Hi Kalyan.
We can easily write out all the possibilities and just identify how many there are with an "odd man" and divide that by the total number of possibilities:
HHHH 
HTHH 
THHH 
TTHH 
HHHT 
HTHT 
THHT 
TTHT 
HHTH 
HTTH 
THTH 
TTTH 
HHTT 
HTTT 
THTT 
TTTT 
So you can just read off the probability in this case, but what about a more general approach where the numbers are too big to use this simplistic approach?
Since each toss (and coin) is "fair", there is an equal likelihood of it turning out as heads or tails. This means there are two equal possibilities. Let s = 2.
Each toss is entirely independent of the others. That means the outcome on one toss has no bearing on the outcome of another toss. There are four independent tosses, so let n = 4.
Now the number of different fourtoss outcomes is the number of choices for each toss multiplied together. This is s^{n }.
That gives us the total (because 2^{4 } = 16, just as in our table). Now we can take a look at the events we really want to count: how many have an "odd man" out. We can interpret this as meaning that everyone except one person gets the same outcome.
Let's say that n1 people get heads. How many choices does the last toss have, if it cannot be the same?
It could be anyone who gets it, so that means there are n possible people who didn't throw heads. How many nonheads outcomes can one toss have? s1.
That means the number of fourtoss outcomes with everyone getting heads except for one person is n(s1). But this is only one of the s cases, so really there are s times this many oddman out situations. That makes the total number of equallylikely outcomes ns(s1). The probability, P, is this divided by the total s^{n}:
Let's check this with the original fourcoins question:
So the formula we came up with works for this situation. Let's test it on one other situation. Let's say 3 people roll fair sixsided dice (one each). What is the likelihood of an "odd man" situation?
In this case, n = 3 people and s = 6 sides, so we predict:
Is this accurate? The numbers are still small enough for us to write out the possibilities:
111  112  113  114  115  116  121  122  123  124  125  126  131  132  133  134  135  136  141  142  143  144  145  146  151  152  153  154  155  156  161  162  163  164  165  166 
211  212  213  214  215  216  221  222  223  224  225  226  231  232  233  234  235  236  241  242  243  244  245  246  251  252  253  254  255  256  261  262  263  264  265  266 
 311  312  313  314  315  316  321  322  323  324  325  326  331  332  333  334  335  336  341  342  343  344  345  346  351  352  353  354  355  356  361  362  363  364  365  366 
411  412  413  414  415  416  421  422  423  424  425  426  431  432  433  434  435  436  441  442  443  444  445  446  451  452  453  454  455  456  461  462  463  464  465  466 
 511  512  513  514  515  516  521  522  523  524  525  526  531  532  533  534  535  536  541  542  543  544  545  546  551  552  553  554  555  556  561  562  563  564  565  566 
611  612  613  614  615  616  621  622  623  624  625  626  631  632  633  634  635  636  641  642  643  644  645  646  651  652  653  654  655  656  661  662  663  664  665  666 

If you add up the yellows, you'll find 15/36 of each table is yellow. This agrees with the formula we made.
Hope this helps!
Stephen La Rocque.
