• The "Kebab Palace Casino" in Madsdale 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).
Determine the player's probability of winning.
I like Steve Herman's method the best, but I had already worked this out before I saw it. Consider the general case of S-sided dice.
Recall the formulae for summing n or n^2 from 1 to N:
The sum from 1 to N of n = (N)(N+1)/2
The sum from 1 to N of n^2 = (N)(N+1)(2N+1)/6
Summing a constant from 1 to N = N*constant
The dice have "S" sides, the player rolls number 'n'. Player wins if house's smaller roll was from 1 to (n-1) AND house's larger roll was from (n+1) to S. But multiply by 2 because of the order of house's high and low rolls. And player rolls 'n' with prob = 1/S.
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Player rolls 'n' wins with probability:
2 * [(n-1)/S * (S-n)/S] * (1/S)
= (2/S^3)(n-1)(S-n)
= (2/S^3)(-n^2 + (S+1)n - S)
to sum this over all n, replace by corresponding value for adding up all n or n^2 from 1 to S
= (2/S^3)(-(S)(S)(2S+1)/6 + (S+1)(S)(S+1)/2 - S^2)
= (2/S^2)(-(S+1)(2S+1)/6 + (S+1)(S+1)/2 - S)
= (2/S^2)(-(2S^2 + 3S + 1)/6 + (S^2+2S+1)/2 - S)
= (2/S^2)(-(2S^2 + 3S + 1)/6 + (S^2 + 1)/2 )
= (1/3)(1/S^2)(-(2S^2 + 3S + 1) + 3(S^2 + 1) )
= (1/3)(1/S^2)( S^2 - 3S + 2 )
= (1/3)( 1 - 3/S + 2/S^2 )
= 57/200 for S = 20
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Posted by Larry
on 2023-04-04 12:30:18 |
Let me count the ways.
