The old advanced math topic got closed (in the middle of me typing a post, actually) so I'm starting a new one. Please let's stay on topic this time

So, to start things off, (and retype what I was writing) I have a problem I need some help with (from some old competition math thing):
Suppose you flip a fair coin 6 times. What is the probability you won't get 2 tails in a row?

As you can see, probability is not my strong point
So I'm unsure of how to approach this. I'm afraid that if I just go through a bunch of scenarios and add up the combinations I'll double count something or not count it at all.
So, can anyone explain how to solve this and keep track of things accurately?
Thanks

At first I estimated that the chance for what you were describing was about 42% (I did some math) but I soon gave up because I don't really know anything about probability either.

I just hope this thread doesn't get derailed again… I don't know if you should have re-made it (that normally is frowned upon).

AonymousGuy wrote:

At first I estimated that the chance for what you were describing was about 42% (I did some math) but I soon gave up because I don't really know anything about probability either.

I just hope this thread doesn't get derailed again… I don't know if you should have re-made it (that normally is frowned upon).

I remade it because I liked the old one and I wanted to talk about some actual math. Plus I had that problem that I was going to post on the old topic but it got closed while I was typing.
The really annoying thing about the last topic was no matter how much I said “guys: back on topic please” but people just kept posting off topic. Hopefully people won't derail this topic.

So, anyone have a solution to my problem? Anyone?

I don't know anything about probability

For flipping a coin, there are 2^6=64 possible roll sequences, which sounds hard. However, making sure we don't get two tails in a row just means making only one tails at a time. There are far fewer of these, and I'd probably just list them and brute force it, though there's probably a better way.
THHHHH
HTHHHH
HHTHHH
HHHTHH
HHHHTH
HHHHHT
THTHHH
HTHTHH
HHTHTH
HHHTHT
THHTHH
HTHHTH
HHTHHT
THHHTH
HTHHHT
THHHHT
THTHTH
HTHTHT
HHHHHH

So the probability is 19/64.

I said 36th before, but it was a mistake. I also forgot the HHHHHH case.

scratcher7_13 wrote:

For flipping a coin, there are 2^6=64 possible roll sequences, which sounds hard. However, making sure we don't get two tails in a row just means making only one tails at a time. There are far fewer of these, and I'd probably just list them and brute force it, though there's probably a better way.
THHHHH
HTHHHH
HHTHHH
HHHTHH
HHHHTH
HHHHHT
THTHHH
HTHTHH
HHTHTH
HHHTHT
THHTHH
HTHHTH
HHTHHT
THHHTH
HTHHHT
THHHHT
THTHTH
HTHTHT

So the probability is 18/36.

AKA 1/2

Because you know that a tails will be followed immediately by a heads or the end in the possible sequences in which you don't get consecutive tails, you can just brute-force half of the domain and then shift it. So imagine that you have a strange coin in which one side gives you 2 heads and the other gives you 1 heads and 1 tails, and flip it 3 times. There are only 8 different possibilities for this;


Now you can shift it once to get the remaining 7 possibilities.


The probability is thus 8+7=15 sequences in 64 actual possibilities, or a probability of 15/64.

nwhi wrote:

(snip)

Whoops, can't edit, forgot to say. Note that taking the result of the strange coin's flip and choosing a random one of the two results in a 1/4 probability of getting tails. I dunno why 6 is the magic number that brings 16/64 (1/4) and 15/64 so close together but hey. The probability is (2^(n/2+1)-1)/2^n.

Lol you guys have different answers. The idea of grouping combos with a “strange coin” is good though. So we know there are 64 possibilities total. If we had a 3 sided die that gave HH, HT, and TH, and we could roll it 3 times. I'll call the sides 1, 2, and 3 respectively. 2 and 3 cannot go in that order so we have
That makes 16 combos, so 16/64 is 1/4. So is it 1/4? Did I miss any combos or add one that doesn't work?
Thanks for the help guys, btw

Great, you made it so I wanna make a probability calculator.

See my previous post again, I forgot one possibility, and I also said 36ths instead of 64ths. Anyway, it's fixed now, and I get 19/64.

MegaApuTurkUltra wrote:

Lol you guys have different answers. The idea of grouping combos with a “strange coin” is good though. So we know there are 64 possibilities total. If we had a 3 sided die that gave HH, HT, and TH, and we could roll it 3 times. I'll call the sides 1, 2, and 3 respectively. 2 and 3 cannot go in that order so we have

111
222
333
112
121
113
131
132
No 123!
221
212
213
Skip 231, 223, and 232
312
321
313
331
332
Skip 323

That makes 16 combos, so 16/64 is 1/4. So is it 1/4? Did I miss any combos or add one that doesn't work?
Thanks for the help guys, btw

yeah xD

I accidentally calculated mine such that you can repeat them and still not get consecutive Ts - so i excluded ones that ended and started on a T.

You left out ‘211’, ‘311’, ‘133’, ‘322’, and ‘122’ according to a quick Python program I wrote – so the probability is actually 21 in 64. Wat.

nwhi wrote:

MegaApuTurkUltra wrote:

Lol you guys have different answers. The idea of grouping combos with a “strange coin” is good though. So we know there are 64 possibilities total. If we had a 3 sided die that gave HH, HT, and TH, and we could roll it 3 times. I'll call the sides 1, 2, and 3 respectively. 2 and 3 cannot go in that order so we have

111
222
333
112
121
113
131
132
No 123!
221
212
213
Skip 231, 223, and 232
312
321
313
331
332
Skip 323

That makes 16 combos, so 16/64 is 1/4. So is it 1/4? Did I miss any combos or add one that doesn't work?
Thanks for the help guys, btw

yeah xD

I accidentally calculated mine such that you can repeat them and still not get consecutive Ts - so i excluded ones that ended and started on a T.

You left out ‘211’, ‘311’, ‘133’, ‘322’, and ‘122’ according to a quick Python program I wrote – so the probability is actually 21 in 64. Wat.

Ah see I'm always doing that. So 21/64 should be the answer!

yay!

On graphing calculators there's usually a probability calculator. On TIs it's under math > prb > nPr / nCr. Writing it all out is the least efficient way to do it.

echs wrote:

On graphing calculators there's usually a probability calculator. On TIs it's under math > prb > nPr / nCr. Writing it all out is the least efficient way to do it.

Perhaps. But it also is the most visual approach, which promotes better understanding of the solution.

However, if you want to use a calculator, that's a good approach too - so long as you understand and use the formuli properly. I've seen a lot of people that jump to the calculator too soon though, without really understanding what they are doing. That seldom results in a correct answer

Paddle2See wrote:

echs wrote:

On graphing calculators there's usually a probability calculator. On TIs it's under math > prb > nPr / nCr. Writing it all out is the least efficient way to do it.

Perhaps. But it also is the most visual approach, which promotes better understanding of the solution.

However, if you want to use a calculator, that's a good approach too - so long as you understand and use the formuli properly. I've seen a lot of people that jump to the calculator too soon though, without really understanding what they are doing. That seldom results in a correct answer

Yep, and if this is in competition math chances are (pun intended) there won't be calculators allowed. I needed to understand how to approach this problem correctly, not just what the answer is.

Anyways thanks for all your help guys!

Since the last thing discussed in the old topic was proofs, I'd like to start that up again.

Try proving that two angles that are congruent and supplementary are both 90 degrees. (I'd like to see how you do it - this theorem was the first I successfully solved individually.)

This was also what I would have posted on the old topic had it not been closed.

Deerleg wrote:

Since the last thing discussed in the old topic was proofs, I'd like to start that up again.

Try proving that two angles that are congruent and supplementary are both 90 degrees. (I'd like to see how you do it - this theorem was the first I successfully solved individually.)

This was also what I would have posted on the old topic had it not been closed.

Ah! I'm not in Geometry anymore! Don't make me look at my old postulates and theorems!

But it shouldn't be that hard - something like angle 1 + angle 2 = 180 degrees, and angle 1 = angle 2, so angle 1 + angle 1 = 180 degrees, so 2(angle 1) = 180 degrees, so angle 1 = 90 degrees, so angle 2 = 90 degrees, so angle 1 and angle 2 = 90 degrees.

Deerleg wrote:

Since the last thing discussed in the old topic was proofs, I'd like to start that up again.

Try proving that two angles that are congruent and supplementary are both 90 degrees. (I'd like to see how you do it - this theorem was the first I successfully solved individually.)

This was also what I would have posted on the old topic had it not been closed.

I don't really remember much about proofs hehe. How'd I do?

Let a = measure of angle 1, and b = measure of angle 2.
a = b (Given - congruent)
a + b = 180° (Given - supplementary)

a + a = 180° (substitution)
a = 90° (simplify)
b = 90° (due to a = b)

Conclusion: a = b = 90°

Edit - Ninja'd again eheh. I think aonymous and I used the same solution. We did systems of equations earlier this week so this method of solving is still fresh in my mind, I suppose