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Suppose we have a roulette wheel with $38$ total slots ($18$ black/red, $2$ neither). The ball lands in one slot selected uniformly at random, independent of all previous spins. An $\$x$ bet on "black" pays $\$2x$ if the ball lands on black, and $\$0$ otherwise.

If you bet $1 on black for 100 consecutive spins, how much money will you end up with in expectation?

You walk up to the Roulette table with $\$15$ and intend to walk away with $\$16$ using the following betting strategy. You will first bet $\$1$ on black. If you win you have $\$16$ and walk away, otherwise you have $\$14$ and bet $\$2$ on black on the next spin. If you win that bet you walk away, otherwise you bet $\$4$ on the next spin. If you win that bet you walk away, otherwise you bet $\$8$ on the next spin and walk away, either with $\$16$ if you win or $\$0$ if you lose.

What's the probability you will walk away with $\$16$?

How much money are you expected to walk away with?

My biggest problem with these kinds of questions is figuring out how to choose the random variable to calculate the expected values. I understand the formula for calculating the expected value, but translating a problem into those terms has been giving me a hard time.

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3 Answers 3

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Let's first look at the first question:

If you bet $1 on black for 100 consecutive spins, how much money will you end up with in expectation?

So you want to know what your final return will be, at the end of 100 spins. Call this $R$. That is just giving it a name, but what is your final return? You can see that it is the sum of the returns from each bet. So let the return on the $i$th bet be $R_i$, then note that $R = R_1 + R_2 + \dots + R_{100}$. So the expected return is $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}]$ by linearity of expectation.

So to calculate $E[R]$, we'll be done if we calculate each $E[R_i]$. Let's try to calculate a particular $E[R_i]$. You bet $1$ dollar, and you get back $2$ if the ball lands on black, and $0$ if it doesn't. In other words, you gain $1$ dollar if it lands on black, and lose $1$ dollar if it doesn't. The probability of the former is $18/38$, and that of the latter is $20/38$. In other words, $R_i$ is $1$ with probability $18/38$, and $-1$ with probability $20/38$, so the expected value of $R_i$ is $E[R_i] = \frac{18}{38}(1) + \frac{20}{38}(-1) = \frac{-2}{38}$. Now, as this is the same for each $R_i$, we have $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}] = \left(\frac{-2}{38}\right)100 \approx -5.26$.


For the second question, let the amount you walk away with be $W$. Let $p = 18/38$, the probability that your bet on black succeeds. There are $5$ possible outcomes:

  • you win your first bet: probability $p$
  • you lose your first bet, and win your second: probability $(1-p)p$
  • you lose your first two bets, and win the third: probability $(1-p)^2p$
  • you lose your first three bets, and win the fourth: probability $(1-p)^3p$
  • you lose all four bets: probability $(1-p)^4$

In the first four outcomes, you walk away with $16$ dollars, so the probability of that happening (let's call it $q$) is $q = p + (1-p)p + (1-p)^2p + (1-p)^3p = 1 - (1-p)^4 = 1 - (20/38)^4 \approx 0.92$.

[More simply, you could think of it as just two outcomes: (a) that you win some bet, which has probability $q = 1 - (1-p)^4$, and (b) that you win no bet (lose all bets), which has probability $(1-p)^4$.]

In other words, $W$ is $16$ with probability $q$, and $0$ with probability $1-q$. So the expected amount of money you walk away with is therefore $E[W] = q(16) + (1-q)0 = (1-(1-p)^4)16 \approx 14.77$.

[Aside: Note that this is less than the $15$ you came in with. This shows that you can't win in expectation even with your clever betting strategy; a consequence of the optional stopping theorem.]

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  • $\begingroup$ Spotted a small error: It should be E[Ri] = 18/38 * 1 + 20/38 * -1 $\endgroup$
    – nknj
    Jul 22, 2015 at 15:52
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Probability of getting a black on any particular spin is 18 in 38, ie $p = \frac{18}{38} = \frac{9}{19} = 0.474$. Let $X$ be the random variable 'number of blacks in 100 consecutive spins'. Then the expected value of X is $E(X) = 100p = \frac{900}{19} = 47.37$, that is, you expect to win 47.37 times in 100 spins. Since you get $\$$2 for each $\$$1 bet, you will therefore expect to walk away with $\$2\times 47.37 = \$94.74$.

For the second part of your question, the only way you can lose is if for four consecutive spins, you get non-blacks each time. So what is the probability of losing 4 times in a row? Well, for 1 spin, the probability of not getting black is $1-p = \frac{20}{38} = \frac{10}{19} = 0.526$ so for 4 consecutive spins, the probability of not getting any black is $(\frac{10}{19})^{4} = \frac{10,000}{130,321} = 0.0767$. That is, you have about a 7.7% chance of losing with this betting strategy. That is, you have 92.3% chance of walking away with $\$$16.

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  • $\begingroup$ Can you elaborate on why the expected value is just $100p$? Particularly, how does the linearity of expectations apply here? If we let $X_i = 1$ if we land on black and $X_i = 0$ if we don't land on black, how can we calculate the expected value then? $\endgroup$ Nov 14, 2012 at 10:53
  • $\begingroup$ Also, for the second part, how would we calculate the "amount of money you are expected to walk away with"? $\endgroup$ Nov 14, 2012 at 10:55
  • $\begingroup$ Expected value of a bet = wager x probability. $\endgroup$
    – theo
    Nov 14, 2012 at 11:02
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    $\begingroup$ If you bet 1 dollar 100 times on black you will win about 47.37 times and thus leave the casino with 94.74 of your original 100 dollars. $\endgroup$ Nov 14, 2012 at 11:03
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    $\begingroup$ The answer given is -$(2/38)*(100), so I believe this is incorrect. $\endgroup$ Nov 14, 2012 at 11:20
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1.

38 total slots (18 black/red, 2 neither)

P(lands on black) = 18/38 = 9/19.

An \$x bet on "black" pays \$2x if the ball lands on black, and \$0 otherwise.

E(one \$1 bet on black) = \$(2*P(lands on black) + 0*P(¬land on black)) = \$2*9/19 = \$18/19

E(a hundred \$1 bets on black) = 100*E(one \$1 bet on black) = 100*\$18/19 ≈ \$94.74 (rounding up)

2.

15 = 1 + 2 + 4 + 8

P(Wins) = 1-P(¬Win) = 1-9/19 = 10/19

P(Wins once, 4 tries) = 1-P(loses 4 times out of 4 tries) = 1-(10/19)^4 = 1 - $\frac{10,000}{130,321}$ = $\frac{120,321}{130,321}$ ≈ 92.3% (rounding down/truncating).

E(Strategy) = P(Success)*Payout - Cost of Playing = $\frac{120,321}{130,321}$*\$16 - \$15 ≈ -\$0.2277

The house always wins means the expected value of playing is always negative.

Playing most lotteries/gambling is usually considered 'irrational' because the expected value of winning is less than the cost of a ticket/playing.

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