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/sci/ - Science & Math

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12779102 No.12779102 [Reply] [Original]

You have 10 coins. A coin has a side that is "heads", and a side that is "tails". You can flip the coins one at a time. You can flip a selected coin until it lands on tails, starting with the first coin and ending with the tenth coin.

After flipping the tenth coin, you can record the results using the below notation.
F=number of flips
T=number of tails
H=number of heads

What is the probability of the following scenarios?
F=10
T=10
H=0

F=20
T=10
H=10

F=30
T=10
H=20

>> No.12779150

>>12779102
1) 1/2^10
2) The last one has to land on tails, so the answer is (1/2^20) * (19 choose 9)
3) (1/2^30) * (29 choose 9)

>> No.12779215

>>12779150
Incorrect. Try again.

>> No.12779592

1) 1/2^10

2) (1/2^20)*(1 + 10*9 + 10c8*(8c2 + 8) + 10c7*(7c3 + 7c2*2 + 7) + 10c6*(6c4 + 6c3*3 + 6c2*3 + 6) + 10c5*(1+ 5*4 + 5c3*6 + 5c2*4 + 5) + 10c4*(8 + 4*8 + 4c2*5 + 4) + 10c3*(12 + 3*6 + 3) + 10c2*(7+2) + 10)

3) ok this is getting a bit tedious

>> No.12779628

>>12779592
You get an F for not completing the assignment.

>> No.12779699

Can someone tell me if this is antifragile?

I'm a brainlet and I don't really know what all of your answers mean. I just want to know where the peak of the bell curve is relative to the three scenarios.
In other words, using my coin flipping algorithm, are you more likely to end up with a majority tails scenario or a majority heads scenario?
Or is the most likely scenario the second scenario(half tails and half heads)?

>> No.12779708

>>12779628
Wait till you see you ratemyprofessor grade.

>> No.12780102

>>12779102
someone please answer this >>12779699

>> No.12780108

>>12779699 >>12780102
What do you mean with "antifragile"?

This math is called combinatorics, how to calculate the numbers of possible scenarios in this.

You always get 10tails. Most of of the time you get more than 10heads so yea the third scenario with more heads than tails is the most "typical". The exact probabilities are quite tedious to calculate as >>12779592 shows. On the other hand this problem is quite easy to simulate on computers to calculate experimental probability curves.

>> No.12780204

>>12779699
Probability = (H+9)! / (9! H! 2^(H+10))

The peak of the curve is in between 8 and 9 heads. The expected value is 10 heads, as it should be with a fair coin.

>> No.12780212

>>12780108
>Most of of the time you get more than 10heads
Nope, with a fair coin that's impossible. The expected value must be 10 heads so will you get more than 10 less than half of the time.

>> No.12780217

>>12780108
>The exact probabilities are quite tedious to calculate
LOL no. It's just (H+9 choose 9)/2^(H+10)

>> No.12780226

>>12779102
This problem is equivalent to flipping one coin until we got 10 tails. Is it not?

>>12780108
It's me again. Now that I tried myself even though I'm rusty with combinatorics.

1) 10 events all tails P = 0.5^10 = 0.098% chance (as others said)

2) to get twenty rolls we must have 20th event tails and 19 before is calculated with binomial distribution. so
P = 0.5 * 19choose9 * 0.5^9 *0.5*10 = 18% chance. (which simplifies to what first anon says)

3) 29 events 19 times we get heads and last event we get tails
P = 29choose20 *0.5^19 * 0.5^10 * 0.5 = 2.8% chance. (which simplifies to what first anon says)

So the very first answer >>12779150 was right x).

>>12780217
Yeah. Sorry I answered before thinking aaaaand sorry to everyone I answered a thread with "sci can't answer this" topic.

>> No.12780233

>>12780226
>This problem is equivalent to flipping one coin until we got 10 tails. Is it not?
yes

>> No.12780256

10 * 1/2 ^ 10 for the first

>> No.12780258

>>12780233
Which is the same as asking how many ways there are to get 9 tails in H+9 flips (the H+10th flip must be the 10th tail). One you realize that the problem is trivial. Trying to count the number of ways to choose heads is the intuitive but wrong technique.