Asked by John

If you toss a fair coin and you toss until a head is followed by a tail. What is the probability that at least 5 tosses are needed for this to occur?
Answered by Scott
you are looking for a consecutive H-T pair

possible pairs are
... H-H, H-T, T-H, T-T

there are 4 consecutive pairs in 5 tosses

the probability of H-T is .25 for any pair of tosses, so the probability of NOT H-T is .75

the probability of 3 consecutive NOT H-T pairs is ... .75^3

there are 32 possible outcomes (2^5) for 5 consecutive tosses
... you can list them to check the answer
Answered by John
When I made a tree diagram for 4 tosses of a coin... I had 6 branches where you tossed until a head was followed by a tail. That was 6 branches out of 16. Then I added another toss and came up with one more for the sample set. HHHHT, so that would be 1 out of 32 produce the head then tail. So the P(at least 5 flips) is 1 - P(less than 5 flips)
1 - (6/16 + 1/32)
I think I am on the right track. Please advise.
Answered by John
Or... do we not break up where the tree ended (as in my 1- solution), but do 1 - 7/32 which would then be 25/32 for the probability of it taking atleast 5 times for a head followed by a tail to occur?
Answered by Reiny
This is a bit more complicated than it first appears
Let A be the event of ... HT

case 1. it happens in 2 tosses
HT**
Prob(A on 2 tosses) = 1(1/4)

case 2. (3 Tosses)
THT, HHT,
Prob(A on 3 tosses) = 2(1/8) = 1/4

case 3. (4 tosses)
TTHT, THHT, HHHT
prob(A on 4 tosses) = 3(1/16) = 3/16

Of course prob(A for an infinite number of tosses) = 1

so Prob( A as stated) = 1 - (1/4+1/4+3/16) = 5/16



Look over my cases and make sure I got them all

e.g. in case 3, we can't count outcomes like HTHT, since we would no longer toss after HT, so I did not list them.
Answered by John
The "atleast 5 tosses" part means 1-P(50rless tosses) so I think we also need the fifth toss that has 1 in 32 ways to get HT, that being TTTHT
So that has to go into the subtracting bracket as well... I think : ) Please advise.
Answered by Reiny
"at least 5 tosses" to means:
5 tosses, or 6 tosses, or 7 tosses, etc

so we <b>don't</b> want 2, 3, or 4 tosses.

agree?

btw, 5 tosses would be quite a few more than TTTHT.
What about HHHHT, THHHT, TTTHT, etc

Answered by Scott
HHHHH ... 1/32
HHHHT ... 1/32
HHHT ... 1/16
HHT ... 1/8
HT ... 1/4
TTTTT ... 1/32
TTTTH ... 1/32
TTTHH , TTTHT ... 1/32 , 1/32
TTHHH , TTHHT , TTHT ... 1/32 , 1/32, 1/16
THHHH , THHHT , THHT , THT
... 1/32 , 1/32 , 1/16 , 1/8

these are the 16 ways to toss a coin until a head is followed by a tail, or 5 tosses are reached; along with their probabilities

the probability that an H-T pair occurs with FEWER than 5 tosses is ... 11/16

this agrees with Reiny's result

sorry about my 1st effort

Answered by John
Awesome! Now I truly understand the scenerio. Thanks so much for your wise and timely replies and solutions : )
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