The Dice Game - Probabilities.

The Dice Game

Three players A, B, C play a game with a single dice.

The rules of the game are:

Player A ALWAYS goes first.

A rolls the dice. If the dice lands showing a 1 then A wins the game. If A does not throw a 1 then B has a turn.

B rolls the dice. If the dice lands showing a 2 or a 3 then B wins the game. If B does not throw a 2 or 3, then C has a turn.

C rolls the dice. If the dice lands showing a 4 or a 5 or a 6 then C wins the game. If C does not throw a 4 or a 5 or a 6 then A starts again.

The procedure continues until there is a winner.

INVESTIGATE ANY OR ALL OF

The probabilities of each A, B, or C winning the game.

Who will be the most likely winner?

The most likely length of the game in terms of the number of rolls of the dice to produce a winner.

I’ve decided to investigate question 1 out of the three.

Player A has to throw a 1 to win the game. This means that in terms of probability, the probability that A will win is going to be (1/6). The probability that A will loose is (5/6) because:

1 – (1/6) = (5/6)

Player B has to throw a 2 or a 3 to win the game. This means that the probability that B will win would be (2/6), because:

(1/6) + (1/6) = (2/6)

The probability that B will loose would be (4/6) because that is the remainder of 1:

1 – (2/6) = (4/6)

Player C has to throw a 4 or a 5 or a 6 to win the game. This means that the probability that C will win would be (3/6) because:

(1/6) + (1/6) + (1/6) = (3/6)

The probability that C will loose the game would be (3/6), because if you have the value that C will win, and take away the probability that C will win from that, you will get that result:

1 – (3/6) = (3/6)

From this information, I create a tree diagram like so:

Player A

Using this tree diagram, I can now try and find out the probability that A will win.

To begin with, P(A) wins on his first go would be (1/6). However, if A doesn’t win on his first attempt, then to get a second go, A has to have lost, B has to have lost and C has to have lost in order for A to have another chance. In essence, what is being said is:

“A will win on his first go, OR A will have lost and B would have lost and C would have lost and then A will have another go.”

To put this into a sum, using the probability rules of AND = ×, OR = +, I can begin to make this series:

P(A) = (1/6) + [(5/6) × (4/6) × (3/6) × (1/6)] …

For A to try and win the third time if he looses the other two times, he would have lost twice, B would have lost twice and C would have lost twice before having another go, so if I dictated what would have happened for A to have a third go, this is what would have happened:

“A would have lost and B would have lost and C would have lost and A would have lost and B would have lost and C would have lost and then A would have another turn.”

To put this into terms of probability, the sum for A to have a third go would be this:

“P(A) on third go = [(5/6) × (4/6) × (3/6)]² × (1/6)

There is one problem with trying to find out the probability of A winning, which is that the instructions do not specify when A will win the game. This means that you could go on indefinitely saying that A might not have won the time before but might win the next time. However, if I multiply the probability that nobody wins in a round by P(A), I will have a series like this:

[(5/6) × (4/6) × (3/6) × P(A)] = [(5/6) × (4/6) × (3/6) × (1/6)] + [(5/6) × (4/6) × (3/6)]² × (1/6) …

The first term of this series would be [(5/6) × (4/6) × (3/6) × (1/6)] because that is the same as multiplying P(A) by the probability that nobody wins in a round, which is the expression on the left side of the equals sign. If I line these series up one under the other, I will have something that looks like this:

P(A) = (1/6) + [(5/6) × (4/6) × (3/6) × (1/6)] + {[(5/6) × (4/6) × (3/6)]² × (1/6)} + …

[P(A) × (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6) × (3/6) × (1/6)] + {[(5/6) × (4/6) × (3/6)]² × (1/6)} + {[(5/6) × (4/6) × (3/6)]³ × (1/6)} + …

I noticed that if I take away [(P(A) × (5/6) × (4/6) × (3/6)] from P(A), then it would cancel out the two coloured terms. This means that you would not have to take into account the question of when player A will win. You would end up with the values in bold if that procedure was carried out, and it would look like this:

P(A) – ...

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[P(A) × (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6) × (3/6) × (1/6)] + {[(5/6) × (4/6) × (3/6)]² × (1/6)} + {[(5/6) × (4/6) × (3/6)]³ × (1/6)} + …

P(A) – [(P(A) × (5/6) × (4/6) × (3/6)] = (1/6) + {[(5/6) × (4/6) × (3/6)]³ × (1/6)}

The last value that is left over is not necessarily going to be [((5/6) × (4/6) × (3/6))³ × (1/6)] because since both series go on infinitely, if you take [(5/6) × (4/6) × (3/6) × P(A)] away from P(A), then you will cancel out every term that is the same, but always the second series will have that “extra” term at the end. However, this value is going to be so small, that it can be ignored since that value will be very close to zero, and it will not make any difference to the probability of A winning. Leaving it out will leave this:

P(A) – [(P(A) × (5/6) × (4/6) × (3/6)] = (1/6)

Now, to find the probability that A will win, I have to rearrange this formula slightly.

P(A) – [(P(A) × (5/6) × (4/6) × (3/6)] can be simplified to give this:

[P(A)][1 – (5/6) × (4/6) × (3/6)] = (1/6)

Now if I take [1 – (5/6) × (4/6) × (3/6)] and change it over to the other side, I can finally gain an answer for the probability that A will win:

P(A) = (1/6) = (1/6) = (3/13)

(1 – (5/6) × (4/6) × (3/6)) (13/18)

Now I am going to find out the probability that B would win.

For B to have a first chance of trying to win, A has to have lost, and then B would have a go, so the first term for the series P(B) is:

P(B) = (5/6) × (2/6)

For B to have a second go, A would have had to have lost, then B would have had to have lost, then C would have had to have lost, then A would have had to have lost again, before B can have another go. This long-winded phrase will end up as this:

P(B) 2nd go = (5/6) × (4/6) × (3/6) × (5/6) × (2/6)

For B to have a third go, it is an extended version of the 2nd go: A would have lost, B would have lost, C would have lost, then back to A but he lost, B would have lost again, C would have lost again, A would lose again and then (finally!) B would have a 3rd go. This phrase would look like this:

P(B) 3rd go = [(5/6) × (4/6) × (3/6)]² × (5/6) × (2/6)

Now, if I put all the values into a long series for P(B), this is what it would look like:

P(B) = [(5/6) × (2/6)] + [(5/6)² × (4/6) × (3/6) × (2/6)] + {[(5/6) × (4/6) × (3/6)]² × (5/6) × (2/6)}…

The problem is the same as before in that I don’t know when B will win, so I have to cancel out the times when B will not win to find out the probability that B will win. To do this, I have to multiply the probability that everyone will lose with the probability that B will win. This would lead to a series like so:

P(B) × (5/6) × (4/6) × (3/6) = [(5/6)² × (4/6) × (3/6) × (2/6)] + {[(5/6) × (4/6) × (3/6)]² × (5/6) × (2/6)} + {[(5/6) × (4/6) × (3/6)]³ × (5/6) × (2/6)}…

With this, I now have to take that series away from the series of P(B), and I will be left with the following:

P(B) – [P(B) × (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6)] + {[(5/6) × (4/6) × (3/6)]³ × (5/6) × (2/6)}

The term that is printed in bold is the infinitely small term, and can actually be ignored in this circumstance because it will make no difference to the answer since it is so close to zero. Leaving it out will leave this:

P(B) – [P(B) × (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6)]

Now, if I simplify it then I can get this:

[P(B)][1 – (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6)]

As before, I shall now take over the [1 – (5/6) × (4/6) × (3/6)] and will be able to obtain an answer:

P(B) = (5/6) × (2/6) = (5/18) = (5/13)

(1 – (5/6) × (4/6) × (3/6)) (13/18)

I shall now find out what the probability is for C winning the game.

For C to have a first go, A has to have lost, B has to have lost and then C would have a go. “Translated” into the language of probability, what is being said here is:

“A has lost and B has lost and then C will have a go.”

This means that the first term for the series P(C) is:

P(C) = (5/6) × (4/6) × (3/6)

The second term would be found like this:

“A has lost and B has lost and C has lost and A has lost and B has lost and C will have another go.”

So basically that would mean that the second term of the series would be:

P(C) second go = [(5/6) × (4/6) × (3/6)]²

The third term is also found in the same fashion, so that A would have lost three times, B would have lost three times, and C would have lost twice and would have his third go. That would altogether mount up to the third term, which would be:

P(C) third go = [(5/6) × (4/6) × (3/6)]³

Altogether, the series would mount up like this:

P(C) = [(5/6) × (4/6) × (3/6)] + [(5/6) × (4/6) × (3/6)]² + [(5/6) × (4/6) × (3/6)]³…

However, like the other probability series, it is not known when C will win, so I have to create another series to cancel out all the times that C would not win. If I multiply the probability that they all lose by the probability that C will win, I will get a series that looks like this:

P(C) × (5/6) × (4/6) × (3/6) = [(5/6) × (4/6) × (3/6)]² + [(5/6) × (4/6) × (3/6)]³ + [(5/6) × (4/6) × (3/6)]…

If I take this series away from the probability that C will win, then I will eliminate all of times that C will not win, and be left with the probability that C will win. This is what the long series will look like:

P(C) – [P(C) × (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6) × (3/6)] + [(5/6) × (4/6) × (3/6)]

The term in bold would be the infinitely small term that would be found if you continued with the series to infinity, and is too small a value to make any difference, so it can be ignored. The left side can be simplified down to:

[P(C)][1 – (5/6) × (4/6) × (3/6)] = [(5/6) × (4/6) × (3/6)]

Now the term [1 – (5/6) × (4/6) × (3/6)] can be changed over to the other side to gain an answer. Here is the worked answer:

P(C) = (5/6) × (4/6) × (3/6) = (5/18) = (5/13)

[1 – (5/6) × (4/6) × (3/6)] (13/18)

Now I’m going to try and find out the formulae for these probabilities. To do this, I have to change the probability tree around slightly, by replacing the fraction values with letters. This is the new probability tree that I am going to use to find out the formulae:

Win

P(A)

Lose

So, replacing the probability fractions with letters gives you this series if I changed the P(A) series:

P(A) = x + (ywqx) + x(ywq)²…

This is what the series basically would be, and like the numerical version, it will continue to infinity. As with the numerical version, it is not stated when A will win the game, so I have to use the same method as before to cancel out all the times that A will not win. For finding out the probabilities of A, B and C winning, I had to create a series that would cancel out all the times that they would not win individually, and I can change that series into algebra too. This new series looked like this numerically and had been taken away from P(A) to cancel out the probabilities that A would not win:

[P(A) × (5/6) × (4/6) × (3/6) ] = [(5/6) × (4/6) × (3/6) × (1/6)] + [(5/6) × (4/6) × (3/6)]² × (1/6) …

Changed into algebra, this series would look like this:

[P(A)](ywq) = (ywqx) + x(ywq)² + x(ywq)³…

Using this by taking this series away from P(A) I can get an answer. If I take away [P(A)](ywq) from P(A) I get this:

P(A) – [P(A)](ywq) = x + x(ywq)³

The term in bold print is the very last term that would occur in this subtraction. However, that term represents the infinitely small fraction and can be ignored because it will make no difference to the answer. So I would be left with this:

P(A) – [P(A)](ywq) = x

Now I can simplify the left hand side of the equation to:

[P(A)][1 – ywq] = x

Now I can take the [1 – ywq] term over to the other side and that leaves the formula for finding out P(A):

P(A) = x

(1 – ywq)

To get the formula for P(B) I have to use the same method and change the fractions into letters. The series for P(B) is would be this in letter form:

P(B) = (yz) + (y²wqz) + (y³w²q²z)…

Like the P(A) series, this series will continue on into infinity. The problem is that I don’t know when B will win, so I have to cancel out the times B will not win to leave the probability that B will win. By multiplying the probability that B will win with the probability that they all lose, I ended up with this series:

P(B) × (5/6) × (4/6) × (3/6) = [(5/6)² × (4/6) × (3/6) × (2/6)] + {[(5/6) × (4/6) × (3/6)]² × (5/6) × (2/6)} + {[(5/6) × (4/6) × (3/6)]³ × (5/6) × (2/6)}…

Changing all of this into the algebraic version will give you this:

[P(B)](ywq) = (y²wqz) + (y³w²q²z) + (y w³q³z)…

Now, if I take [P(B)](ywq) away from P(B), this is what would happen:

P(B) – [P(B)](ywq) = yz + (y w³q³z)

It has cancelled out all of the times that B would not win. The type in bold is the infinitely small value again, and as before can be ignored because it will not make any difference to the actual solution. I can simplify the left hand side also, leaving me with this:

[P(B)][1 – ywq] = yz

Now with this I can take [1 – ywq] over and get left with the formula for finding P(B):

P(B) = yz

(1 – ywq)

To obtain the formula for P(C), I have to use the same method as twice before. Here is what the series for P(C) would be:

P(C) = (ywr) + (y²w²qr) + (y³w³q²r)…

Again, I don’t know when C will win, so I have to use the same method as before to get rid of all the times that C will not win. If I multiply the probability that C will win with the probability that everyone loses, then I will get a series that looks like this numerically:

P(C) × (5/6) × (4/6) × (3/6) = [(5/6) × (4/6) × (3/6)]² + [(5/6) × (4/6) × (3/6)]³ + [(5/6) × (4/6) × (3/6)]…

Changing all these values into the algebraic form, I will get this:

[P(C)](ywq) = (y²w²qr) + (y³w³q²r) + (y w q³r)…

By taking [P(C)](ywq) away from P(C), I will be left with this:

P(C) – [P(C)](ywq) = (ywr) + (y w q³r)

The term in bold is the infinitely small term and can be ignored since it will make no difference to the answer since it is so small. The terms on the left side of the equation can be simplified down to look like the following, and I have also in steps simplified it down until I had the formula for P(C):

[P(C)][1 – ywq] = ywr

P(C) = ywr

(1 – ywq)

So, all in all, I have the three formulae to find out the solutions to P(A), P(B) and P(C). However, there must be a way to simplify that down even further so I can use just one formula to find out these probabilities.

Basically what these series are made of in their long series form is a number that is continuously multiplied by another number that grows by the number of times you increase its power. All this is then added together. I’m going to call this number “d”. The number that has an index I’m going to call “k”. This new series I will call “B”, and it would look like this compared to an algebraic series taken from before:

P(C) = (ywr) + (y²w²qr) + (y³w³q²r)…

B = d + dk + dk²…

The problem here is that the index number would go on infinitely, so there is a formula for the index. This is going to be kª , because k is at the power 0 on the first term, and 0 is one less than 1. So, this means that the sequence would look like this:

B = d + dk + dk²…+ dkª

The problem with this investigation was that it did not indicate when the players were going to win. I had to create another series by multiplying the probability that a player would win with the probability that all three players would lose. An example of this series algebraically is this:

[P(C)](ywq) = (y²w²qr) + (y³w³q²r) + (y w q³r)

If I change this series around into the newer algebraic version, it will look like this:

Bk = dk + dk² + dk³…+ dkª + dkª

I would then have to take the new series away from the “probability that the player would win” series, simplified it and then obtain the answer. I am now going to do the same thing for this new set of series, and from this I should hopefully obtain the formula for this whole investigation:

B – Bk = d – dkª

B(1 – k) = d(1 – kª)

B = d(1 – kª)

(1 – k)

This is the formula for the investigation. It is called “the sum of a geometric series”. This investigation was based on geometric series, but in this case I was able to obtain an answer. In this investigation, k stands for (5/6) × (4/6) × (3/6), which sums up to a total that is smaller than 1. When you give the fraction a power, you make the fraction smaller the bigger the power number is. If you make the index as large as infinity, then the fraction will get smaller and smaller. This value is so close to zero that it can almost be counted as zero, because the value is so infinitely small. That is why I ignored the infinitely small term; it would make no difference to the answers. kª in this case is 0. 1 – 0 = 1. d × 1 = d. With these sums, the formula would look like this:

B = d

(1 – k)

This is the full algebraic solution to the investigation; the formulae that I found for each of the probabilities look somewhat like this. But what would happen if k was not a fraction? Well, any number that is larger than 1 will grow very large when you give the number a power. If the power were infinitely big, then the number would be infinitely big too. With an infinitely big number, you cannot obtain an answer, because infinity is an endless number. What I mean is 1 – (k to the power of infinity) is immeasurable. The reason why in this investigation I was able to obtain an answer to the probability series was because of k being smaller than 1, i.e. a fraction, I was able to obtain an answer, and the number is a fraction because you use fractions in probability.

This concludes my investigation.