Okay , see you need to understand that it is economics section therefore I do think a coin probability of landing on one side does not truly display the economic situation .
I still would like to answer your question
See Probability= Favourable outcome/total number of outcomes .
Now it's 1/2 , which is for one bet
For 10 bets , getting the same side will be , 1/2*1/2*1/2*1/2-------- 10 times - = 1/1024
If you are going to consider only one flip then you have to use different method .
If it's just second flip :-
1*1/2*1*1--------- = idea is you have to calculate it for the flip you want. But at the same time you have to write 1 for the flip you are not considering.
Lemme know if you understand otherwise I will try and elaborate it more .
Topic: Probability problem heads or tails ... (Read 260 times)
OP needs to go back and learn some basics of probability theory.
If you flip a coin 10 times what is the probability of winning all 10 bets ? Formula ?
Of course, we all understand that theoretically heads or tails can fall out with equal probability. However, heads can fall out in 100% of cases, as well as tails. I believe that such an event cannot be predicted with 100% certainty.
The more times you repeat the experiment, the closer the results get to 50/50. Have a look at the law of large numbers. Each individual coin toss gives a random result, but those random results average closer and closer to 50% with each additional coin toss you make. This is how casinos work. They can make a big loss on an individual game, but they know the overall percentage over many games averages out in their favour.
Of course in theory the outcome of an individual coin toss is pre-determined and dependent on spin speed and direction, air resistance, etc - but that's a separate issue.
If you flip a coin 10 times what is the probability of winning all 10 bets ? Formula ?
Player514 gives the correct answer above.
If an example helps, let's consider all possible outcomes, based on a single coin toss being 50/50 heads or tails.
1 coin toss:
H
T
2 outcomes, 50% chance of each (1/2)
2 coin tosses:
HH
HT
TH
TT
4 outcomes (2^2), 25% chance of each (1/4). So 25% chance of HH, 50% chance of one head one tail (TH outcome plus HT outcome), 25% chance of TT.
3 coin tosses:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
8 outcomes (2^3), 12.5% chance of each. So for example chance of getting exactly 2 heads one tail is 37.5% ( = 3/8 = 3 outcomes out of the 8 total), in bold above.
The maths might appear complicated, but is actually quite straightforward. Have a look at this for an example of calculating probability of winning a lottery.
The chance of the 10th coin being heads is 50%.
But the chance of all 9 preceding coins being heads is (1/2)^9, or 1/512.
So the 10th coin being heads makes your overall sequence ten consecutive heads, so the formula becomes (1/2)^10, or 1/1024.
If an example helps, let's consider all possible outcomes, based on a single coin toss being 50/50 heads or tails.
1 coin toss:
H
T
2 outcomes, 50% chance of each (1/2)
2 coin tosses:
HH
HT
TH
TT
4 outcomes (2^2), 25% chance of each (1/4). So 25% chance of HH, 50% chance of one head one tail (TH outcome plus HT outcome), 25% chance of TT.
3 coin tosses:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT
8 outcomes (2^3), 12.5% chance of each. So for example chance of getting exactly 2 heads one tail is 37.5% ( = 3/8 = 3 outcomes out of the 8 total), in bold above.
The maths might appear complicated, but is actually quite straightforward. Have a look at this for an example of calculating probability of winning a lottery.
so it's 1/1024 chance ...
Yes.The chance of the 10th coin being heads is 50%.
But the chance of all 9 preceding coins being heads is (1/2)^9, or 1/512.
So the 10th coin being heads makes your overall sequence ten consecutive heads, so the formula becomes (1/2)^10, or 1/1024.
The coin has exactly 2 sides. Tossing a coin every time, you have exactly 2 options that can fall out: obverse or reverse. Accordingly, every time you throw a coin, you get 2 equally probable options.
100%: 2 = 50% for each option, in any of the selected shots.
100%: 2 = 50% for each option, in any of the selected shots.
Two faced a character in the Batman series uses a toss coin to make a decision to kill someone, so even if he wants too kill Joker he has no choice but to abide by what the result of the tossed coin, so for me it's 50-50% that you will win unless you are good at tossing coin and you can manipulate the results.
This thread belongs to the gambling discussion I guess.
But anyway I'd like to give my opinion, this is a very simple question but we can find a lot of information about its probability, at first look, you can tell that the probability is 50/50 but actually it could be not because depending on the form of the coin, the head and the tail is not always equal, or the other side could be heavier.
you might be interested with reading this article - https://www.ripleys.com/weird-news/coin-toss-or-not/
But anyway I'd like to give my opinion, this is a very simple question but we can find a lot of information about its probability, at first look, you can tell that the probability is 50/50 but actually it could be not because depending on the form of the coin, the head and the tail is not always equal, or the other side could be heavier.
you might be interested with reading this article - https://www.ripleys.com/weird-news/coin-toss-or-not/
If i flip a coin 10 times, what is the probability of getting heads on the 10th flip ? or on the second flip ? what is the math formula ?
It is still 50%, because it only has 2 sides of money. Every time you fling it up, it has only two possibilities. even if you roll it 1000 times it still has the same probability.I don't know why it is still in discussion but definitely, there's only one answer here. Obviously, 50% as you already explained it. Maybe the OP is looking for other answers that he wants to hear. But sorry, there's only one answer here though - still 50% no matter how many flips you will have. If he wants, go back to your probability class.
The coin has exactly 2 sides. Tossing a coin every time, you have exactly 2 options that can fall out: obverse or reverse. Accordingly, every time you throw a coin, you get 2 equally probable options.
100%: 2 = 50% for each option, in any of the selected shots.
100%: 2 = 50% for each option, in any of the selected shots.
The coin has exactly 2 sides. Tossing a coin every time, you have exactly 2 options that can fall out: obverse or reverse. Accordingly, every time you throw a coin, you get 2 equally probable options.
100%: 2 = 50% for each option, in any of the selected shots.
100%: 2 = 50% for each option, in any of the selected shots.
This is more like solving a mathematical issue. It is either 10/10 or 2/10 at the second flip. More on the other option is like 2:10 because at the second flip, you could turn it up or at the 10th time.
it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
I think your question is different from what you're asking then. Are you asking: what is the probability I can win $1024 on the 10th round?
This is different from the coin question. If you're asking the chance of winning a coin flip back to back 10 times, then you calculate as such:
On each round you have a 1/2 chance of winning or doubling your initial bet. This is because it's a coin flip.
However, now you need to keep in mind that you need to win all of these in a row. As in, you need a coin flip to be heads on the first, second, third, etc. flips. This means out of 10 flips, all 10 must be heads to double your money.
In the case that we have 10 flips and each flip has a 1/2 chance of giving you a payout, we have (1/2) * (1/2) * (1/2) ... ten times, or (1/2) ^ 10 which is a 0.0009765625 chance of doubling your money 10 times in a row.
That's a pretty small chance. You might end up spending more than what you would win.
Nope was thinking of something else ...
rate of return on investment (roi) 2% / month
kinder garden method:
1.02
1.0404
1.061208
1.08243216
1.104080803
1.126162419
1.148685667
1.17165938
1.195092568
1.218994419 final balance
1/1.218994419 = 0.820348301
1- 0.820348301 = 0.179651699
17.965169946 % chance to lose money on investment
Using simple compound interes formula 1×(1+0.02)^10 = 1.21899442
....
17.965170013% chance to lose money on investment
I'm not really sure what you're doing with this. ROI is not a chance based thing. Getting 2% per month is telling you how much you're increasing by, not the chance of you increasing/decreasing your wealth. I don't think this logic is fully sound, but again I'm not too sure about what you were trying to get at with this.
If i flip a coin 10 times, what is the probability of getting heads on the 10th flip ? or on the second flip ? what is the math formula ?
It is still 50%, because it only has 2 sides of money. Every time you fling it up, it has only two possibilities. even if you roll it 1000 times it still has the same probability. 
It's still a probability based formula. I guess there is no known formula for telling precisely which side of the it will be top or down. So maths people basically created a probability based formula for the problem. All the factors that make a coin flip to a particular side has to be known first to create a definite formula.
it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
I think your question is different from what you're asking then. Are you asking: what is the probability I can win $1024 on the 10th round?
This is different from the coin question. If you're asking the chance of winning a coin flip back to back 10 times, then you calculate as such:
On each round you have a 1/2 chance of winning or doubling your initial bet. This is because it's a coin flip.
However, now you need to keep in mind that you need to win all of these in a row. As in, you need a coin flip to be heads on the first, second, third, etc. flips. This means out of 10 flips, all 10 must be heads to double your money.
In the case that we have 10 flips and each flip has a 1/2 chance of giving you a payout, we have (1/2) * (1/2) * (1/2) ... ten times, or (1/2) ^ 10 which is a 0.0009765625 chance of doubling your money 10 times in a row.
That's a pretty small chance. You might end up spending more than what you would win.
Nope was thinking of something else ...
rate of return on investment (roi) 2% / month
kinder garden method:
1.02
1.0404
1.061208
1.08243216
1.104080803
1.126162419
1.148685667
1.17165938
1.195092568
1.218994419 final balance
1/1.218994419 = 0.820348301
1- 0.820348301 = 0.179651699
17.965169946 % chance to lose money on investment
Using simple compound interes formula 1×(1+0.02)^10 = 1.21899442
....
17.965170013% chance to lose money on investment
it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
I think your question is different from what you're asking then. Are you asking: what is the probability I can win $1024 on the 10th round?
This is different from the coin question. If you're asking the chance of winning a coin flip back to back 10 times, then you calculate as such:
On each round you have a 1/2 chance of winning or doubling your initial bet. This is because it's a coin flip.
However, now you need to keep in mind that you need to win all of these in a row. As in, you need a coin flip to be heads on the first, second, third, etc. flips. This means out of 10 flips, all 10 must be heads to double your money.
In the case that we have 10 flips and each flip has a 1/2 chance of giving you a payout, we have (1/2) * (1/2) * (1/2) ... ten times, or (1/2) ^ 10 which is a 0.0009765625 chance of doubling your money 10 times in a row.
That's a pretty small chance. You might end up spending more than what you would win.
If i flip a coin 10 times, what is the probability of getting heads on the 10th flip ? or on the second flip ? what is the math formula ?
Each coin flip is independent. Flipping one coin 10 times is the same as flipping 10 different coins at the same time and looking at all of the results. We know that a single coin will land on heads 50% of the time and tails on the other 50%. That's for 1 coin flip. Now, as we said before, each coin flip is independent of each other. That means, the second, third, fourth, etc. coin has the same chance as the first to be either heads or tails.
The 1st coin flip is 50%, the second is 50% and the tenth is also 50%. It's all the same as one single coin flip.
it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
I think your question is different from what you're asking then. Are you asking: what is the probability I can win $1024 on the 10th round?
This is different from the coin question. If you're asking the chance of winning a coin flip back to back 10 times, then you calculate as such:
On each round you have a 1/2 chance of winning or doubling your initial bet. This is because it's a coin flip.
However, now you need to keep in mind that you need to win all of these in a row. As in, you need a coin flip to be heads on the first, second, third, etc. flips. This means out of 10 flips, all 10 must be heads to double your money.
In the case that we have 10 flips and each flip has a 1/2 chance of giving you a payout, we have (1/2) * (1/2) * (1/2) ... ten times, or (1/2) ^ 10 which is a 0.0009765625 chance of doubling your money 10 times in a row.
nice, finally a good answer ... sry if question was wrong ...i am not a english native speaker
so it's 1/1024 chance ... thank you for answer
No problem!
If i flip a coin 10 times, what is the probability of getting heads on the 10th flip ? or on the second flip ? what is the math formula ?
Each coin flip is independent. Flipping one coin 10 times is the same as flipping 10 different coins at the same time and looking at all of the results. We know that a single coin will land on heads 50% of the time and tails on the other 50%. That's for 1 coin flip. Now, as we said before, each coin flip is independent of each other. That means, the second, third, fourth, etc. coin has the same chance as the first to be either heads or tails.
The 1st coin flip is 50%, the second is 50% and the tenth is also 50%. It's all the same as one single coin flip.
it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
I think your question is different from what you're asking then. Are you asking: what is the probability I can win $1024 on the 10th round?
This is different from the coin question. If you're asking the chance of winning a coin flip back to back 10 times, then you calculate as such:
On each round you have a 1/2 chance of winning or doubling your initial bet. This is because it's a coin flip.
However, now you need to keep in mind that you need to win all of these in a row. As in, you need a coin flip to be heads on the first, second, third, etc. flips. This means out of 10 flips, all 10 must be heads to double your money.
In the case that we have 10 flips and each flip has a 1/2 chance of giving you a payout, we have (1/2) * (1/2) * (1/2) ... ten times, or (1/2) ^ 10 which is a 0.0009765625 chance of doubling your money 10 times in a row.
nice, finally a good answer ... sry if question was wrong ...i am not a english native speaker
so it's 1/1024 chance ... thank you for answer
If i flip a coin 10 times, what is the probability of getting heads on the 10th flip ? or on the second flip ? what is the math formula ?
Each coin flip is independent. Flipping one coin 10 times is the same as flipping 10 different coins at the same time and looking at all of the results. We know that a single coin will land on heads 50% of the time and tails on the other 50%. That's for 1 coin flip. Now, as we said before, each coin flip is independent of each other. That means, the second, third, fourth, etc. coin has the same chance as the first to be either heads or tails.
The 1st coin flip is 50%, the second is 50% and the tenth is also 50%. It's all the same as one single coin flip.
it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
I think your question is different from what you're asking then. Are you asking: what is the probability I can win $1024 on the 10th round?
This is different from the coin question. If you're asking the chance of winning a coin flip back to back 10 times, then you calculate as such:
On each round you have a 1/2 chance of winning or doubling your initial bet. This is because it's a coin flip.
However, now you need to keep in mind that you need to win all of these in a row. As in, you need a coin flip to be heads on the first, second, third, etc. flips. This means out of 10 flips, all 10 must be heads to double your money.
In the case that we have 10 flips and each flip has a 1/2 chance of giving you a payout, we have (1/2) * (1/2) * (1/2) ... ten times, or (1/2) ^ 10 which is a 0.0009765625 chance of doubling your money 10 times in a row.
If i flip a coin 10 times, what is the probability of getting heads on the 10th flip ? or on the second flip ? what is the math formula ?
Each coin flip is independent. Flipping one coin 10 times is the same as flipping 10 different coins at the same time and looking at all of the results. We know that a single coin will land on heads 50% of the time and tails on the other 50%. That's for 1 coin flip. Now, as we said before, each coin flip is independent of each other. That means, the second, third, fourth, etc. coin has the same chance as the first to be either heads or tails.
The 1st coin flip is 50%, the second is 50% and the tenth is also 50%. It's all the same as one single coin flip.
it shore looks like that but something is not logical
bet 1 | $2
bet 2 | $4
bet 3 | $8
bet 4 | $16
bet 5 | $32
bet 6 | $64
bet 7 | $128
bet 8 | $256
bet 9 | $512
bet 10 | $1024
On a 10 bets game like this you can win $1024 or you could lose your $1
So it does not make sense that each event is independent ... and last bet probability of getting heads is 50%
My instinct / gut is telling me that 50% chance for last bet is false ...
What i suspect is that the probability of winning last bet decreases the more you bet ...so the probability of getting heads decreases also ...
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