The Loaded Dice

Brown

Penultimate Amazing
Brown
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Note: the following is based upon a discussion of the work of Colin Blyth in Martin Gardner's "The Colossal Book of Mathematics." I have phrased the discussion in terms of dice, but the same analysis can apply to a number of other items or activities. Gardner discusses the matter using spinners.

Imagine that you have three six-sided dice. The object of the game is to roll all three dice, and the die that shows the highest number wins. As we shall see, it is possible for any of the three dice to win, and ties are not possible.

Die A is a normal six-sided die, but it has the number 3 on every face. As a result, Die A rolls a value of 3 one hundred percent of the time.

Die B is a bit more complicated. Die B has the number 6 on one face, the number 4 on another face, and the number 2 on four faces. In addition, Die B is loaded so that the number 6 comes up twenty-two percent of the time, and the number 4 comes up twenty-two percent of the time. Of course, this means that the number 2 comes up fifty-six percent of the time.

Die C has the number 5 on three faces and the number 1 on three faces. But Die C is slightly loaded, so that the number 1 comes up fifty-one percent of the time, and the number 5 comes up forty-nine percent of the time.

In summary, the probabilities of the numbers coming up are:
Code: 
  Die [b]A[/b]              Die [b]B[/b]              Die [b]C[/b] 
3    1.00          2    0.56          1    0.51
                   4    0.22          5    0.49
                   6    0.22
When all three dice are rolled, which of these three dice has the best chance of showing the highest number?

If we compare Die A to Die B, we see that 56 percent of the time, Die B will show a 2. Because Die A always shows a 3, Die A will beat Die B 56 percent of the time.

By similar reasoning, Die A will beat Die C 51 percent of the time.

If we compare Die B to Die C, we see that Die B is heavily favored to beat Die C. If Die B rolls a 6, which happens 22 percent of the time, Die B wins, regardless of what number comes up on Die C. And if Die B rolls a 4 or a 2, which happens 78 percent of the time, Die B wins if Die C rolls a 1, which happens more than half the time. So Die B is favored over Die C by 0.22 + (0.78)(0.51) = 0.6178; meaning that Die B will beat Die C 61.78 percent of the time.

Thus, it would appear that if you had to bet on one of the dice to win, Die A would be the best bet, and Die C would be the worst.

But actually, the best bet is Die C, and the worst is Die A. I will omit the calculations at this time, so that you can see whether you can convince yourself that this conclusion is correct.
 
Brown said:

But actually, the best bet is Die C, and the worst is Die A. I will omit the calculations at this time, so that you can see whether you can convince yourself that this conclusion is correct.
Click to expand...

Cute problem. I just crunched the numbers myself and it appears that your conclusion is correct.
 
I was going to post that Dragonrock's analysis was correct, but that analysis has been (temporarily) withheld so that others may have a go at this subject.

If you don't like doing math, try doing a simulation on a computer. The program is very easy to write. Have the computer simulate, say, 10,000 tosses and see which die does the best. For fun, also keep track of the head-to-head matchups with each roll. Does Die A really beat Die B most of the time, and does Die A really beat Die C most of the time, and does Die B really beat Die C more than 61 percent of the time?
 
For those of you chewing on the problem, consider this:

You are a manager of a baseball team. You are coming up to the final two innings of a very important game and you need to select a relief pitcher who will face at least six batters. You have three relief pitchers at your disposal, all of them basically equal in every respect except in the number of strikeouts they can be expected to throw.

When you put Albert into the game, he is "Mr. Reliable," always striking out three batters. When you put Benny into the game, he strikes out two batters 56 percent of the time, he strikes out four batters 22 percent of the time, and he strikes out six batters 22 percent of the time. When Carl pitches, he can be expected to strike out five batters 49 percent of the time, and only one batter 51 percent of the time.

Unfortunately, Albert's wife is pregnant, so Albert has rushed off to the hospital because his wife is in labor. The only players you can choose to play are Benny and Carl.

You are, of course, planning to put Benny in the game. 61.78 percent of the time, Benny will strike out more batters than Carl. Benny is obviously the best choice.

But just before the eighth inning is about to begin, Albert suddently walks into the dugout in uniform! ("False alarm," he explains about his wife's delivery.) Is Carl now obviously the best choice, merely because Albert is present??
 
Brown said:
For those of you chewing on the problem, consider this:

You are a manager of a baseball team. You are coming up to the final two innings of a very important game and you need to select a relief pitcher who will face at least six batters. You have three relief pitchers at your disposal, all of them basically equal in every respect except in the number of strikeouts they can be expected to throw.

When you put Albert into the game, he is "Mr. Reliable," always striking out three batters. When you put Benny into the game, he strikes out two batters 56 percent of the time, he strikes out four batters 22 percent of the time, and he strikes out six batters 22 percent of the time. When Carl pitches, he can be expected to strike out five batters 49 percent of the time, and only one batter 51 percent of the time.

Unfortunately, Albert's wife is pregnant, so Albert has rushed off to the hospital because his wife is in labor. The only players you can choose to play are Benny and Carl.

You are, of course, planning to put Benny in the game. 61.78 percent of the time, Benny will strike out more batters than Carl. Benny is obviously the best choice.

But just before the eighth inning is about to begin, Albert suddently walks into the dugout in uniform! ("False alarm," he explains about his wife's delivery.) Is Carl now obviously the best choice, merely because Albert is present??
Click to expand...
Well, in this case you don't have enough information, since you aren't just comparing them to each other, you are comparing them to the other team. You stated that they are basically equal except for the number of strikeouts, so I'll add the following info to the the baseball problem:

With the current score difference you note that, no matter which of the three pitchers they face the opposing team will
win 75% of the time if struck out once,
win 60% of the time if struck out twice,
win 45% of the time if struck out thrice,
win 30% of the time if struck out four time,
win 15% of the time if struck out five times,
win 0% of the time if struck out six times.

Then you'd expect Albert to get a win 55% of the time, Benny about 60% of the time and Carl about 57%, so if Benny is there he is always the best player give these numbers. So in the baseball case you aren't comparing performace against eachother but against an opposing team.

If the first problem was rephrased so the object was to roll over a certain number instead of roll higher than the opponents then the prefered choice becomes unambiguous without the conundrum presented in the current form.

Walt
 
A quick simulation gives me the following results:

A vs. B vs. C:
A wins 30% of the time
B wins 32% of the time
C wins 38% of the time

A vs. B:
A wins 56% of the time
B wins 44% of the time

A vs. C
A wins 51% of the time
C wins 49% of the time

B vs. C
B wins 62% of the time
C wins 38% of the time


Very interesting. :)
 
Die A Die B Die C
3 1.00 2 0.56 1 0.51
4 0.22 5 0.49
6 0.22

I don't think this is counterintuitive or requires any "number crunching" much less simulations.

Die B beats Die A 44% of the time and beats Die C almost 22% of the time. Die C beats Die B almost 78% of the time and beats Die A 49% of the time. This is my intuitive solution and having typed it I can see that it is mathematically wrong. Too lazy to figure it out (makes drinky-drinky gesture).

Actually, having gone back and put in two "almost"s I'm happy with it.
 
6 possible outcomes, with the result specified below

3, 2, 1 - A
3, 2, 5 - C
3, 4, 1 - B
3, 4, 5 - C
3, 6, 1 - B
3, 6, 5 - B

just multiply out and see which is most likely

C has the Hgighest unmber 38% of the time
B 33% of the time
A 29% of the time

(with rounding)
 
Brown said:
Imagine that you have three six-sided dice. The object of the game is to roll all three dice, and the die that shows the highest number wins. As we shall see, it is possible for any of the three dice to win, and ties are not possible.

Die A is a normal six-sided die, but it has the number 3 on every face. As a result, Die A rolls a value of 3 one hundred percent of the time...
Click to expand...
This is a slightly simpler version of the Non-Transitive Dice discovery. The idea behind non-transitive dice is that: If 2 players are asked to each roll one of (say) 4 dice, the person who chooses his die second will ALWAYS have an edge, for example

Die 1 has an edge against Die 2
Die 2 has an edge against Die 3
Die 3 has an edge against Die 4
Die 4 has an edge against Die 1

So, based on which die the first person chooses, the second person can always choose a die with better performance.

You can buy 3 and 4 die non-transtivie dice here

Some other non-transitive sets can be found here
 
Brown said:
Code: 
  Die [b]A[/b]              Die [b]B[/b]              Die [b]C[/b] 
3    1.00          2    0.56          1    0.51
                   4    0.22          5    0.49
                   6    0.22
Click to expand...

Isn't it pretty obvious?

A wins iff B comes up 2 and C comes up 1. = .56*.51 of the time (~29%).
B wins if it comes up 4 and c comes up 1 or it comes up 6 = .22*.51+.22 (~33%)
C wins the rest of the time, which is clearly more that 33%
 
Exactly my point.... it's pretty obvious, although in my case it was right solution, wrong math.
 
The computations above are basically correct.

A can win only when B shows a 2 and C shows a 1, which occurs 0.56 * 0.51 = 0.2856, or 28.56 percent of the time.

When B shows a 6, which happens 22 percent of the time, he is invincible. B can also win if B shows a 4 and C shows a 1. So B wins 0.22 + (0.22 * 0.51) = 0.3322, or 33.22 percent of the time.

The only way for C to win is to show a 5, which happens 49 percent of the time. When C shows a 5, C is guaranteed to beat A, and will beat B if B shows something other than a 6. C wins 0.49 * 0.78 = 0.3822, or 38.22 percent of the time. A and B each win less than one-third of the time, while C wins significantly more than one-third of the time.

There are other ways to calculate these numbers, of course.

This problem can be applied to a number of scenarios. Martin Gardner discusses the problem in terms of pies and drug effectiveness. A physician may prefer prescribing drug A over drug C, but suddenly, drug B comes along and the physician suddenly finds that drug C is preferable to drug A!
 
Re: Re: The Loaded Dice

Iconoclast said:
This is a slightly simpler version of the Non-Transitive Dice discovery. The idea behind non-transitive dice is that: If 2 players are asked to each roll one of (say) 4 dice, the person who chooses his die second will ALWAYS have an edge, for example

Die 1 has an edge against Die 2
Die 2 has an edge against Die 3
Die 3 has an edge against Die 4
Die 4 has an edge against Die 1

So, based on which die the first person chooses, the second person can always choose a die with better performance.
Click to expand...
Good observation. There is certainly a lot of similarity between Blyth's paradox and non-transitive games, in that the results seem so counterintuitive. Martin Gardner, in "The Colossal Book of Mathematics," discusses other non-transitive games, including a surprising non-transitive coin-flipping game.

It seems to me that part of the appeal of Blyth's paradox is that it appears to be transitive. A is preferable to B, A is preferable to C, and B is preferable to C. But these preferences only hold up in head-to-head competition. When all three dice compete at once, the die that performs the worst in head-to-head competition is actually the best, and the die that performs the best in head-to-head competition is actually the worst.
 
This discussion reminds me of looking at data, say weights and heights. When looking at weights and heights individually, the values 250lbs and 5'4" probably wouldn't stand out. However, if the pair (250, 5'4") are from the same person, then that point certainly does stand out on the graph.
 
I guess the answer is that A and B are likely to win in the same scenarios (only when C rolls a 1), so they have to share the wins between themselves.
 
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