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So, this is basically a bet. Someone, say Tim, chooses a number from 1 to 6 and then I roll 3 dice. if Tim's number shows up in exactly 1 die, the bet is tied and I give him back the money, if the number shows up in exactly 2 dice, i pay him 2x the amount he bet, if the number shows up in exactly 3 dice, i pay him 3x the amount he bet, and if the given number doesn't show up in any die, he loses the money.

My question is, what are the probabilities for each possible scenario?

Thank you very much

  • 0
    My question is; Have you studied **Binomial Distributions** recently? – 2017-02-14

1 Answers 1

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(a) Probability of $3$ is $^3C_3(\frac{1}{6})^3$.

(b) Probability of $2$ is $^3C_2(\frac{1}{6})^2(\frac{5}{6})$

(c) Probability of $1$ is $^3C_1(\frac{1}{6})(\frac{5}{6})^2$

(d) Probability of $0$ is $^3C_0(\frac{5}{6})^3$

Expected return is (d)-(b)-2(a) times the stake.

You can expect to come out up over a series of bets, but I'll leave the arithmetic to you.

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    It might be more useful if you replaced the threes with $3 \choose 2$ and $3 \choose 1$ to show where the threes come from, and then do the same for (a) and (d) to show why the coefficient is 1. – 2017-02-14