sign.

6) Conclusion
The expected value of the sample sum after 10 plays is −4, which matches the provided answer: -4.

Suppose I am playing a game. I bet $2. There is a 40% chance that I win my $2 back plus another $1 There is a 20% chance that I win $2 back (break even) There is a 40% chance that I lose my $2 bet. Using the box model and R, calculate the EV of the sample sum of playing this game 10 times.数值题

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