Which is more likely

[gazuty]

You have an evenly weighted coin (or a football team with three matches coming up which are all 50/50).

In the long run which pattern will arise more often:

A: HTT / WLL

B: HTH / WLW

C: they are the same Gazuty you told us the coin was evenly weighted/ team chance was 50/50.

D: Gazuty how does this improve me on betfair?

(Of course I need your reasoning if you are to get the glory and fame).

[poklius]

They both have equal 12.5% chance to occur, so yeah.. D?

[PeterLe]

I wont spoil it, but I think you've been reading a daniel kahneman book about a certain type of fallacy perhaps?

[gazuty]

They both have equal 12.5% chance to occur, so yeah.. D?

No they don't have an equal 12.5% chance.

I wont spoil it, but I think you've been reading a daniel kahneman book about a certain type of fallacy perhaps?

I have read thinking fast and slow but this came out of a ted talk I watched.

.......

And the answer is you are more likely to see HTT than HTH.

I'll explain later so people can ponder why that might by. The reading is not intuitive if you spend 2 seconds with ones brain saying each toss is 50/50. Remember you are looking for occurrences of this pattern in a sequence. Spend a little longer and the answer will be apparent.

Pattern recognition is (for me) a key to success. Humans are very good at it.

[Tanden]

I think the answer depends on what rules you use to count the occurrences.

The original wording, ie "In the long run which pattern will arise more often:"

places no restrictions on which occurrences can be counted and so there will be
roughly equal numbers of HTT & HTH (and of the other 6 possible combinations of 3 coin tosses).

If I was to change the rules slightly and say you can only count an occurrence if it appears at least 10 coin tosses after the previous occurrence then we would obviously have fewer occurrences of all combinations but would they all still be equal?

[gazuty]

The original wording, ie "In the long run which pattern will arise more often:"

places no restrictions on which occurrences can be counted and so there will be
roughly equal numbers of HTT & HTH (and of the other 6 possible combinations of 3 coin tosses).

I am asking if you keep flipping a coin, which sequence will you see more often HTT or HTH.

They do not have an equal chance of appearing. HTT will appear more often than HTH.

If you think about it you will see why. Don't necessarily focus on the probability of a successful pattern and that will bring you closer to the reason.

[marksmeets302]

I'm intrigued, but can you explain the question a bit more? A coin has only two sides. What is the mapping of the landings to the characters? For instance, what sequence of landings would HTT represent?

[gazuty]

I'm intrigued, but can you explain the question a bit more? A coin has only two sides. What is the mapping of the landings to the characters? For instance, what sequence of landings would HTT represent?

HTT is heads, tails, tails

but you might see flips come out like this.

HHHTHTTTHTHTHTHHHTTHHHTTHTHTHTH

I have just highlighted the first occurrence of the HTH sequence.

Of course the H at the end in that sequence also happens to go on to a HTT sequence which is what we are looking for. I typed this sequence so it is not produced by coin toss.

However, I've almost given away now why if a fair coin with no bias is tossed, you will find more occurrences of HTT than HTH.

[marksmeets302]

Sorry, was a bit confused by the W and L (even though PeterLe pointed that out already, more coffee is needed).

If you write down a tree representing H and T sequences like this

Code: Select all

     
          H
      H       T
   H    T   H    T
 H  T  H T H T  H T

Then you see these are the sequences containing HTT:
HHTT, HTTH, HTTT, THTT
and these for HTH:
HHTH, HTHH, HTHT, THTH, THTT

So unless I need even more coffee, I'd say HTH is more likely to occur in a random sequence of H and Ts ?

[gazuty]

You need more coffee. Have a look again.

I'm going to sleep but I'll post up the solution in the morning unless it is solved overnight.

I'm having trouble falling asleep. Here's the solution

We have two contestants. Marksmeets going for HTH and Gazuty going for HTT.

Kate Upton is tossing the coin.

Every time we reach HT the crowd is on the edge of their seats. It's 50/50.

But let's think from the losers perspective.

If Gazuty loses (because Kate flipped a H), Gazuty is a minimum 2 flips away from a victory. When he loses Gazuty has the starting H of the next possible winning sequence.

If Markmeets loses (because Kate flipped a T), Markmeets is a minimum 3 flips away from victory.

And folks that's the ball game because in the long run that produces more victories for Gazuty.

[Tanden]

Hi marksmeets302

There is a slight mistake in your sequences containing HTH

You have included THTT which, of course, doesn't contain HTH

So your example comes up with the same answer for both

[Tanden]

Hi again gazuty

Interesting thread and I actually agree with you but am wondering if the original wording sounds too much like the following scenario...

Toss a coin and write down the resulting sequence and then circle every occurrence of HTT and HTH.

If you did that, I say you would find the same number of occurrences.

However, if you didn't have a whole sequence to examine after it occurred but, instead, you had one person counting the HTT's as they happened and another counting the HTH's as they happened there would be fewer HTH's.

I still think even with that description there is room for ambiguity but I am trying not to give too much of the game away myself.

[gazuty]

Ive edited an earlier post with the solution. Disputations welcome (but I am very confident).

[xitian]

If Gazuty loses (because Kate flipped a H), Gazuty is a minimum 2 flips away from a victory. When he loses Gazuty has the starting H of the next possible winning sequence.

If Markmeets loses (because Kate flipped a T), Markmeets is a minimum 3 flips away from victory.

I was originally thinking along these lines too. But in the first case, you are BOTH only 2 flips away from victory. And in the second case, you are BOTH still 3 flips away from victory.

I don't see how that makes either one more likely, unless I've missed something?

[xitian]

I was kinda curious so I ran a quick monte carlo simulation in Excel:
1. Generate a column of 1 / 0 using RAND()
2. Next column checks for last three being 101
3. Next column checks for last three being 100
4. Filled down 30,000 rows (as that's all I could be bothered to do!)
5. Refreshed sheet about 30 times to generate a total of around 1,000,000 flips.

Totals:
101 - 127760
100 - 127267
A discrepancy of around 0.3%.

What was the TED talk that you mentioned earlier? Perhaps either we've misunderstood it, or we're misinterpreting the point.