Probability - Calculate probability of win on multiple bets
Clash Royale CLAN TAG#URR8PPP
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I've got this situation. There's a game where I can bet on the result of a random generated number that could be $1$ or $2$. In fact, I've got $50%$ probability that the number will be $1$ and $50%$ probability that it will be $2$.
But what if I bet $10$ times, and I win if I get $1$ just one time? I mean, I could get $2$ four times but then get $1$ and win. Is it true that, if in that $50%$ of lose probability there's another bet with $50%$ probability of win, the final probability is $75%$?
Thanks!
probability
edited Mar 27 '17 at 19:03
Andre.J
698415
asked Mar 27 '17 at 18:40
DamiToma
1061
add a comment |Â
up vote
0
down vote
favorite
I've got this situation. There's a game where I can bet on the result of a random generated number that could be $1$ or $2$. In fact, I've got $50%$ probability that the number will be $1$ and $50%$ probability that it will be $2$.
But what if I bet $10$ times, and I win if I get $1$ just one time? I mean, I could get $2$ four times but then get $1$ and win. Is it true that, if in that $50%$ of lose probability there's another bet with $50%$ probability of win, the final probability is $75%$?
Thanks!
probability
edited Mar 27 '17 at 19:03
Andre.J
698415
asked Mar 27 '17 at 18:40
DamiToma
1061
What do you mean by "final probability"? It is true that you have a 75% chance of winning by the end of two turns.
– Harambe
Mar 27 '17 at 18:47
This is not clear. The probability of guessing wrong ten times in a row is $frac 12^10approx 0.000977$. Is that what you are asking?
– lulu
Mar 27 '17 at 18:48
@Shanye2020 That's exactly what I mean. Thanks!
– DamiToma
Mar 27 '17 at 18:50
@lulu Yes I'm asking that. Sorry for my english, I'm italiano. Thanks!
– DamiToma
Mar 27 '17 at 18:50
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I've got this situation. There's a game where I can bet on the result of a random generated number that could be $1$ or $2$. In fact, I've got $50%$ probability that the number will be $1$ and $50%$ probability that it will be $2$.
But what if I bet $10$ times, and I win if I get $1$ just one time? I mean, I could get $2$ four times but then get $1$ and win. Is it true that, if in that $50%$ of lose probability there's another bet with $50%$ probability of win, the final probability is $75%$?
Thanks!
probability
edited Mar 27 '17 at 19:03
Andre.J
698415
asked Mar 27 '17 at 18:40
DamiToma
1061
I've got this situation. There's a game where I can bet on the result of a random generated number that could be $1$ or $2$. In fact, I've got $50%$ probability that the number will be $1$ and $50%$ probability that it will be $2$.
But what if I bet $10$ times, and I win if I get $1$ just one time? I mean, I could get $2$ four times but then get $1$ and win. Is it true that, if in that $50%$ of lose probability there's another bet with $50%$ probability of win, the final probability is $75%$?
Thanks!
probability
edited Mar 27 '17 at 19:03
Andre.J
698415
asked Mar 27 '17 at 18:40
DamiToma
1061
edited Mar 27 '17 at 19:03
Andre.J
698415
edited Mar 27 '17 at 19:03
Andre.J
698415
edited Mar 27 '17 at 19:03
Andre.J
698415
698415
asked Mar 27 '17 at 18:40
DamiToma
1061
asked Mar 27 '17 at 18:40
DamiToma
1061
asked Mar 27 '17 at 18:40
DamiToma
1061
1061
What do you mean by "final probability"? It is true that you have a 75% chance of winning by the end of two turns.
– Harambe
Mar 27 '17 at 18:47
This is not clear. The probability of guessing wrong ten times in a row is $frac 12^10approx 0.000977$. Is that what you are asking?
– lulu
Mar 27 '17 at 18:48
@Shanye2020 That's exactly what I mean. Thanks!
– DamiToma
Mar 27 '17 at 18:50
@lulu Yes I'm asking that. Sorry for my english, I'm italiano. Thanks!
– DamiToma
Mar 27 '17 at 18:50
add a comment |Â
What do you mean by "final probability"? It is true that you have a 75% chance of winning by the end of two turns.
– Harambe
Mar 27 '17 at 18:47
This is not clear. The probability of guessing wrong ten times in a row is $frac 12^10approx 0.000977$. Is that what you are asking?
– lulu
Mar 27 '17 at 18:48
@Shanye2020 That's exactly what I mean. Thanks!
– DamiToma
Mar 27 '17 at 18:50
@lulu Yes I'm asking that. Sorry for my english, I'm italiano. Thanks!
– DamiToma
Mar 27 '17 at 18:50
What do you mean by "final probability"? It is true that you have a 75% chance of winning by the end of two turns.
– Harambe
Mar 27 '17 at 18:47
What do you mean by "final probability"? It is true that you have a 75% chance of winning by the end of two turns.
– Harambe
Mar 27 '17 at 18:47
This is not clear. The probability of guessing wrong ten times in a row is $frac 12^10approx 0.000977$. Is that what you are asking?
– lulu
Mar 27 '17 at 18:48
This is not clear. The probability of guessing wrong ten times in a row is $frac 12^10approx 0.000977$. Is that what you are asking?
– lulu
Mar 27 '17 at 18:48
@Shanye2020 That's exactly what I mean. Thanks!
– DamiToma
Mar 27 '17 at 18:50
@Shanye2020 That's exactly what I mean. Thanks!
– DamiToma
Mar 27 '17 at 18:50
@lulu Yes I'm asking that. Sorry for my english, I'm italiano. Thanks!
– DamiToma
Mar 27 '17 at 18:50
@lulu Yes I'm asking that. Sorry for my english, I'm italiano. Thanks!
– DamiToma
Mar 27 '17 at 18:50
add a comment |Â
1 Answer
1
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up vote
0
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The probability to guess wrong 10 out of 10 when every single event has a 50 % prob is 100 ÷ (2×2×2×2×2×2×2×2×2×2)= 0.09765625. Not 0,000977
But if you place a bett independently one after the other then the stats are higher. If the first event is lets say black the next event has the same probability to be black 50%. So so for 10 wrong guess is 0.097 % for 9 wrong is 100÷(2×2×2×2×2×2×2×2×2)=0.1953125
When you are at the last pickk after you where wrong 9 times you have 50-50 % chance. Sorry for my bad english
answered Aug 21 at 9:22
eftimie budescu
1
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The probability to guess wrong 10 out of 10 when every single event has a 50 % prob is 100 ÷ (2×2×2×2×2×2×2×2×2×2)= 0.09765625. Not 0,000977
But if you place a bett independently one after the other then the stats are higher. If the first event is lets say black the next event has the same probability to be black 50%. So so for 10 wrong guess is 0.097 % for 9 wrong is 100÷(2×2×2×2×2×2×2×2×2)=0.1953125
When you are at the last pickk after you where wrong 9 times you have 50-50 % chance. Sorry for my bad english
answered Aug 21 at 9:22
eftimie budescu
1
add a comment |Â
up vote
0
down vote
The probability to guess wrong 10 out of 10 when every single event has a 50 % prob is 100 ÷ (2×2×2×2×2×2×2×2×2×2)= 0.09765625. Not 0,000977
But if you place a bett independently one after the other then the stats are higher. If the first event is lets say black the next event has the same probability to be black 50%. So so for 10 wrong guess is 0.097 % for 9 wrong is 100÷(2×2×2×2×2×2×2×2×2)=0.1953125
When you are at the last pickk after you where wrong 9 times you have 50-50 % chance. Sorry for my bad english
answered Aug 21 at 9:22
eftimie budescu
1
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The probability to guess wrong 10 out of 10 when every single event has a 50 % prob is 100 ÷ (2×2×2×2×2×2×2×2×2×2)= 0.09765625. Not 0,000977
But if you place a bett independently one after the other then the stats are higher. If the first event is lets say black the next event has the same probability to be black 50%. So so for 10 wrong guess is 0.097 % for 9 wrong is 100÷(2×2×2×2×2×2×2×2×2)=0.1953125
When you are at the last pickk after you where wrong 9 times you have 50-50 % chance. Sorry for my bad english
answered Aug 21 at 9:22
eftimie budescu
1
The probability to guess wrong 10 out of 10 when every single event has a 50 % prob is 100 ÷ (2×2×2×2×2×2×2×2×2×2)= 0.09765625. Not 0,000977
But if you place a bett independently one after the other then the stats are higher. If the first event is lets say black the next event has the same probability to be black 50%. So so for 10 wrong guess is 0.097 % for 9 wrong is 100÷(2×2×2×2×2×2×2×2×2)=0.1953125
When you are at the last pickk after you where wrong 9 times you have 50-50 % chance. Sorry for my bad english
answered Aug 21 at 9:22
eftimie budescu
1
answered Aug 21 at 9:22
eftimie budescu
1
answered Aug 21 at 9:22
eftimie budescu
1
answered Aug 21 at 9:22
eftimie budescu
1
1
add a comment |Â
add a comment |Â
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What do you mean by "final probability"? It is true that you have a 75% chance of winning by the end of two turns.
– Harambe
Mar 27 '17 at 18:47
This is not clear. The probability of guessing wrong ten times in a row is $frac 12^10approx 0.000977$. Is that what you are asking?
– lulu
Mar 27 '17 at 18:48
@Shanye2020 That's exactly what I mean. Thanks!
– DamiToma
Mar 27 '17 at 18:50
@lulu Yes I'm asking that. Sorry for my english, I'm italiano. Thanks!
– DamiToma
Mar 27 '17 at 18:50