A wealthy mathematician WG offered you a proposition you cannot refuse: participate for one hour exactly in one of the following games:
a. Throw four regular fair dice and if at least one six appears you get a dollar, otherwise you pay a dollar.
b. Keep executing throws of two fair dice until both show a six - you win a dollar , but if 25 throws were unsuccessful you lose a dollar and begin a new round of 25 or less (when 2 sixes appeared earlier).
If the time limit expires within the “25 throw” sequence the current run is aborted.
WG thinks that in both cases the chances are about fifty-fifty
1. What would be your choice assuming you were given 60 seconds to make up your mind?
2.Solve analytically what would be the expected winnings in each of the cases.
3. Get the answers by simulating both cases.
Assume that a throw lasts 10 seconds and the payouts are executed in no time.
Please compare the results of 1, 2, 3.
1. My choice would be a. It takes less time to play a game, so if there's an advantage to each that's about the same your winnings are higher. Of course, if each is disadvantageous this is reversed, but I'd say both are slightly advantageous.
2. The probability that none of 4 shows a 6 is (5/6)^4 ~= .482253086419752. The probability of a win is therefore 1 - .482253086419752 = .517746913580248. In one hour you play this game 360 times, winning 186 times (expected value just above this, 186.39) and losing 174 times, for a net gain of $12.
In game b, each throw has 1/36 probability of getting double 6. The probability that none of 25 tries achieves this is (35/36)^25 ~= .494468453761619 so the probability of a win is .505531546238381. But now we need to know the expected length of the game (though it's certainly more than in game a, as that's just one throw).
The probability of only 1 throw is 1/36; of 2 throws 35/36 * 1/36; 3 throws, (35/36)^2 * (1/36), etc.
The expected number of throws for the game is 5.83742432054107 + .494468453761619 * 25, the first factor being the expected value of the length of the game if its within the first 25 throws; the second is the contribution to the expected length from the losses. These 18.2 throws take 182 seconds you get in 19.8 games. You win 10.0 times and lose 9.8 times, for a net gain of about 20 cents.
3. This program simulates 1000 hour-long sessions:
tot=0; tot2=0;
for tr=1:1000
winnings=0;
for t=1:360
for die=1:4
d=randi(6); if d==6 break, end
end
if d==6
winnings=winnings+1;
else
winnings=winnings-1;
end
end
winnings2=0;
t=0;
while t<360
for toss=1:25
t=t+1; if t>3600, break, end
d1=randi(6); d2=randi(6);
if d1==6 && d2==6
winnings2=winnings2+1;
break
end
end
if (d1~=6 || d2~=6) && toss==25
winnings2=winnings2-1;
end
end
disp([winnings winnings2])
tot=tot+winnings; tot2=tot2+winnings2;
end
disp([tot/1000 tot2/1000])
A partial list of the results (each line a new 1-hour trial) and the averages at the bottom:
game a game b
6 -4
4 -5
6 -2
-10 -2
44 -3
-10 1
8 1
-20 -3
4 1
8 -5
30 -3
-4 9
28 0
10 -7
8 7
12 -2
0 -3
10 8
2 8
44 -1
-16 2
-18 8
26 -5
8 -3
26 2
28 -6
4 -8
-12 -1
14 2
2 -1
6 -1
-12 -7
0 3
. . .
-18 3
64 2
14 -4
8 -1
-18 6
12 -4
20 1
22 7
22 12
22 3
8 -4
-4 10
62 4
16 -5
14 11
24 11
-10 7
20 -4
Obviously there was considerable variation among hour-long trials.
The average for the thousand hour-long trials of game a: 11.998 and game b: 0.304
The average gain over 100,000 hour-long trials was 12.8034 and 0.23522 respectively.
My error in part 2 was in rounding 186.39 to 186. The expected value is really 186.39 - 173.61 = 12.78.
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Posted by Charlie
on 2021-05-28 11:12:11 |
computer solution
