Mixing
Candy Crush - Expected extra moves
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Candy Crush have recently introduced a new "feature" where by watching an ad you can get to spin a wheel to get extra moves (you used to have to "pay" for it).
This option pops up if you fail to complete the level in the allotted moves.
You spin a wheel which has 8 possible outcomes. 4 of them give you extra moves and 4 do not.
Out of the 4 that do, one gives you 15 extra moves and three give you five.
If you still fail to complete the level you will again get the same "offer" and so on.
Assuming that the probability of hitting any of the outcomes is the same and that I never manage to finish the level, what would be my expected number of extra moves on any one level?
TBH, I have no idea even where to start.
Edit:
If you do NOT win extra moves then the level is over and you have failed.
statistics expected-value
asked Jan 20 at 11:38
theblitztheblitz
1033
$endgroup$
add a comment |
$begingroup$
Candy Crush have recently introduced a new "feature" where by watching an ad you can get to spin a wheel to get extra moves (you used to have to "pay" for it).
This option pops up if you fail to complete the level in the allotted moves.
You spin a wheel which has 8 possible outcomes. 4 of them give you extra moves and 4 do not.
Out of the 4 that do, one gives you 15 extra moves and three give you five.
If you still fail to complete the level you will again get the same "offer" and so on.
Assuming that the probability of hitting any of the outcomes is the same and that I never manage to finish the level, what would be my expected number of extra moves on any one level?
TBH, I have no idea even where to start.
Edit:
If you do NOT win extra moves then the level is over and you have failed.
statistics expected-value
asked Jan 20 at 11:38
theblitztheblitz
1033
$endgroup$
$begingroup$
I don't understand what you mean by "I never manage to finish the level". This would imply it is possible to achieve infinite extra moves by simply repeatedly failing the level - hence the expected value is infinite.
$endgroup$
– Peter Foreman
Jan 20 at 11:45
1
$begingroup$
Maybe it wasn't clear but if you don't win extra moves then your game is over.
$endgroup$
– theblitz
Jan 20 at 11:46
add a comment |
$begingroup$
Candy Crush have recently introduced a new "feature" where by watching an ad you can get to spin a wheel to get extra moves (you used to have to "pay" for it).
This option pops up if you fail to complete the level in the allotted moves.
You spin a wheel which has 8 possible outcomes. 4 of them give you extra moves and 4 do not.
Out of the 4 that do, one gives you 15 extra moves and three give you five.
If you still fail to complete the level you will again get the same "offer" and so on.
Assuming that the probability of hitting any of the outcomes is the same and that I never manage to finish the level, what would be my expected number of extra moves on any one level?
TBH, I have no idea even where to start.
Edit:
If you do NOT win extra moves then the level is over and you have failed.
statistics expected-value
asked Jan 20 at 11:38
theblitztheblitz
1033
$endgroup$
Candy Crush have recently introduced a new "feature" where by watching an ad you can get to spin a wheel to get extra moves (you used to have to "pay" for it).
This option pops up if you fail to complete the level in the allotted moves.
You spin a wheel which has 8 possible outcomes. 4 of them give you extra moves and 4 do not.
Out of the 4 that do, one gives you 15 extra moves and three give you five.
If you still fail to complete the level you will again get the same "offer" and so on.
Assuming that the probability of hitting any of the outcomes is the same and that I never manage to finish the level, what would be my expected number of extra moves on any one level?
TBH, I have no idea even where to start.
Edit:
If you do NOT win extra moves then the level is over and you have failed.
statistics expected-value
statistics expected-value
asked Jan 20 at 11:38
theblitztheblitz
1033
asked Jan 20 at 11:38
theblitztheblitz
1033
edited Jan 20 at 11:47
theblitz
asked Jan 20 at 11:38
theblitztheblitz
1033
asked Jan 20 at 11:38
theblitztheblitz
1033
asked Jan 20 at 11:38
theblitztheblitz
1033
1033
$begingroup$
I don't understand what you mean by "I never manage to finish the level". This would imply it is possible to achieve infinite extra moves by simply repeatedly failing the level - hence the expected value is infinite.
$endgroup$
– Peter Foreman
Jan 20 at 11:45
1
$begingroup$
Maybe it wasn't clear but if you don't win extra moves then your game is over.
$endgroup$
– theblitz
Jan 20 at 11:46
add a comment |
$begingroup$
I don't understand what you mean by "I never manage to finish the level". This would imply it is possible to achieve infinite extra moves by simply repeatedly failing the level - hence the expected value is infinite.
$endgroup$
– Peter Foreman
Jan 20 at 11:45
1
$begingroup$
Maybe it wasn't clear but if you don't win extra moves then your game is over.
$endgroup$
– theblitz
Jan 20 at 11:46
$begingroup$
I don't understand what you mean by "I never manage to finish the level". This would imply it is possible to achieve infinite extra moves by simply repeatedly failing the level - hence the expected value is infinite.
$endgroup$
– Peter Foreman
Jan 20 at 11:45
$begingroup$
I don't understand what you mean by "I never manage to finish the level". This would imply it is possible to achieve infinite extra moves by simply repeatedly failing the level - hence the expected value is infinite.
$endgroup$
– Peter Foreman
Jan 20 at 11:45
1
1
$begingroup$
Maybe it wasn't clear but if you don't win extra moves then your game is over.
$endgroup$
– theblitz
Jan 20 at 11:46
$begingroup$
Maybe it wasn't clear but if you don't win extra moves then your game is over.
$endgroup$
– theblitz
Jan 20 at 11:46
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The expected number of extra moves is $7.5$. Assume that the expected value is $E(X) in mathbb{R}$. We then know that we either get $+5$ extra moves and then continue to achieve $E(X)$ extra moves (as expected) or we get $+15$ extra moves and then achieve $E(X)$ extra moves each with $frac{3}{8}$ and $frac{1}{8}$ chances respectively. Thus we have: $$E(X) = frac{3}{8} (5+E(X)) + frac{1}{8} (15+E(X))$$
$$therefore E(X) = frac{15}{2} = 7.5$$
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
$endgroup$
1
$begingroup$
Wow. Once someone shows you it really looks so simple. :)
$endgroup$
– theblitz
Jan 20 at 12:15
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
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oldest
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active
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votes
$begingroup$
The expected number of extra moves is $7.5$. Assume that the expected value is $E(X) in mathbb{R}$. We then know that we either get $+5$ extra moves and then continue to achieve $E(X)$ extra moves (as expected) or we get $+15$ extra moves and then achieve $E(X)$ extra moves each with $frac{3}{8}$ and $frac{1}{8}$ chances respectively. Thus we have: $$E(X) = frac{3}{8} (5+E(X)) + frac{1}{8} (15+E(X))$$
$$therefore E(X) = frac{15}{2} = 7.5$$
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
$endgroup$
1
$begingroup$
Wow. Once someone shows you it really looks so simple. :)
$endgroup$
– theblitz
Jan 20 at 12:15
add a comment |
$begingroup$
The expected number of extra moves is $7.5$. Assume that the expected value is $E(X) in mathbb{R}$. We then know that we either get $+5$ extra moves and then continue to achieve $E(X)$ extra moves (as expected) or we get $+15$ extra moves and then achieve $E(X)$ extra moves each with $frac{3}{8}$ and $frac{1}{8}$ chances respectively. Thus we have: $$E(X) = frac{3}{8} (5+E(X)) + frac{1}{8} (15+E(X))$$
$$therefore E(X) = frac{15}{2} = 7.5$$
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
$endgroup$
1
$begingroup$
Wow. Once someone shows you it really looks so simple. :)
$endgroup$
– theblitz
Jan 20 at 12:15
add a comment |
$begingroup$
The expected number of extra moves is $7.5$. Assume that the expected value is $E(X) in mathbb{R}$. We then know that we either get $+5$ extra moves and then continue to achieve $E(X)$ extra moves (as expected) or we get $+15$ extra moves and then achieve $E(X)$ extra moves each with $frac{3}{8}$ and $frac{1}{8}$ chances respectively. Thus we have: $$E(X) = frac{3}{8} (5+E(X)) + frac{1}{8} (15+E(X))$$
$$therefore E(X) = frac{15}{2} = 7.5$$
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
$endgroup$
The expected number of extra moves is $7.5$. Assume that the expected value is $E(X) in mathbb{R}$. We then know that we either get $+5$ extra moves and then continue to achieve $E(X)$ extra moves (as expected) or we get $+15$ extra moves and then achieve $E(X)$ extra moves each with $frac{3}{8}$ and $frac{1}{8}$ chances respectively. Thus we have: $$E(X) = frac{3}{8} (5+E(X)) + frac{1}{8} (15+E(X))$$
$$therefore E(X) = frac{15}{2} = 7.5$$
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
answered Jan 20 at 12:01
Peter ForemanPeter Foreman
2,7421214
2,7421214
1
$begingroup$
Wow. Once someone shows you it really looks so simple. :)
$endgroup$
– theblitz
Jan 20 at 12:15
add a comment |
1
$begingroup$
Wow. Once someone shows you it really looks so simple. :)
$endgroup$
– theblitz
Jan 20 at 12:15
1
1
$begingroup$
Wow. Once someone shows you it really looks so simple. :)
$endgroup$
– theblitz
Jan 20 at 12:15
$begingroup$
Wow. Once someone shows you it really looks so simple. :)
$endgroup$
– theblitz
Jan 20 at 12:15
add a comment |
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$begingroup$
I don't understand what you mean by "I never manage to finish the level". This would imply it is possible to achieve infinite extra moves by simply repeatedly failing the level - hence the expected value is infinite.
$endgroup$
– Peter Foreman
Jan 20 at 11:45
1
$begingroup$
Maybe it wasn't clear but if you don't win extra moves then your game is over.
$endgroup$
– theblitz
Jan 20 at 11:46