Diciest Probability!

Probability Level 5

Satvik proposes a game of dice to Agnishom.

He says, "There are $2$ standard six-sided dice and the aim is to have the highest total of two dice. First you will throw $2$ dice. If you are happy with your total, that's great!

If you aren't happy with the total you have thrown, you can roll both dice again. However if you roll a second time, then whatever total you get, you must keep.

Now I will roll both dice. I have to accept the total I throw the first time, however if we both have rolled the same total then I win."

Given optimal strategy by Agnishom, what is the probability for Agnishom winning this game?

The answer is 0.5674.

2 solutions

Satyen Nabar
Aug 28, 2015

There are a total of 36 outcomes of rolling two dice. The probabilities for the rolls are

2, 12 = 1/36

3,11 -- 2/36

4, 10 -- 3/36

5, 10 -- 4/36

6, 8 -- 5/36

7 --- 6/36

Formulating Optimal Strategy - - -

If Agnishom rolls a 6 and keeps it, he can win only if Satvik rolls a 2, 3, 4, 5 with a 10/36 = 0.27 chance. Remember Satvik also wins if he rolls a 6. Thus its obvious that any roll 6 and below, Agnishom must reroll.

If Agnishom rolls a 8, he wins if Satvik rolls a 2, 3, 4, 5, 6, 7 with 21/36 = 0.58 chance. Thus its obvious that any roll of 8 or higher Agnishom must keep.

Now lets see what is optimal strategy if Agnishom rolls a 7. Agnishom wins if Satvik rolls a 2, 3, 4, 5, 6 with 15/36 = 0.4166 chance. Can we improve that by re-rolling ?

If you roll a second time, then you have to take the sum of the probabilities of each roll multiplied by the probability of winning with that roll. This gives --

2 : 1/36 * 0 = 0

3 : 2/36 * 1/36 = 2/1296

4: 3/36 * 3/36 = 9/1296

5: 4/36 * 6/36 = 24/1296

6: 5/36 * 10/36 = 50/1296

7: 6/36 * 15/36 = 90/1296

8: 5/36 * 21/36 = 105/1296

9: 4/36 * 26/36 = 104/1296

10: 3/36 * 30/36 = 90/1296

11: 2/36 * 33/36 = 66/1296

12: 1/36 * 35/36 = 35/1296

We add all these probabilities to get 575/1296 = 0.443 chance which has improved our chances of winning if we re-roll. So now our strategy is clear.

Any roll 7 or below, Agnishom must re-roll. And keep any roll 8 and above.

Take the probability for each roll for the first throw of the dice multiplied by the probability of winning after that first roll. Any number 8 or higher has the same probabilities calculated above.

That's 105 + 104 + 90 + 66 +35 = 400/1296 = 0. 3086 chance of winning with the first roll.

Any number 7 or lower will be re-rolled, and from above we know that the probability of winning on the second roll is 575/1296. The probability of rolling 7 or lower is 21/36. This gives a probability for 2-7 roll when you re-roll at 21/36 * 575/1296 = 12075/46656 = 0.2588.

Total probability of winning on first and second roll is 0.3086 + 0.2588 = 0.567.

Frank Petiprin
Dec 14, 2017

import random
#Monte Carlo Solution using PYTHONISTA
bet,winsat,winagn = 1,0,0
#Number of games to be played
end = 50000000
#Optimal Strategy for Agn. Determined by experimentation
optstra,die = 7, 6
random.seed(9002)
for game in range(1, end + 1):
    ra = random.randint(1, die + 1)
    ba = random.randint(1, die + 1)
    suma = ra + ba
#If true agn throws again.
    if (( suma <= optstra )):
        ra = random.randint(1, die + 1)
        ba = random.randint(1, die + 1)
        suma = ra + ba
    rs = random.randint(1, die + 1)
    bs = random.randint(1, die + 1)
    sums = rs + bs
    if (sums >= suma):
        winsat += bet
    else:
        winagn += bet
print()
print('Agn =', winagn,'Sat = ', winsat, optstra, game)
avgagn = winagn/(winagn+winsat)
avgsat = winsat/(winagn+winsat)
print('Agn = ',avgagn,'Sat = ',avgsat)
input('stop program')
#Three Program runs Two 100,000 PLAYS and One 50,000,000 PLAYS

Agn = 57505 Sat =  42495 7 100000
Agn =  0.57505 Sat =  0.42495
stop program

Agn = 57330 Sat =  42670 7 100000
Agn =  0.5733 Sat =  0.4267
stop program

Agn = 28636801 Sat =  21363199 7 50000000
Agn =  0.57273602 Sat =  0.42726398
stop program

Diciest Probability!

The answer is 0.5674.

2 solutions

0 pending reports