Brain Teasers
Kebab Palace Casino
Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.Probability
Just to be different, the "Kebab Palace Casino" in Madadia has dice with 20 sides.
The "house" rolls two 20-sided dice and the "player" rolls one 20-sided die. If the player rolls a number on his die between the two numbers the house rolled, then the player wins. Otherwise, the house wins (including ties).
What is the player's probability of winning?
The "house" rolls two 20-sided dice and the "player" rolls one 20-sided die. If the player rolls a number on his die between the two numbers the house rolled, then the player wins. Otherwise, the house wins (including ties).
What is the player's probability of winning?
Answer
It is easily seen that if any two of the three dice agree that the house wins. The probability that this does not happen is 19*18/(20*20).If the three numbers are different, the probability of winning is 1/3.
So the chance of winning is 19*18/(20*20*3) = 3*19/200 = 57/200.
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Comments
Actually, the odds of being between the numbers isn't 1/3, as you could also tie.
I made a spreadsheet to solve this one:
Column A: Number of digits apart the house's two dice are (from 0 to 19)
Column B: Ways you can win given these possibilities (row 1: 0, rows 2-20: 0-1.
Column C: Number of ways you can come up with the house dice in this fashion:
(from 20 ways to 1 way)
Column D: Total number of winning possibilities (Column B times Column C)
A B C D
Both are same 0 20 0
1 digit apart 0 19 0
2 digits apart 1 18 18
3 digits apart 2 17 34
4 digits apart 3 16 48
5 digits apart 4 15 60
6 digits apart 5 14 70
7 digits apart 6 13 78
8 digits apart 7 12 84
9 digits apart 8 11 88
10 digits apart 9 10 90
11 digits apart 10 9 90
12 digits apart 11 8 88
13 digits apart 12 7 84
14 digits apart 13 6 78
15 digits apart 14 5 70
16 digits apart 15 4 60
17 digits apart 16 3 48
18 digits apart 17 2 34
19 digits apart 18 1 18
total: 1140
Compare this to the total number of ways to roll the dice (20*20*20), which is 8000, and you get 1140/8000, or 14.25%.
I made a spreadsheet to solve this one:
Column A: Number of digits apart the house's two dice are (from 0 to 19)
Column B: Ways you can win given these possibilities (row 1: 0, rows 2-20: 0-1.
Column C: Number of ways you can come up with the house dice in this fashion:
(from 20 ways to 1 way)
Column D: Total number of winning possibilities (Column B times Column C)
A B C D
Both are same 0 20 0
1 digit apart 0 19 0
2 digits apart 1 18 18
3 digits apart 2 17 34
4 digits apart 3 16 48
5 digits apart 4 15 60
6 digits apart 5 14 70
7 digits apart 6 13 78
8 digits apart 7 12 84
9 digits apart 8 11 88
10 digits apart 9 10 90
11 digits apart 10 9 90
12 digits apart 11 8 88
13 digits apart 12 7 84
14 digits apart 13 6 78
15 digits apart 14 5 70
16 digits apart 15 4 60
17 digits apart 16 3 48
18 digits apart 17 2 34
19 digits apart 18 1 18
total: 1140
Compare this to the total number of ways to roll the dice (20*20*20), which is 8000, and you get 1140/8000, or 14.25%.
I used pen and paper, so it is possible that human error is involved! sorry
I used a spreadsheet also, recording all 400 possible combinations that the house could throw, and determining how many numbers I could roll to win with each combination. I got the same answer as Mad. Quax, I think you accounted for only half the possibles by examining only the difference between the 2 die. You did not consider that it doesn't matter which die has the higher number.
Why'd you have to make this so complicated
I agree with Mad and Gizzer. I did a spreadsheet and came up with the same answer they did.
Excellent question (and it is definitely 28.5% -- just to be sure, I ran all 8000 3-roll cases in Excel...)
I guess I got a wrong answer again. I couldn't follow my own logic and math, so I sure ain't gonna try to understand someone else's.
I'm getting dumber by the minute.
The given solution 57/200 is elegant and correct.
I took a similar approach with different math. There are 20^3 = 8000 ways to roll three 20-sided dice. There are 20C3 = 1140 ways to choose three different values. For each combination of three dice, there are two permutations where the player wins (where the player has middle and the house has low/high or high/low).
This gives a probability of (2 x 1140)/8000 = 2280/8000 = 57/200.
I took a similar approach with different math. There are 20^3 = 8000 ways to roll three 20-sided dice. There are 20C3 = 1140 ways to choose three different values. For each combination of three dice, there are two permutations where the player wins (where the player has middle and the house has low/high or high/low).
This gives a probability of (2 x 1140)/8000 = 2280/8000 = 57/200.
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