All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
Coin tossing (Posted on 2004-06-11) Difficulty: 3 of 5
I threw a coin n times, and never got three tails in a row. I calculated the odds of this event, and found out they were just about even; 50%-50%. How many times did I throw the coin?

A second question: what were the chances of having not gotten three heads in a row either?

See The Solution Submitted by Federico Kereki    
Rating: 3.6667 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution simulations Comment 25 of 25 |

The following program simulates 100,000 trials each at n = 3 through 12, and reports the results when, as in the stated condition, the tossing has continued until that many tosses have been made, under the assumption that as soon as three heads or three tails in a row has been reached, tossing ends.

DEFDBL A-Z
OPEN "ctsim1.txt" FOR OUTPUT AS #2
RANDOMIZE TIMER
FOR n = 3 TO 12
  PRINT #2, n
  bothCt = 0: neitherCt = 0: only3TCt = 0: only3HCt = 0
  FOR trial = 1 TO 100000
   had3T = 0: had3H = 0
   headCt = 0: tailCt = 0
   FOR toss = 1 TO n
     heads = INT(2 * RND(1))
     IF heads THEN
       headCt = headCt + 1
       tailCt = 0
     ELSE
       headCt = 0
       tailCt = tailCt + 1
     END IF
     t = toss ' hold for later, as NEXT leaves incremented by 1
     IF tailCt = 3 THEN had3T = 1: EXIT FOR
     IF headCt = 3 THEN had3H = 1: EXIT FOR
   NEXT
   IF t = n THEN
     IF had3T THEN
      IF had3H THEN
       bothCt = bothCt + 1
      ELSE
       only3TCt = only3TCt + 1
      END IF
     ELSE
      IF had3H THEN
       only3HCt = only3HCt + 1
      ELSE
       neitherCt = neitherCt + 1
      END IF
     END IF
   END IF
  NEXT trial
  PRINT #2, "        hadTTT   noTTT"
  PRINT #2, "hadHHH"; : PRINT #2, USING " ####### #######"; bothCt; only3HCt
  PRINT #2, " noHHH"; : PRINT #2, USING " ####### #######"; only3TCt; neitherCt
  PRINT #2, "      "; : PRINT #2, USING " ####### #######"; only3TCt + bothCt; neitherCt + only3HCt
  PRINT #2,
NEXT n
CLOSE

The results are

 3 
        hadTTT   noTTT
hadHHH       0   12519
 noHHH   12487   74994
         12487   87513
 4 
        hadTTT   noTTT
hadHHH       0    6256
 noHHH    6264   62246
          6264   68502
 5 
        hadTTT   noTTT
hadHHH       0    6174
 noHHH    6277   50138
          6277   56312
 6 
        hadTTT   noTTT
hadHHH       0    4635
 noHHH    4707   40765
          4707   45400
 7 
        hadTTT   noTTT
hadHHH       0    3909
 noHHH    4021   32619
          4021   36528
 8 
        hadTTT   noTTT
hadHHH       0    3203
 noHHH    3137   26671
          3137   29874
 9 
        hadTTT   noTTT
hadHHH       0    2465
 noHHH    2609   21382
          2609   23847
 10 
        hadTTT   noTTT
hadHHH       0    2003
 noHHH    2048   17395
          2048   19398
 11 
        hadTTT   noTTT
hadHHH       0    1670
 noHHH    1665   13945
          1665   15615
 12 
        hadTTT   noTTT
hadHHH       0    1347
 noHHH    1405   11445
          1405   12792

As expected, when n=3, all trials proceeded to completion of the three tosses, and in 1/8 of the cases TTT was achieved and in another 1/8 of the cases HHH was achieved.

With 100,000 trials, a 50-50 probability of getting the observed results (going all the way to n tosses, and getting no TTT's) would give 50,000 such results. That is most closely matched when t=5, with 56,312 of that set of events happening and when t=6, with 45,400 of the trials reaching 6 tosses with no TTT. In the latter case, of the 45,400 reaching 6 tosses with no TTT, 40,765 also had no HHH, for a conditional prob of 89.79%.

If on the other hand we always allow the tossing to continue until n tosses have been reached, this opens the possibility that you will get both TTT and HHH.  The simulation program is the same as the above but without the EXIT FOR's.  The results are:

 3 
        hadTTT   noTTT
hadHHH       0   12263
 noHHH   12437   75300
         12437   87563
 4 
        hadTTT   noTTT
hadHHH       0   18811
 noHHH   18721   62468
         18721   81279
 5 
        hadTTT   noTTT
hadHHH       0   24835
 noHHH   25148   50017
         25148   74852
 6 
        hadTTT   noTTT
hadHHH    3104   28210
 noHHH   27900   40786
         31004   68996
 7 
        hadTTT   noTTT
hadHHH    6272   30440
 noHHH   30680   32608
         36952   63048
 8 
        hadTTT   noTTT
hadHHH   10163   31680
 noHHH   31602   26555
         41765   58235
 9 
        hadTTT   noTTT
hadHHH   14363   31988
 noHHH   32151   21498
         46514   53486
 10 
        hadTTT   noTTT
hadHHH   18763   31808
 noHHH   32097   17332
         50860   49140
 11 
        hadTTT   noTTT
hadHHH   23481   31569
 noHHH   30996   13954
         54477   45523
 12 
        hadTTT   noTTT
hadHHH   28055   30167
 noHHH   30451   11327
         58506   41494

In this case, it is n=10 that has the closest to a 50-50 split between hadTTT and noTTT.  Since it always goes to completion of the n tosses, all 100,000 are shown.  Of the 49,140 that had no TTT in 10 tosses, 17,332 also had no HHH, for a conditional prob. of 35.37%.


  Posted by Charlie on 2004-06-17 11:12:25
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (8)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information