Craps with 3 Die

Probability Level 3

Kenji and Manbu want to play a different game of craps with three dice. Kenji's dice are all 6-sided die while Manbu has one each of a 4-, 6-, and 8-sided die. Manbu reasons with Kenji that his dice will have an equal advantage as (4+6+8)/3 = 18/3=6. So, Kenji agrees to play.

Whoever rolls a 7 or 11 with their three dice on the first roll wins. Whoever rolls a 3, 4, or 18 on the first roll craps out. Does Kenji or Manbu have a greater probability of winning the first roll?

Manbu has advantage Kenji has advantage Both have equal chance

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2 solutions

Chew-Seong Cheong
Feb 11, 2020

Kenji

  • Total number of outcomes of Kenji's three dice, N k = 6 × 6 × 6 = 216 N_k = 6 \times 6 \times 6 = 216
  • Number of ways to roll a 7, N k 7 = 15 N_{k7} = 15
  • Number of ways to roll an 11, N k 11 = 27 N_{k11} = 27
  • Number of ways to roll a 3, N k 3 = 1 N_{k3} = 1
  • Number of ways to roll a 4, N k 4 = 3 N_{k4} = 3
  • Number of ways to roll an 18, N k 18 = 1 N_{k18} = 1
  • Probability of rolling a win, p k = 15 + 27 216 = 7 36 p_k = \frac {15+27}{216} = \frac 7{36}
  • Probability of being crapped, q k = 1 + 3 + 1 216 = 5 216 q_k = \frac {1+3+1}{216} = \frac 5{216}

Manbu

  • Total number of outcomes of Manbu's three dice, N m = 4 × 6 × 8 = 192 N_m = 4 \times 6 \times 8 = 192
  • Number of ways to roll a 7, N m 7 = 14 N_{m7} = 14
  • Number of ways to roll an 11, N m 11 = 23 N_{m11} = 23
  • Number of ways to roll a 3, N m 3 = 1 N_{m3} = 1
  • Number of ways to roll a 4, N m 4 = 3 N_{m4} = 3
  • Number of ways to roll an 18, N m 18 = 1 N_{m18} = 1
  • Probability of rolling a win, p m = 14 + 23 192 = 37 192 p_m = \frac {14+23}{192} = \frac {37}{192}
  • Probability of being crapped, q m = 1 + 3 + 1 192 = 5 192 q_m = \frac {1+3+1}{192} = \frac 5{192}

Therefore,

  • Probability that Kenji wins, P k = p k + q m = 7 36 + 5 192 = 381 1728 P_k = p_k + q_m = \frac 7{36} + \frac 5{192} = \frac {381}{1728}
  • Probability that Manbu wins, P m = p m + q k = 37 192 + 5 216 = 373 1728 P_m = p_m + q_k = \frac {37}{192} + \frac 5{216} = \frac {373}{1728}
  • Kenji has advantage
0 Helpful 0 Interesting 0 Brilliant 0 Confused

The last part of this is not correct. The probability P k P_k that Kenji wins is not p k + q m p_k + q_m in any scenario that makes sense. True, the question does not make clear what is really going on.

  • One interpretation simply asks which is greater, p k p_k or p m p_m ? In other words, who has the better chance of winning outright in a single round.
  • You are thinking that they play competitively, both throwing at the same time. Thus Kenji wins in the first round if either he wins outright and Manbu does not (otherwise it presumably is a tie) or else if he does not either win or crap out and Manbu does crap out. The probability of this is p k ( 1 − p m ) + ( 1 − p k − q k ) q m p_k(1-p_m) + (1 - p_k - q_k)q_m , which is not equal to p k + q m p_k + q_m .
  • Alternatively, we could ask who has the greater chance of winning (noncompetitively). Kenji's chance of winning is p k p k + q k \frac{p_k}{p_k + q_k} , while Manbu's is defined similarly.
  • Finally, we could simply ask who has the greater chance of winning if they play competively (allowing for the fact that they could draw). This makes Kenji's chance of winning equal to p k ( 1 − p m ) + ( 1 − p k − q k ) q m p k ( 1 − p m ) + ( 1 − p k − q k ) q m + p m ( 1 − p k ) + ( 1 − p m − q m ) q k \frac{p_k(1 - p_m) + (1 - p_k - q_k)q_m}{p_k(1 - p_m) + (1 - p_k - q_k)q_m + p_m(1 - p_k) + (1 - p_m-q_m)q_k}

Mark Hennings - 1 year, 4 months ago

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Thanks for the explanation.

Chew-Seong Cheong - 1 year, 4 months ago

0 Helpful 0 Interesting 0 Brilliant 0 Confused

The question asks for the probabilities of winning on the first roll. Thus crapping out or a null result are both cases of not winning. Thus all we need to consider is that fact that 7 36 > 37 192 \tfrac{7}{36} > \tfrac{37}{192} to deduce that Kenji has a better chance of winning on the first roll.

The fact that Kenji also has a smaller chance of crapping out reinforces this probability differential, and makes it even more likely that Kenji will win overall. Kenji's overall chance of winning is 42 47 \tfrac{42}{47} , while Manbu's is 37 42 \tfrac{37}{42} .

Mark Hennings - 1 year, 4 months ago

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