Mixing
Help in understanding a dice problem
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I am a novice student in probability and I thought of this problem recently. Would like to clarify on some concepts in probability.
A and B bet $n$ marbles each to play a game with a six-sided dice with the following rules:
- A earns $1$ point when the number rolled leaves a remainder $0$ when divided by $3$.
- B earns $1$ point when the number rolled leaves a remainder $1$ when divided by $3$.
- No point is earned when the number rolled leaves a remainder $2$ when divided by $3$.
- The person with the highest number of points takes all the marbles, otherwise the marbles will be shared equally.
(a) Suppose die is rolled a total of $4$ times, in which the first two rolls are $3$ and $6$, how should the marbles be distributed fairly?
(b) If the dice is continued to be rolled a sufficiently large number of times, how should the marbles be distributed fairly in the end?
For (a), there is a total of $36$ outcomes in which A is a winner in $32$ outcomes and both draw in $4$ outcomes. So A should receive $frac{32}{36} times 2n + frac{1}{2}times frac{4}{36} times 2n$ marbles.
For (b). I think that expected number of rolls with $3$ or $6$ will be equal to expected number of rolls with $1$ or $4$ since each number is equally likely to occur on each roll. Hence the outcome will be the same as the outcome in (a).
probability
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
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add a comment |
$begingroup$
I am a novice student in probability and I thought of this problem recently. Would like to clarify on some concepts in probability.
A and B bet $n$ marbles each to play a game with a six-sided dice with the following rules:
- A earns $1$ point when the number rolled leaves a remainder $0$ when divided by $3$.
- B earns $1$ point when the number rolled leaves a remainder $1$ when divided by $3$.
- No point is earned when the number rolled leaves a remainder $2$ when divided by $3$.
- The person with the highest number of points takes all the marbles, otherwise the marbles will be shared equally.
(a) Suppose die is rolled a total of $4$ times, in which the first two rolls are $3$ and $6$, how should the marbles be distributed fairly?
(b) If the dice is continued to be rolled a sufficiently large number of times, how should the marbles be distributed fairly in the end?
For (a), there is a total of $36$ outcomes in which A is a winner in $32$ outcomes and both draw in $4$ outcomes. So A should receive $frac{32}{36} times 2n + frac{1}{2}times frac{4}{36} times 2n$ marbles.
For (b). I think that expected number of rolls with $3$ or $6$ will be equal to expected number of rolls with $1$ or $4$ since each number is equally likely to occur on each roll. Hence the outcome will be the same as the outcome in (a).
probability
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
$endgroup$
$begingroup$
I don't understand when the game ends (and then clause 4 is invoked). In (a), it seems to be that it ends after 4 throws. But when does it end in (b)? The only interpretation I can come up with is that it ends after a predetermined number $k$ (and we had $k=4$ in (a)), and the question is how the fair distribution should be for $k to infty$.
$endgroup$
– Ingix
Jan 22 at 18:05
add a comment |
$begingroup$
I am a novice student in probability and I thought of this problem recently. Would like to clarify on some concepts in probability.
A and B bet $n$ marbles each to play a game with a six-sided dice with the following rules:
- A earns $1$ point when the number rolled leaves a remainder $0$ when divided by $3$.
- B earns $1$ point when the number rolled leaves a remainder $1$ when divided by $3$.
- No point is earned when the number rolled leaves a remainder $2$ when divided by $3$.
- The person with the highest number of points takes all the marbles, otherwise the marbles will be shared equally.
(a) Suppose die is rolled a total of $4$ times, in which the first two rolls are $3$ and $6$, how should the marbles be distributed fairly?
(b) If the dice is continued to be rolled a sufficiently large number of times, how should the marbles be distributed fairly in the end?
For (a), there is a total of $36$ outcomes in which A is a winner in $32$ outcomes and both draw in $4$ outcomes. So A should receive $frac{32}{36} times 2n + frac{1}{2}times frac{4}{36} times 2n$ marbles.
For (b). I think that expected number of rolls with $3$ or $6$ will be equal to expected number of rolls with $1$ or $4$ since each number is equally likely to occur on each roll. Hence the outcome will be the same as the outcome in (a).
probability
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
$endgroup$
I am a novice student in probability and I thought of this problem recently. Would like to clarify on some concepts in probability.
A and B bet $n$ marbles each to play a game with a six-sided dice with the following rules:
- A earns $1$ point when the number rolled leaves a remainder $0$ when divided by $3$.
- B earns $1$ point when the number rolled leaves a remainder $1$ when divided by $3$.
- No point is earned when the number rolled leaves a remainder $2$ when divided by $3$.
- The person with the highest number of points takes all the marbles, otherwise the marbles will be shared equally.
(a) Suppose die is rolled a total of $4$ times, in which the first two rolls are $3$ and $6$, how should the marbles be distributed fairly?
(b) If the dice is continued to be rolled a sufficiently large number of times, how should the marbles be distributed fairly in the end?
For (a), there is a total of $36$ outcomes in which A is a winner in $32$ outcomes and both draw in $4$ outcomes. So A should receive $frac{32}{36} times 2n + frac{1}{2}times frac{4}{36} times 2n$ marbles.
For (b). I think that expected number of rolls with $3$ or $6$ will be equal to expected number of rolls with $1$ or $4$ since each number is equally likely to occur on each roll. Hence the outcome will be the same as the outcome in (a).
probability
probability
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
asked Jan 22 at 17:27
Alexy VincenzoAlexy Vincenzo
2,1803926
2,1803926
$begingroup$
I don't understand when the game ends (and then clause 4 is invoked). In (a), it seems to be that it ends after 4 throws. But when does it end in (b)? The only interpretation I can come up with is that it ends after a predetermined number $k$ (and we had $k=4$ in (a)), and the question is how the fair distribution should be for $k to infty$.
$endgroup$
– Ingix
Jan 22 at 18:05
add a comment |
$begingroup$
I don't understand when the game ends (and then clause 4 is invoked). In (a), it seems to be that it ends after 4 throws. But when does it end in (b)? The only interpretation I can come up with is that it ends after a predetermined number $k$ (and we had $k=4$ in (a)), and the question is how the fair distribution should be for $k to infty$.
$endgroup$
– Ingix
Jan 22 at 18:05
$begingroup$
I don't understand when the game ends (and then clause 4 is invoked). In (a), it seems to be that it ends after 4 throws. But when does it end in (b)? The only interpretation I can come up with is that it ends after a predetermined number $k$ (and we had $k=4$ in (a)), and the question is how the fair distribution should be for $k to infty$.
$endgroup$
– Ingix
Jan 22 at 18:05
$begingroup$
I don't understand when the game ends (and then clause 4 is invoked). In (a), it seems to be that it ends after 4 throws. But when does it end in (b)? The only interpretation I can come up with is that it ends after a predetermined number $k$ (and we had $k=4$ in (a)), and the question is how the fair distribution should be for $k to infty$.
$endgroup$
– Ingix
Jan 22 at 18:05
add a comment |
1 Answer
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I'm not sure if I understood properly the meaning of "distributing marbles fairly"!
What I see, for example in case "a)", is that, as you say:
- A will win with probability $P(A)=frac{32}{36} = frac{8}{9}$;
- They will draw with probability $P(D)= frac{1}{9}$;
- B cannot win.
Given this, what we could say is that the expected number of marbles won by A will be the one you computed.
So, I'll assume computing the "fair distribution of marbles" here stands for computing the "average number of marbles won by the players".
Enough with the pedantry! So now, case "b)": let's try to change the value $N$ of the total draws and see what happen to probabilities:
$N=2 longrightarrow$ there's no uncertainty at all; A wins;
$$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=3 longrightarrow$ there's one uncertain roll, but A will win regardless of its outcome; $$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=4 longrightarrow$ B cannot win, but he can hope for a draw! As you computed: $$P(A)=frac{8}{9};quad P(B)=0;quad P(D)=frac{1}{9}$$
$N=5 longrightarrow$ B can now win! All outcomes are possible: in order to B to win he has to get all three points for himself: so $P(B)=frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$.
Conversely, we have a draw if two rolls are for b and one is wasted; this can be done in three distinct ways: (b-b-w), (b-w-b), and (w-b-b), and each one of these three has probability $frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$ to occur. So, the total probability of a draw will be $P(D)=frac{1}{27}cdot 3 = frac{3}{27}$.
Now, explicit computation of $P(A)$ would be slightly more tedious, but fortunately we already know what would be the result, due to the fact that the sum of the three probabilities must be 1! So: $P(A) = 1-P(B)-P(C) = frac{23}{27}$.
Summarizing, for N=5:
$$P(A)=frac{23}{27};quad P(B)=frac{1}{27};quad P(D)=frac{3}{27}$$
We see that things actually change! In particular, we managed to give a positive probability to previously impossible outcomes, by allowing more rolls.
nb: as @ingix pointed out in the comments, what follows is wrong. I will now put the wrong portion of the answer between two horizontal lines, then below I will give a correct version.
---------------------------------------------------------------
WRONG VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B, as well as a draw, both become more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the three events become absolutely the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof; but you can already convince yourself, for example by looking at how the ratio ${P(D)}/{P(A)} $ varies with N, that we'll eventually approach equiprobability...).
So the expected values of winned marbles would be $$left(frac{2n}{3},frac{2n}{3},frac{2n}{3}right) $$
---------------------------------------------------------------
CORRECT VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B becomes more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the two events "A wins" and "B wins" become exactly the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof).
What we could ask ourselves now is: what happens to $P(D)$ as N grows?
The correct answer, which, again, can be proven, is: it gets more and more improbable (as @Ingix pointed out in the comments: would you believe that after 100000 rolls A and B would have the same exact number of points?).
Fortunately, for our problem we really don't care about P(D), given that we accept that, when $N$ is very large, $P(A)approx P(B)$; let's see why:
suppose $P(A)=P(B)=rho$; then, thanks to normalization, we have $P(D)=1-P(A)-P(B)=1-2rho$. Now, the expected number of marbles won by A is:
$$ rhocdot 2n + (1-2rho)cdot n = 2nrho + n - 2nrho = n$$
irrespectively of the value of $rho$!
So the expected values of winned marbles for large N will be $$left(n,nright) $$
answered Jan 22 at 18:40
Rocco Rocco
15410
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1
$begingroup$
The probability for a draw absolutely does not become equal to the other 2 outcomes. Just use your 'big number' argument again: Would you bet that after 100,000 dice throws both players have exactly the same score? Surely not. As $N$ increases, the probability of an exact draw approaches 0, and the probability that a given player wins approches $frac12$ (no matter how much the initial bias, like here were player $A$ starts with 2 points ahead). So the fair distribution for a large enough $N$ is each player gets $n$ marbles.
$endgroup$
– Ingix
Jan 22 at 18:58
1
$begingroup$
Thanks for your insights. I am not sure if I interpret you correctly: If we continue throwing the dice for a sufficiently large number of times, $P($3 or 6 is rolled$)$ ,$P($1 or 4 is rolled$)$ and $P($2 or 5 is rolled$)$ approaches $frac{1}{3}$ each. Hence A and B will win approx same number of points?
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– Alexy Vincenzo
Jan 23 at 13:18
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@Ingix oh my, I'm so ashamed I got that SO wrong ahah! Thanks for pointing that out, you're clearly right, I'll change my answer right away!
$endgroup$
– Rocco
Jan 23 at 15:04
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@AlexyVincenzo P(3 or 6 is rolled) ,P(1 or 4 is rolled) and P(2 or 5 is rolled) are ALWAYS 1/3 each, due to the fact that "die don't have memory" :). What approaches uniformity is the probability of a global victory of A, which in my answer I call P(A), with respect to a global victory of B. That's because, as you "zoom out", the initial advantage of A looks more and more little with respect to the total number of rolls.
$endgroup$
– Rocco
Jan 23 at 15:38
1
$begingroup$
It's a matter of semantics if you consider a difference of 1,000 points big or small when doing a million throws. In other words, in my world $N$ and $N+sqrt{N}$ are roughly the same number of points.
$endgroup$
– Ingix
Jan 23 at 17:34
|
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I'm not sure if I understood properly the meaning of "distributing marbles fairly"!
What I see, for example in case "a)", is that, as you say:
- A will win with probability $P(A)=frac{32}{36} = frac{8}{9}$;
- They will draw with probability $P(D)= frac{1}{9}$;
- B cannot win.
Given this, what we could say is that the expected number of marbles won by A will be the one you computed.
So, I'll assume computing the "fair distribution of marbles" here stands for computing the "average number of marbles won by the players".
Enough with the pedantry! So now, case "b)": let's try to change the value $N$ of the total draws and see what happen to probabilities:
$N=2 longrightarrow$ there's no uncertainty at all; A wins;
$$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=3 longrightarrow$ there's one uncertain roll, but A will win regardless of its outcome; $$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=4 longrightarrow$ B cannot win, but he can hope for a draw! As you computed: $$P(A)=frac{8}{9};quad P(B)=0;quad P(D)=frac{1}{9}$$
$N=5 longrightarrow$ B can now win! All outcomes are possible: in order to B to win he has to get all three points for himself: so $P(B)=frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$.
Conversely, we have a draw if two rolls are for b and one is wasted; this can be done in three distinct ways: (b-b-w), (b-w-b), and (w-b-b), and each one of these three has probability $frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$ to occur. So, the total probability of a draw will be $P(D)=frac{1}{27}cdot 3 = frac{3}{27}$.
Now, explicit computation of $P(A)$ would be slightly more tedious, but fortunately we already know what would be the result, due to the fact that the sum of the three probabilities must be 1! So: $P(A) = 1-P(B)-P(C) = frac{23}{27}$.
Summarizing, for N=5:
$$P(A)=frac{23}{27};quad P(B)=frac{1}{27};quad P(D)=frac{3}{27}$$
We see that things actually change! In particular, we managed to give a positive probability to previously impossible outcomes, by allowing more rolls.
nb: as @ingix pointed out in the comments, what follows is wrong. I will now put the wrong portion of the answer between two horizontal lines, then below I will give a correct version.
---------------------------------------------------------------
WRONG VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B, as well as a draw, both become more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the three events become absolutely the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof; but you can already convince yourself, for example by looking at how the ratio ${P(D)}/{P(A)} $ varies with N, that we'll eventually approach equiprobability...).
So the expected values of winned marbles would be $$left(frac{2n}{3},frac{2n}{3},frac{2n}{3}right) $$
---------------------------------------------------------------
CORRECT VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B becomes more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the two events "A wins" and "B wins" become exactly the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof).
What we could ask ourselves now is: what happens to $P(D)$ as N grows?
The correct answer, which, again, can be proven, is: it gets more and more improbable (as @Ingix pointed out in the comments: would you believe that after 100000 rolls A and B would have the same exact number of points?).
Fortunately, for our problem we really don't care about P(D), given that we accept that, when $N$ is very large, $P(A)approx P(B)$; let's see why:
suppose $P(A)=P(B)=rho$; then, thanks to normalization, we have $P(D)=1-P(A)-P(B)=1-2rho$. Now, the expected number of marbles won by A is:
$$ rhocdot 2n + (1-2rho)cdot n = 2nrho + n - 2nrho = n$$
irrespectively of the value of $rho$!
So the expected values of winned marbles for large N will be $$left(n,nright) $$
answered Jan 22 at 18:40
Rocco Rocco
15410
$endgroup$
1
$begingroup$
The probability for a draw absolutely does not become equal to the other 2 outcomes. Just use your 'big number' argument again: Would you bet that after 100,000 dice throws both players have exactly the same score? Surely not. As $N$ increases, the probability of an exact draw approaches 0, and the probability that a given player wins approches $frac12$ (no matter how much the initial bias, like here were player $A$ starts with 2 points ahead). So the fair distribution for a large enough $N$ is each player gets $n$ marbles.
$endgroup$
– Ingix
Jan 22 at 18:58
1
$begingroup$
Thanks for your insights. I am not sure if I interpret you correctly: If we continue throwing the dice for a sufficiently large number of times, $P($3 or 6 is rolled$)$ ,$P($1 or 4 is rolled$)$ and $P($2 or 5 is rolled$)$ approaches $frac{1}{3}$ each. Hence A and B will win approx same number of points?
$endgroup$
– Alexy Vincenzo
Jan 23 at 13:18
$begingroup$
@Ingix oh my, I'm so ashamed I got that SO wrong ahah! Thanks for pointing that out, you're clearly right, I'll change my answer right away!
$endgroup$
– Rocco
Jan 23 at 15:04
$begingroup$
@AlexyVincenzo P(3 or 6 is rolled) ,P(1 or 4 is rolled) and P(2 or 5 is rolled) are ALWAYS 1/3 each, due to the fact that "die don't have memory" :). What approaches uniformity is the probability of a global victory of A, which in my answer I call P(A), with respect to a global victory of B. That's because, as you "zoom out", the initial advantage of A looks more and more little with respect to the total number of rolls.
$endgroup$
– Rocco
Jan 23 at 15:38
1
$begingroup$
It's a matter of semantics if you consider a difference of 1,000 points big or small when doing a million throws. In other words, in my world $N$ and $N+sqrt{N}$ are roughly the same number of points.
$endgroup$
– Ingix
Jan 23 at 17:34
|
show 6 more comments
$begingroup$
I'm not sure if I understood properly the meaning of "distributing marbles fairly"!
What I see, for example in case "a)", is that, as you say:
- A will win with probability $P(A)=frac{32}{36} = frac{8}{9}$;
- They will draw with probability $P(D)= frac{1}{9}$;
- B cannot win.
Given this, what we could say is that the expected number of marbles won by A will be the one you computed.
So, I'll assume computing the "fair distribution of marbles" here stands for computing the "average number of marbles won by the players".
Enough with the pedantry! So now, case "b)": let's try to change the value $N$ of the total draws and see what happen to probabilities:
$N=2 longrightarrow$ there's no uncertainty at all; A wins;
$$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=3 longrightarrow$ there's one uncertain roll, but A will win regardless of its outcome; $$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=4 longrightarrow$ B cannot win, but he can hope for a draw! As you computed: $$P(A)=frac{8}{9};quad P(B)=0;quad P(D)=frac{1}{9}$$
$N=5 longrightarrow$ B can now win! All outcomes are possible: in order to B to win he has to get all three points for himself: so $P(B)=frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$.
Conversely, we have a draw if two rolls are for b and one is wasted; this can be done in three distinct ways: (b-b-w), (b-w-b), and (w-b-b), and each one of these three has probability $frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$ to occur. So, the total probability of a draw will be $P(D)=frac{1}{27}cdot 3 = frac{3}{27}$.
Now, explicit computation of $P(A)$ would be slightly more tedious, but fortunately we already know what would be the result, due to the fact that the sum of the three probabilities must be 1! So: $P(A) = 1-P(B)-P(C) = frac{23}{27}$.
Summarizing, for N=5:
$$P(A)=frac{23}{27};quad P(B)=frac{1}{27};quad P(D)=frac{3}{27}$$
We see that things actually change! In particular, we managed to give a positive probability to previously impossible outcomes, by allowing more rolls.
nb: as @ingix pointed out in the comments, what follows is wrong. I will now put the wrong portion of the answer between two horizontal lines, then below I will give a correct version.
---------------------------------------------------------------
WRONG VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B, as well as a draw, both become more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the three events become absolutely the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof; but you can already convince yourself, for example by looking at how the ratio ${P(D)}/{P(A)} $ varies with N, that we'll eventually approach equiprobability...).
So the expected values of winned marbles would be $$left(frac{2n}{3},frac{2n}{3},frac{2n}{3}right) $$
---------------------------------------------------------------
CORRECT VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B becomes more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the two events "A wins" and "B wins" become exactly the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof).
What we could ask ourselves now is: what happens to $P(D)$ as N grows?
The correct answer, which, again, can be proven, is: it gets more and more improbable (as @Ingix pointed out in the comments: would you believe that after 100000 rolls A and B would have the same exact number of points?).
Fortunately, for our problem we really don't care about P(D), given that we accept that, when $N$ is very large, $P(A)approx P(B)$; let's see why:
suppose $P(A)=P(B)=rho$; then, thanks to normalization, we have $P(D)=1-P(A)-P(B)=1-2rho$. Now, the expected number of marbles won by A is:
$$ rhocdot 2n + (1-2rho)cdot n = 2nrho + n - 2nrho = n$$
irrespectively of the value of $rho$!
So the expected values of winned marbles for large N will be $$left(n,nright) $$
answered Jan 22 at 18:40
Rocco Rocco
15410
$endgroup$
1
$begingroup$
The probability for a draw absolutely does not become equal to the other 2 outcomes. Just use your 'big number' argument again: Would you bet that after 100,000 dice throws both players have exactly the same score? Surely not. As $N$ increases, the probability of an exact draw approaches 0, and the probability that a given player wins approches $frac12$ (no matter how much the initial bias, like here were player $A$ starts with 2 points ahead). So the fair distribution for a large enough $N$ is each player gets $n$ marbles.
$endgroup$
– Ingix
Jan 22 at 18:58
1
$begingroup$
Thanks for your insights. I am not sure if I interpret you correctly: If we continue throwing the dice for a sufficiently large number of times, $P($3 or 6 is rolled$)$ ,$P($1 or 4 is rolled$)$ and $P($2 or 5 is rolled$)$ approaches $frac{1}{3}$ each. Hence A and B will win approx same number of points?
$endgroup$
– Alexy Vincenzo
Jan 23 at 13:18
$begingroup$
@Ingix oh my, I'm so ashamed I got that SO wrong ahah! Thanks for pointing that out, you're clearly right, I'll change my answer right away!
$endgroup$
– Rocco
Jan 23 at 15:04
$begingroup$
@AlexyVincenzo P(3 or 6 is rolled) ,P(1 or 4 is rolled) and P(2 or 5 is rolled) are ALWAYS 1/3 each, due to the fact that "die don't have memory" :). What approaches uniformity is the probability of a global victory of A, which in my answer I call P(A), with respect to a global victory of B. That's because, as you "zoom out", the initial advantage of A looks more and more little with respect to the total number of rolls.
$endgroup$
– Rocco
Jan 23 at 15:38
1
$begingroup$
It's a matter of semantics if you consider a difference of 1,000 points big or small when doing a million throws. In other words, in my world $N$ and $N+sqrt{N}$ are roughly the same number of points.
$endgroup$
– Ingix
Jan 23 at 17:34
|
show 6 more comments
$begingroup$
I'm not sure if I understood properly the meaning of "distributing marbles fairly"!
What I see, for example in case "a)", is that, as you say:
- A will win with probability $P(A)=frac{32}{36} = frac{8}{9}$;
- They will draw with probability $P(D)= frac{1}{9}$;
- B cannot win.
Given this, what we could say is that the expected number of marbles won by A will be the one you computed.
So, I'll assume computing the "fair distribution of marbles" here stands for computing the "average number of marbles won by the players".
Enough with the pedantry! So now, case "b)": let's try to change the value $N$ of the total draws and see what happen to probabilities:
$N=2 longrightarrow$ there's no uncertainty at all; A wins;
$$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=3 longrightarrow$ there's one uncertain roll, but A will win regardless of its outcome; $$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=4 longrightarrow$ B cannot win, but he can hope for a draw! As you computed: $$P(A)=frac{8}{9};quad P(B)=0;quad P(D)=frac{1}{9}$$
$N=5 longrightarrow$ B can now win! All outcomes are possible: in order to B to win he has to get all three points for himself: so $P(B)=frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$.
Conversely, we have a draw if two rolls are for b and one is wasted; this can be done in three distinct ways: (b-b-w), (b-w-b), and (w-b-b), and each one of these three has probability $frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$ to occur. So, the total probability of a draw will be $P(D)=frac{1}{27}cdot 3 = frac{3}{27}$.
Now, explicit computation of $P(A)$ would be slightly more tedious, but fortunately we already know what would be the result, due to the fact that the sum of the three probabilities must be 1! So: $P(A) = 1-P(B)-P(C) = frac{23}{27}$.
Summarizing, for N=5:
$$P(A)=frac{23}{27};quad P(B)=frac{1}{27};quad P(D)=frac{3}{27}$$
We see that things actually change! In particular, we managed to give a positive probability to previously impossible outcomes, by allowing more rolls.
nb: as @ingix pointed out in the comments, what follows is wrong. I will now put the wrong portion of the answer between two horizontal lines, then below I will give a correct version.
---------------------------------------------------------------
WRONG VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B, as well as a draw, both become more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the three events become absolutely the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof; but you can already convince yourself, for example by looking at how the ratio ${P(D)}/{P(A)} $ varies with N, that we'll eventually approach equiprobability...).
So the expected values of winned marbles would be $$left(frac{2n}{3},frac{2n}{3},frac{2n}{3}right) $$
---------------------------------------------------------------
CORRECT VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B becomes more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the two events "A wins" and "B wins" become exactly the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof).
What we could ask ourselves now is: what happens to $P(D)$ as N grows?
The correct answer, which, again, can be proven, is: it gets more and more improbable (as @Ingix pointed out in the comments: would you believe that after 100000 rolls A and B would have the same exact number of points?).
Fortunately, for our problem we really don't care about P(D), given that we accept that, when $N$ is very large, $P(A)approx P(B)$; let's see why:
suppose $P(A)=P(B)=rho$; then, thanks to normalization, we have $P(D)=1-P(A)-P(B)=1-2rho$. Now, the expected number of marbles won by A is:
$$ rhocdot 2n + (1-2rho)cdot n = 2nrho + n - 2nrho = n$$
irrespectively of the value of $rho$!
So the expected values of winned marbles for large N will be $$left(n,nright) $$
answered Jan 22 at 18:40
Rocco Rocco
15410
$endgroup$
I'm not sure if I understood properly the meaning of "distributing marbles fairly"!
What I see, for example in case "a)", is that, as you say:
- A will win with probability $P(A)=frac{32}{36} = frac{8}{9}$;
- They will draw with probability $P(D)= frac{1}{9}$;
- B cannot win.
Given this, what we could say is that the expected number of marbles won by A will be the one you computed.
So, I'll assume computing the "fair distribution of marbles" here stands for computing the "average number of marbles won by the players".
Enough with the pedantry! So now, case "b)": let's try to change the value $N$ of the total draws and see what happen to probabilities:
$N=2 longrightarrow$ there's no uncertainty at all; A wins;
$$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=3 longrightarrow$ there's one uncertain roll, but A will win regardless of its outcome; $$P(A)=1;quad P(B)=0;quad P(D)=0$$
$N=4 longrightarrow$ B cannot win, but he can hope for a draw! As you computed: $$P(A)=frac{8}{9};quad P(B)=0;quad P(D)=frac{1}{9}$$
$N=5 longrightarrow$ B can now win! All outcomes are possible: in order to B to win he has to get all three points for himself: so $P(B)=frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$.
Conversely, we have a draw if two rolls are for b and one is wasted; this can be done in three distinct ways: (b-b-w), (b-w-b), and (w-b-b), and each one of these three has probability $frac{2}{6}cdotfrac{2}{6}cdotfrac{2}{6}=frac{1}{27}$ to occur. So, the total probability of a draw will be $P(D)=frac{1}{27}cdot 3 = frac{3}{27}$.
Now, explicit computation of $P(A)$ would be slightly more tedious, but fortunately we already know what would be the result, due to the fact that the sum of the three probabilities must be 1! So: $P(A) = 1-P(B)-P(C) = frac{23}{27}$.
Summarizing, for N=5:
$$P(A)=frac{23}{27};quad P(B)=frac{1}{27};quad P(D)=frac{3}{27}$$
We see that things actually change! In particular, we managed to give a positive probability to previously impossible outcomes, by allowing more rolls.
nb: as @ingix pointed out in the comments, what follows is wrong. I will now put the wrong portion of the answer between two horizontal lines, then below I will give a correct version.
---------------------------------------------------------------
WRONG VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B, as well as a draw, both become more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the three events become absolutely the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof; but you can already convince yourself, for example by looking at how the ratio ${P(D)}/{P(A)} $ varies with N, that we'll eventually approach equiprobability...).
So the expected values of winned marbles would be $$left(frac{2n}{3},frac{2n}{3},frac{2n}{3}right) $$
---------------------------------------------------------------
CORRECT VERSION:
If you kept on computing for growing N (being very careful about "different ways to realize an outcome" as the one we've just seen, and which need to be counted properly - and that's not easy!) you would clearly see that a victory for B becomes more and more probable - or, let's say, less and less improbable, as we add more "room" for B's comeback by adding more draws.
Think about it: would you be more discouraged losing the first two rolls in a match of 5 rolls, or in a match of 50000?
In the limit $Nrightarrow +infty$ you can prove that the probabilities of the two events "A wins" and "B wins" become exactly the same (yet I'm not sure your textbook actually wants you to try it - if you want I can try and sketch a proof).
What we could ask ourselves now is: what happens to $P(D)$ as N grows?
The correct answer, which, again, can be proven, is: it gets more and more improbable (as @Ingix pointed out in the comments: would you believe that after 100000 rolls A and B would have the same exact number of points?).
Fortunately, for our problem we really don't care about P(D), given that we accept that, when $N$ is very large, $P(A)approx P(B)$; let's see why:
suppose $P(A)=P(B)=rho$; then, thanks to normalization, we have $P(D)=1-P(A)-P(B)=1-2rho$. Now, the expected number of marbles won by A is:
$$ rhocdot 2n + (1-2rho)cdot n = 2nrho + n - 2nrho = n$$
irrespectively of the value of $rho$!
So the expected values of winned marbles for large N will be $$left(n,nright) $$
answered Jan 22 at 18:40
Rocco Rocco
15410
edited Jan 23 at 15:34
answered Jan 22 at 18:40
Rocco Rocco
15410
answered Jan 22 at 18:40
Rocco Rocco
15410
answered Jan 22 at 18:40
Rocco Rocco
15410
15410
1
$begingroup$
The probability for a draw absolutely does not become equal to the other 2 outcomes. Just use your 'big number' argument again: Would you bet that after 100,000 dice throws both players have exactly the same score? Surely not. As $N$ increases, the probability of an exact draw approaches 0, and the probability that a given player wins approches $frac12$ (no matter how much the initial bias, like here were player $A$ starts with 2 points ahead). So the fair distribution for a large enough $N$ is each player gets $n$ marbles.
$endgroup$
– Ingix
Jan 22 at 18:58
1
$begingroup$
Thanks for your insights. I am not sure if I interpret you correctly: If we continue throwing the dice for a sufficiently large number of times, $P($3 or 6 is rolled$)$ ,$P($1 or 4 is rolled$)$ and $P($2 or 5 is rolled$)$ approaches $frac{1}{3}$ each. Hence A and B will win approx same number of points?
$endgroup$
– Alexy Vincenzo
Jan 23 at 13:18
$begingroup$
@Ingix oh my, I'm so ashamed I got that SO wrong ahah! Thanks for pointing that out, you're clearly right, I'll change my answer right away!
$endgroup$
– Rocco
Jan 23 at 15:04
$begingroup$
@AlexyVincenzo P(3 or 6 is rolled) ,P(1 or 4 is rolled) and P(2 or 5 is rolled) are ALWAYS 1/3 each, due to the fact that "die don't have memory" :). What approaches uniformity is the probability of a global victory of A, which in my answer I call P(A), with respect to a global victory of B. That's because, as you "zoom out", the initial advantage of A looks more and more little with respect to the total number of rolls.
$endgroup$
– Rocco
Jan 23 at 15:38
1
$begingroup$
It's a matter of semantics if you consider a difference of 1,000 points big or small when doing a million throws. In other words, in my world $N$ and $N+sqrt{N}$ are roughly the same number of points.
$endgroup$
– Ingix
Jan 23 at 17:34
|
show 6 more comments
1
$begingroup$
The probability for a draw absolutely does not become equal to the other 2 outcomes. Just use your 'big number' argument again: Would you bet that after 100,000 dice throws both players have exactly the same score? Surely not. As $N$ increases, the probability of an exact draw approaches 0, and the probability that a given player wins approches $frac12$ (no matter how much the initial bias, like here were player $A$ starts with 2 points ahead). So the fair distribution for a large enough $N$ is each player gets $n$ marbles.
$endgroup$
– Ingix
Jan 22 at 18:58
1
$begingroup$
Thanks for your insights. I am not sure if I interpret you correctly: If we continue throwing the dice for a sufficiently large number of times, $P($3 or 6 is rolled$)$ ,$P($1 or 4 is rolled$)$ and $P($2 or 5 is rolled$)$ approaches $frac{1}{3}$ each. Hence A and B will win approx same number of points?
$endgroup$
– Alexy Vincenzo
Jan 23 at 13:18
$begingroup$
@Ingix oh my, I'm so ashamed I got that SO wrong ahah! Thanks for pointing that out, you're clearly right, I'll change my answer right away!
$endgroup$
– Rocco
Jan 23 at 15:04
$begingroup$
@AlexyVincenzo P(3 or 6 is rolled) ,P(1 or 4 is rolled) and P(2 or 5 is rolled) are ALWAYS 1/3 each, due to the fact that "die don't have memory" :). What approaches uniformity is the probability of a global victory of A, which in my answer I call P(A), with respect to a global victory of B. That's because, as you "zoom out", the initial advantage of A looks more and more little with respect to the total number of rolls.
$endgroup$
– Rocco
Jan 23 at 15:38
1
$begingroup$
It's a matter of semantics if you consider a difference of 1,000 points big or small when doing a million throws. In other words, in my world $N$ and $N+sqrt{N}$ are roughly the same number of points.
$endgroup$
– Ingix
Jan 23 at 17:34
1
1
$begingroup$
The probability for a draw absolutely does not become equal to the other 2 outcomes. Just use your 'big number' argument again: Would you bet that after 100,000 dice throws both players have exactly the same score? Surely not. As $N$ increases, the probability of an exact draw approaches 0, and the probability that a given player wins approches $frac12$ (no matter how much the initial bias, like here were player $A$ starts with 2 points ahead). So the fair distribution for a large enough $N$ is each player gets $n$ marbles.
$endgroup$
– Ingix
Jan 22 at 18:58
$begingroup$
The probability for a draw absolutely does not become equal to the other 2 outcomes. Just use your 'big number' argument again: Would you bet that after 100,000 dice throws both players have exactly the same score? Surely not. As $N$ increases, the probability of an exact draw approaches 0, and the probability that a given player wins approches $frac12$ (no matter how much the initial bias, like here were player $A$ starts with 2 points ahead). So the fair distribution for a large enough $N$ is each player gets $n$ marbles.
$endgroup$
– Ingix
Jan 22 at 18:58
1
1
$begingroup$
Thanks for your insights. I am not sure if I interpret you correctly: If we continue throwing the dice for a sufficiently large number of times, $P($3 or 6 is rolled$)$ ,$P($1 or 4 is rolled$)$ and $P($2 or 5 is rolled$)$ approaches $frac{1}{3}$ each. Hence A and B will win approx same number of points?
$endgroup$
– Alexy Vincenzo
Jan 23 at 13:18
$begingroup$
Thanks for your insights. I am not sure if I interpret you correctly: If we continue throwing the dice for a sufficiently large number of times, $P($3 or 6 is rolled$)$ ,$P($1 or 4 is rolled$)$ and $P($2 or 5 is rolled$)$ approaches $frac{1}{3}$ each. Hence A and B will win approx same number of points?
$endgroup$
– Alexy Vincenzo
Jan 23 at 13:18
$begingroup$
@Ingix oh my, I'm so ashamed I got that SO wrong ahah! Thanks for pointing that out, you're clearly right, I'll change my answer right away!
$endgroup$
– Rocco
Jan 23 at 15:04
$begingroup$
@Ingix oh my, I'm so ashamed I got that SO wrong ahah! Thanks for pointing that out, you're clearly right, I'll change my answer right away!
$endgroup$
– Rocco
Jan 23 at 15:04
$begingroup$
@AlexyVincenzo P(3 or 6 is rolled) ,P(1 or 4 is rolled) and P(2 or 5 is rolled) are ALWAYS 1/3 each, due to the fact that "die don't have memory" :). What approaches uniformity is the probability of a global victory of A, which in my answer I call P(A), with respect to a global victory of B. That's because, as you "zoom out", the initial advantage of A looks more and more little with respect to the total number of rolls.
$endgroup$
– Rocco
Jan 23 at 15:38
$begingroup$
@AlexyVincenzo P(3 or 6 is rolled) ,P(1 or 4 is rolled) and P(2 or 5 is rolled) are ALWAYS 1/3 each, due to the fact that "die don't have memory" :). What approaches uniformity is the probability of a global victory of A, which in my answer I call P(A), with respect to a global victory of B. That's because, as you "zoom out", the initial advantage of A looks more and more little with respect to the total number of rolls.
$endgroup$
– Rocco
Jan 23 at 15:38
1
1
$begingroup$
It's a matter of semantics if you consider a difference of 1,000 points big or small when doing a million throws. In other words, in my world $N$ and $N+sqrt{N}$ are roughly the same number of points.
$endgroup$
– Ingix
Jan 23 at 17:34
$begingroup$
It's a matter of semantics if you consider a difference of 1,000 points big or small when doing a million throws. In other words, in my world $N$ and $N+sqrt{N}$ are roughly the same number of points.
$endgroup$
– Ingix
Jan 23 at 17:34
|
show 6 more comments
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$begingroup$
I don't understand when the game ends (and then clause 4 is invoked). In (a), it seems to be that it ends after 4 throws. But when does it end in (b)? The only interpretation I can come up with is that it ends after a predetermined number $k$ (and we had $k=4$ in (a)), and the question is how the fair distribution should be for $k to infty$.
$endgroup$
– Ingix
Jan 22 at 18:05