Which Comes First, HH or HT?
Steve Chamberlin
John Chandler <j...@a.cs.okstate.edu> wrote:
>If we flip an unbiased coin twice,
>HH and HT each have a probability of 1/4 of occurring.
>(H = Heads, T = Tails)
>
>But, suppose we agree to flip a coin until HH or HT
>occurs in any two consecutive flips.
>The probability of HT occurring first is greater than for HH.
>
>Clifford Konold,
>"Confessions of a Coin Flipper and Would-Be Instructor"
>The American Statistician, 49 (May 1995) 203-209.
For those of us who don't have easy access to the above paper, would
you care to explain why this is true?
A quick thought experiment: neither HH nor HT can occur until an H
comes up once, so we begin by flipping the coin repeatedly until we
obtain an H. The next flip will be either H or T with equal
probability, so HH and HT have an equal probability of occuring first
in a sequence of coin flips. A small C program I wrote to simulate
this agrees: both HH and HT should be equally likely.
Why do you claim that the probability of HT occuring first is greater
than that for HH?
--Steve Chamberlin
--
Steve Chamberlin | s...@mak.com
MaK Technologies | http://www.mak.com/~slc/home.html
Michael Jason Lewis
John Chandler <j...@a.cs.okstate.edu> wrote:
}In article <47llfk$m...@atlantis.mak.com>,
}Steve Chamberlin <s...@atlantis.mak.com> wrote:
}>In article <47lhrj$n...@news.cis.okstate.edu>,
}>John Chandler <j...@a.cs.okstate.edu> wrote:
}>>If we flip an unbiased coin twice,
}>>HH and HT each have a probability of 1/4 of occurring.
}>>(H = Heads, T = Tails)
}>>
}>>But, suppose we agree to flip a coin until HH or HT
}>>occurs in any two consecutive flips.
}>>The probability of HT occurring first is greater than for HH.
}>>
}>>Clifford Konold,
}>>"Confessions of a Coin Flipper and Would-Be Instructor"
}>>The American Statistician, 49 (May 1995) 203-209.
}>
}>comes up once, so we begin by flipping the coin repeatedly until we
}>obtain an H. The next flip will be either H or T with equal
}>probability, so HH and HT have an equal probability of occuring first
}>in a sequence of coin flips. A small C program I wrote to simulate
}>this agrees: both HH and HT should be equally likely.
}
}Running HH versus TH seems to give about a 3:1 advantage
}to TH.
}
}Konold was wrong? The referee(s) let an error like that
}get through?
So it would appear. HH vs. TH is somewhat obvious; if you ever flip a
coin and it's a tail, HH cannot happen without TH happening immediately
before it. Therefore the odds would be 3:1 in favor of TH, since the first
two flips can be HH, TH, TT (leading to TH) or HT (leading to TH).
- Mike
Tony Roberts
Chandler) wrote:
> If we flip an unbiased coin twice,
> HH and HT each have a probability of 1/4 of occurring.
> (H = Heads, T = Tails)
>
> But, suppose we agree to flip a coin until HH or HT
> occurs in any two consecutive flips.
> The probability of HT occurring first is greater than for HH.
>
I do not believe it.
If you compare HH versus TH, then there is a difference and TH is favoured.
Tony
--------------------------------------------------------------------------
Prof A J Roberts
Dept Mathematics & Computing E-mail: arob...@usq.edu.au
Uni of Southern Queensland Phone: (76) 312943
Toowoomba, Queensland 4350 Fax: (76) 312721
Australia WWW: http://www.sci.usq.edu.au
/pub/MC/staff/robertsa/home.html
--------------------------------------------------------------------------
John Chandler
Steve Chamberlin <s...@atlantis.mak.com> wrote:
>In article <47lhrj$n...@news.cis.okstate.edu>,
>>If we flip an unbiased coin twice,
>>HH and HT each have a probability of 1/4 of occurring.
>>(H = Heads, T = Tails)
>>
>>But, suppose we agree to flip a coin until HH or HT
>>occurs in any two consecutive flips.
>>The probability of HT occurring first is greater than for HH.
>>
>>"Confessions of a Coin Flipper and Would-Be Instructor"
>>The American Statistician, 49 (May 1995) 203-209.
>
>you care to explain why this is true?
>
>comes up once, so we begin by flipping the coin repeatedly until we
>obtain an H. The next flip will be either H or T with equal
>probability, so HH and HT have an equal probability of occuring first
>in a sequence of coin flips. A small C program I wrote to simulate
>this agrees: both HH and HT should be equally likely.
>
>than that for HH?
>
>--Steve Chamberlin
>
>
>--
>Steve Chamberlin | s...@mak.com
>MaK Technologies | http://www.mak.com/~slc/home.html
My simulation results agree with Chamberlin's.
Running HH versus TH seems to give about a 3:1 advantage
to TH.
Konold was wrong? The referee(s) let an error like that
get through?
--
John Chandler
j...@a.cs.okstate.edu
Steve Brecher
Chandler) wrote:
> Running HH versus TH seems to give about a 3:1 advantage
> to TH.
stopping condition a = HH
stopping condition b = TH
HHa
HT
HTHb HTT
HTTHb HTTT
HTTTHb HTTTT
HTTTTHb HTTTTT
...
THb
TT
TTHb TTT
TTTHb TTTT
TTTTHb TTTTT
TTTTTHb TTTTTT
...
p(a) = 1/4 = 0.25
p(b) = 1/4 + 2(1/8 + 1/16 + 1/32 + 1/64 + ...) = 0.75
--
st...@brecher.reno.nv.us (Steve Brecher)
Alan Cook
>
> In article <47lhrj$n...@news.cis.okstate.edu>,
> John Chandler <j...@a.cs.okstate.edu> wrote:
> >If we flip an unbiased coin twice,
> >HH and HT each have a probability of 1/4 of occurring.
> >(H = Heads, T = Tails)
> >
> >But, suppose we agree to flip a coin until HH or HT
> >occurs in any two consecutive flips.
> >The probability of HT occurring first is greater than for HH.
> >
> >Clifford Konold,
> >"Confessions of a Coin Flipper and Would-Be Instructor"
> >The American Statistician, 49 (May 1995) 203-209.
>
>A small C program I wrote to simulate
> this agrees: both HH and HT should be equally likely.
>
> Why do you claim that the probability of HT occuring first is greater
> than that for HH?
>
> --Steve Chamberlin
>
> --
> Steve Chamberlin | s...@mak.com
> MaK Technologies | http://www.mak.com/~slc/home.html
The article must have been misquoted, or is in error. In the case of
sequential coin flips Steve is correct, HH and HT have an equal chance of
occurring. In the case of flipping two coins, the original statement is
correct, as there are two ways to make HT (HT and TH), but only one way to
make HH.
-----------------------------------------------------------------------------
Alan V. Cook, N7CEU
DoD#: 0701
Home: avc...@earthlink.net Office: Alan_V...@ccmail.anatcp.rockwell.com
-----------------------------------------------------------------------------
Per Erik Manne
>
> In article <47p58u$e...@news.cis.okstate.edu>,
> }>In article <47lhrj$n...@news.cis.okstate.edu>,
> }>John Chandler <j...@a.cs.okstate.edu> wrote:
> }>>If we flip an unbiased coin twice,
> }>>HH and HT each have a probability of 1/4 of occurring.
> }>>(H = Heads, T = Tails)
> }>>
> }>>But, suppose we agree to flip a coin until HH or HT
> }>>occurs in any two consecutive flips.
> }>>The probability of HT occurring first is greater than for HH.
> }>>
> }>>Clifford Konold,
> }>>"Confessions of a Coin Flipper and Would-Be Instructor"
> }>>The American Statistician, 49 (May 1995) 203-209.
> }>
> }>comes up once, so we begin by flipping the coin repeatedly until we
> }>obtain an H. The next flip will be either H or T with equal
> }>probability, so HH and HT have an equal probability of occuring first
> }>this agrees: both HH and HT should be equally likely.
> }
> }to TH.
> }
> }get through?
>
> coin and it's a tail, HH cannot happen without TH happening immediately
> before it. Therefore the odds would be 3:1 in favor of TH, since the first
> two flips can be HH, TH, TT (leading to TH) or HT (leading to TH).
>
> - Mike
Interestingly, HH ties HT, HT ties TH, but HH loses to TH (in the
long run).
--
Bergen,
Per Manne
p...@hamilton.nhh.no
John Chandler
Tony Roberts <arob...@usq.edu.au> wrote:
>In article <47lhrj$n...@news.cis.okstate.edu>, j...@a.cs.okstate.edu (John
>Chandler) wrote:
>
>> If we flip an unbiased coin twice,
>> HH and HT each have a probability of 1/4 of occurring.
>> (H = Heads, T = Tails)
>>
>> But, suppose we agree to flip a coin until HH or HT
>> occurs in any two consecutive flips.
>> The probability of HT occurring first is greater than for HH.
>>
>
>
>If you compare HH versus TH, then there is a difference and TH is favoured.
>
> Tony
>--------------------------------------------------------------------------
> Prof A J Roberts
> Dept Mathematics & Computing E-mail: arob...@usq.edu.au
> Uni of Southern Queensland Phone: (76) 312943
> Toowoomba, Queensland 4350 Fax: (76) 312721
> Australia WWW: http://www.sci.usq.edu.au
> /pub/MC/staff/robertsa/home.html
>--------------------------------------------------------------------------
You are right.
Obviously you have not seen several of the responses,
in which it developed that:
I quoted correctly from Clifford Konold's article.
The article is wrong.
The probability of first occurrence of HH versus HT is 1/2.
The probability of first occurrence of HH versus TH is 1/4.
The mean number of flips until HH occurs is six.
The mean number of flips until HT (or TH) occurs is four,
not six as one respondent claimed. (Proved by R. Israel,
"verified" by simulation.)
--
John Chandler
j...@a.cs.okstate.edu
Philip Galanter
...
>HT and HH are equally likely to occur first.
>TH is 3:1 to occur before HH.
>No doubt a correction will be published in the journal.
>How did a referee ever let this slip by?
>
>Another question (answered in Konold's references,
>which I haven't had time to look up):
>
> What is the mean number of flips until HH is found?
> What about HT?
>
>The first few terms in the summation of Probability(i)*Length(i)
>are not all equal for corresponding terms in the two sums,
>according to my preliminary scratchwork.
>It looks like Distance(HH)>Distance(HT).
>This would not contradict the fact that Prob(HH)=Prob(HT) for
>first occurrences, but it would provide a nice paradox for students.
>(Distance(HT)=Distance(TH), so obviously Prob of first occurrence
>cannot easily be predicted from Distance.)
Wouldn't Distance(HH)=Distance(HT)?
Those who are visual thinkers (like myself) may find a diagram helpful.
(I am not much of a math formalist, but this must be some kind of
Markov chain or something...let me know if I've miscalculated the mean flips
below)
For HH vrs TH: For HH vrs HT
(start) (start)
/\ /\
/ \ _ / \ _
V V / | V V / |
(H)->(T) | (H)<-(T) |
| |^__| /\ ^__|
| | / \
V V V V
(HH) (TH) (HH) (HT)
For HH vrs. TH: HH is only a 2 flip path with a prob. of .25
TH is therefore a prob of .75
The mean number of flips from start to T is 1.333.
The mean number of flips from T to TH is 2 = 1 +.5 +.25 + ...
The total mean flips from start to TH is 3.333
For HH vrs HT: Both HH and HT paths depart from the same H node
Both have a prob of .5
The number of flips from H to HH (or HT) is 1
The mean number of flips from start to H is:
2 = 1 + .5 + .25 + ...
The total mean flips from start to HH (or HT) is 3
Phil
-----------------------------------------------------------------------
Philip Galanter New York University phone: 212-998-3041
Associate Director 251 Mercer fax: 212-995-4120
for Arts Technology New York, NY 10012 internet: gala...@nyu.edu
N Y U A c a d e m i c C o m p u t i n g F a c i l i t y
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John Chandler
[I replied to this post by e-mail, forwarding some other posts to Phil.]
Directed graphs are the right way to attack this, all right,
but the correct answers for the expected number of flips
are six for HH and four for HT, proved by R. Israel and B. Turlach.
The general method is given in two of the references
in Konold's paper, probably (I haven't seen them yet).
--
John Chandler
j...@a.cs.okstate.edu
Philip Galanter
After more e-mail with Chandler I think it turns out that different
interpretations of the question generate different answers...
CASE 1
Some take the number of flips question to refer to independent trials for the
HH and HT case, and thus use 2 independent graphs.
The HH = 6 and HT = 4 averages are correct if the questions are posed like this:
"Take a coin and flip it. Keep flipping it until you get two heads in
a row. Count how many flips it took. Do this many times. What will the
average number of flips be?
Now take a coin and flip it. Keep flipping it until you get a head
followed by a tail. Count how many flips it took. Do this many times.
What will the average number of flips be?"
CASE 2
I took the number of flips question to refer to the games competitive
context, and thus use 1 graph with both HH and HT outcomes.
My answer was that both HH and HT take 3 flips on the average, and are
equally likely. I think this is true if the question is stated like this:
"Take a coin and flip it. Keep flipping it until you get either two
heads in a row or a head followed by a tail. Count how many flips it took.
Do this many times. Where you ended up with two heads in a row what
was the average number of flips it took? Where you ended up with
a head followed by a tail what was the average number of flips it took?"
I think most readers would find the original context more like CASE 2
...i.e. given the rules of the game what are the relative probabilities of
HH and HT winning, and by the way, what are the average number of flips to
reach each result _in competition_?
Phil
p.s. For those who missed it here is how I would graph the HH HT question.
(start)
/\
/ \ _
V V / |
(H)<-(T) |
/\ ^__|
/ \
V V
(HH) (HT)
If you get an H, and only if you get an H, you will surely get
a "win" on the next flip. Therefore HH & HT are equally likely
_and_ exactly 1 flip away from the _first_ H.
The only remaining question then is how many flips on the average
does it take to get the first H? The answer to this is 2.
(1 + .5 + .25 + ... = 2).
2+1=3
I don't know if this counts as a mathematical proof, but
I'd be interested to see any flaws in the logic.
Douglas Kubler
>In article <aroberts-0...@139.86.144.66>,
>Tony Roberts <arob...@usq.edu.au> wrote:
>>In article <47lhrj$n...@news.cis.okstate.edu>, j...@a.cs.okstate.edu (John
>--
>John Chandler
>j...@a.cs.okstate.edu
(where is the Konold article?)
This topic was handled by M. Gardner in Scientific American more than
20 years ago.
The game was to let a player predict a sequence (length N) of heads
and tails, the opponent can always pick a sequence (length N) that
will occur with greater or equal probabilty. Sort of a
paper-scissors-rock game, the second player always wins in the long
run.
Douglas Kubler
Dave Cromley
>interpretations of the question generate different answers...
Right. I composed the following, so am posting it.
Everybody is right. The reason for this long thread
is that people are commenting on four different situations:
1) P(hh before ht) = .5 with a mean length of 3
P(ht before hh) = .5 with a mean length of 3
2) P(hh before th) = .25 with a mean length of 2
P(th before hh) = .75 with a mean length of 3.333
3) mean length to get a HH is 6
4) mean length to get a HT is 4
MSDOS QBASIC simulater follows:
'--simulates four tail flip situations
' David A. Cromley
CLS : RANDOMIZE TIMER
n100 = 1000
heads = 1: tails = 2
FOR i = 1 TO n100
IF RND < .5 THEN flip = heads: ELSE flip = tails
IF flip = heads THEN PRINT "H"; : ELSE PRINT "T";
'--(hh before ht) and (ht before hh) stats
len1 = len1 + 1
IF flip = heads AND old1 = heads THEN
hhhtcnt = hhhtcnt + 1: hhhtlen = hhhtlen + len1
old1 = 0: len1 = 0
ELSEIF flip = tails AND old1 = heads THEN
hthhcnt = hthhcnt + 1: hthhlen = hthhlen + len1
old1 = 0: len1 = 0
ELSE
old1 = flip
END IF
'--(hh before th) and (th before hh) stats
len2 = len2 + 1
IF flip = heads AND old2 = heads THEN
hhthcnt = hhthcnt + 1: hhthlen = hhthlen + len2
old2 = 0: len2 = 0
ELSEIF flip = heads AND old2 = tails THEN
thhhcnt = thhhcnt + 1: thhhlen = thhhlen + len2
old2 = 0: len2 = 0
ELSE
old2 = flip
END IF
'--(hh) stats
len3 = len3 + 1
IF flip = heads AND old3 = heads THEN
hhcnt = hhcnt + 1: hhlen = hhlen + len3
old3 = 0: len3 = 0
ELSE
old3 = flip
END IF
'--(ht) stats
len4 = len4 + 1
IF flip = heads AND old4 = tails THEN
htcnt = htcnt + 1: htlen = htlen + len4
old4 = 0: len4 = 0
ELSE
old4 = flip
END IF
NEXT i
ed1$ = "##### #.##### #.#####"
PRINT : PRINT "stats after"; n100; "tosses"
PRINT
PRINT "Stats for (hh before ht) and (ht before hh):"
t = hhhtcnt + hthhcnt
PRINT "hh count, percent, average length"
PRINT USING ed1$; hhhtcnt; hhhtcnt / t; hhhtlen / hhhtcnt
PRINT "ht count, percent, average length"
PRINT USING ed1$; hthhcnt; hthhcnt / t; hthhlen / hthhcnt
PRINT
PRINT "Stats for (hh before th) and (th before hh):"
t = hhthcnt + thhhcnt
PRINT "hh count, percent, average length"
PRINT USING ed1$; hhthcnt; hhthcnt / t; hhthlen / hhthcnt
PRINT "th count, percent, average length"
PRINT USING ed1$; thhhcnt; thhhcnt / t; thhhlen / thhhcnt
PRINT
PRINT "Stats for hh:"
PRINT "hh count, average length"
PRINT USING ed1$; hhcnt; hhlen / hhcnt
PRINT
PRINT "Stats for ht:"
PRINT "ht count, average length"
PRINT USING ed1$; htcnt; htlen / htcnt