John played the game 4 times, and tells us he has won a total of $8....

[markosheehan]

A game can be played a charity fundraiser. in the game the player spins an arrow on a wheel and wins the amount shown on the the sector that the arrow stops on. there are 5 different sectors $0,$0,$3,$5,$6 and the arrow is just as likely to stop in one sector as any other sector

(i) john played the game 4 times and tells us he has won a total of $8. in how many different ways could he of done this

 

[Deleted member 4993]

A game can be played a charity fundraiser. in the game the player spins an arrow on a wheel and wins the amount shown on the the sector that the arrow stops on. there are 5 different sectors $0,$0,$3,$5,$6 and the arrow is just as likely to stop in one sector as any other sector

(i) john played the game 4 times and tells us he has won a total of $8. in how many different ways could he of done this

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[Ishuda]

A game can be played a charity fundraiser. in the game the player spins an arrow on a wheel and wins the amount shown on the the sector that the arrow stops on. there are 5 different sectors $0,$0,$3,$5,$6 and the arrow is just as likely to stop in one sector as any other sector

(i) john played the game 4 times and tells us he has won a total of $8. in how many different ways could he of done this

Just to be definitive, I would assume that you mean he received $8 for the 4 spins and won (or lost) the $8 minus the cost of spinning the wheel each time.

 

[pka]

A game can be played a charity fundraiser. in the game the player spins an arrow on a wheel and wins the amount shown on the the sector that the arrow stops on. there are 5 different sectors $0,$0,$3,$5,$6 and the arrow is just as likely to stop in one sector as any other sector
(i) john played the game 4 times and tells us he has won a total of $8. in how many different ways could he of done this

The expansion of \(\displaystyle \left(1+x^3+x^5+x^6\right)^4\), found here, actually answers the exact question as posted.
The term in the expansion \(\displaystyle 12x^8\) tells us that there are twelve ways to get a sum of eight in four tries,
However, that in no way tells us anything about taking in consideration the cost of a spin. See the reply above.
Moreover, that expansion does not help with probabilities in any way.
Let me explain. There are twenty-four ways to get to \(\displaystyle \$11,~(0,0,5,6)\text{ or }(0,3,3,5)\) but each of those can appear in twelve way. BUT the probability of \(\displaystyle (0,0,5,6)\) is different from the probability of \(\displaystyle (0,3,3,5)\).

 

[markosheehan]

The expansion of \(\displaystyle \left(1+x^3+x^5+x^6\right)^4\), found here, actually answers the exact question as posted.
The term in the expansion \(\displaystyle 12x^8\) tells us that there are twelve ways to get a sum of eight in four tries,
However, that in no way tells us anything about taking in consideration the cost of a spin. See the reply above.
Moreover, that expansion does not help with probabilities in any way.
Let me explain. There are twenty-four ways to get to \(\displaystyle \$11,~(0,0,5,6)\text{ or }(0,3,3,5)\) but each of those can appear in twelve way. BUT the probability of \(\displaystyle (0,0,5,6)\) is different from the probability of \(\displaystyle (0,3,3,5)\).

do you know any way of solving the question with factorials i thought the answers was 4×3×2×1 as he if cant get the same prize twice E.g he cant get $3 the first spin and then get it $3 again on the second spin as he would not get $8 in total. and the cost does not matter in the question

 

[Ishuda]

The expansion of \(\displaystyle \left(1+x^3+x^5+x^6\right)^4\), found here, actually answers the exact question as posted.
The term in the expansion \(\displaystyle 12x^8\) tells us that there are twelve ways to get a sum of eight in four tries,
However, that in no way tells us anything about taking in consideration the cost of a spin. See the reply above.
Moreover, that expansion does not help with probabilities in any way.
Let me explain. There are twenty-four ways to get to \(\displaystyle \$11,~(0,0,5,6)\text{ or }(0,3,3,5)\) but each of those can appear in twelve way. BUT the probability of \(\displaystyle (0,0,5,6)\) is different from the probability of \(\displaystyle (0,3,3,5)\).

Wouldn't the 12 be for 1 single sector for zero? Seems to me that if we look at on a sector basis, the correct answer is 24. That is, we do know we must have 1 five, 1 three, and 2 zero's. Since the zero's are distinguishable by sectors should the zeros be distinguishable, e.g. a red zero and and a blue zero? If that is the case and we consider the arrangements different, i.e. (0,0,3,5) is different than (0,0,3,5), we have 24 possibilities. Please note, I'm asking - not telling.

 

[pka]

do you know any way of solving the question with factorials i thought the answers was 4×3×2×1 as he if cant get the same prize twice E.g he cant get $3 the first spin and then get it $3 again on the second spin as he would not get $8 in total. and the cost does not matter in the question

This often happens with people who think that surely there is a mathematics formula to just 'plug-into' and out comes an answer. This particular problem is far more complicated than most can even imagine. It involves modeling with strings using repeated entries. The easiest way to count those is using generating functions. Then using knowledge of independent probabilities. The product 4×3×2×1 really has nothing to do with the answer to this question.

Wouldn't the 12 be for 1 single sector for zero? Seems to me that if we look at on a sector basis, the correct answer is 24. That is, we do know we must have 1 five, 1 three, and 2 zero's. Since the zero's are distinguishable by sectors should the zeros be distinguishable, e.g. a red zero and and a blue zero? If that is the case and we consider the arrangements different, i.e. (0,0,3,5) is different than (0,0,3,5), we have 24 possibilities. Please note, I'm asking - not telling.

Now to continue the answer to both concerns. The string \(\displaystyle 0,0,3,5\) is the only set of four outcomes that adds to \(\displaystyle 8\). Note that I used the word set. No order is implied by sets. That string is four characters long but two are repeated. Thus it can be rearranged in \(\displaystyle \frac{4!}{2!}=12\) ways. In this case the generating function is \(\displaystyle (1+x^3+x^5+x^6)^4\). The coefficients corresponding to the exponents tell us the number of ways to get sums.

Now we turn probabilities. \(\displaystyle \mathcal{P}(0)=\tfrac{2}{5},~\mathcal{P}(3)=\tfrac{1}{5},~\mathcal{P}(5)=\tfrac{1}{5},~\mathcal{P}(6)=\tfrac{1}{5}\).
So the probability of the string \(\displaystyle \mathcal{P}(<3,0,5,0>)=\dfrac{4}{625}\) That for just that one arrangement!
There are twelve possible. Thus \(\displaystyle \mathcal{P}($8)=\dfrac{48}{625}\).

 

[Ishuda]

Wouldn't the 12 be for 1 single sector for zero? Seems to me that if we look at on a sector basis, the correct answer is 24. That is, we do know we must have 1 five, 1 three, and 2 zero's. Since the zero's are distinguishable by sectors should the zeros be distinguishable, e.g. a red zero and and a blue zero? If that is the case and we consider the arrangements different, i.e. (0,0,3,5) is different than (0,0,3,5), we have 24 possibilities. Please note, I'm asking - not telling.

WELL i CAN'T ALWAYS BE CORRECT: Just think of the 1 as a blue zero;
0, 0, 3, 5 == 0, 0, 5, 3 == 0, 1, 3, 5 == 0, 1, 5, 3 == 1, 0, 3, 5 == 1, 0, 5, 3 == 1, 1, 3, 5 == 1, 1, 5, 3

0, 3, 0, 5 == 0, 5, 0, 3 == 0, 3, 1, 5 == 0, 5, 1, 3 == 1, 3, 0, 5 == 1, 5, 0, 3 == 1, 3, 1, 5 == 1, 5, 1, 3

3, 0, 0, 5 == 5, 0, 0, 3 == 3, 0, 1, 5 == 5, 0, 1, 3 == 3, 1, 0, 5 == 5, 1, 0, 3 == 3, 1, 1, 5 == 5, 1, 1, 3

3, 0, 5, 0 == 5, 0, 3, 0 == 3, 0, 5, 1 == 5, 0, 3, 1 == 3, 1, 5, 0 == 5, 1, 3, 0 == 3, 1, 5, 1 == 5, 1, 3, 1

3, 5, 0, 0 == 5, 3, 0, 0 == 3, 5, 0, 1 == 5, 3, 0, 1 == 3, 5, 1, 0 == 5, 3, 1, 0 == 3, 5, 1, 1 == 5, 3, 1, 1

5, 3, 0, 0 == 3, 5, 0, 0 == 5, 3, 0, 1 == 3, 5, 0, 1 == 5, 3, 1, 0 == 3, 5, 1, 0 == 5, 3, 1, 1 == 3, 5, 1, 1
The bold is if there is no difference between the two zero's.