Coin Toss

The total number of possible out comes when tossing one coin and then another two coins is $2 \times 2 \times 2 = 8$

The total number of outcomes in which Mark gets the same number of heads as Dan is $3$ .

1. Mark gets one head, Dan gets a head then a tail
2. Mark gets one head, Dan gets a tail then a head
3. Mark gets a tail, Dan gets a tail and another tail.

The probability of Mark getting the same number of heads as Dan is equal to the number of successful outcomes divided by the number of possible outcomes $= \frac{3}{8}$

We can see that Mark can have 2 possible outcomes of the penny he tossed and corresponding to each of the two outcome of Mark's penny, Dan's can have $2^2=4$ possible outcomes of his tossing of two pennies. Now, then the total no. of paired outcomes of Dan and Mark $=2\times 4 = \boxed{8}$

Now, the number of heads that fullfill the given criteria of getting same no. of heads by both on toss can be $1$ or $0$ as more than this is impossible as Mark tossed one penny and thus max. same no. of heads that can be achieved $=1$

Now, there are 2 outcomes $(H,HT),(H,TH)$ of getting same no. of heads $=1$

And, there is only 1 outcome $(T,TT)$ of getting same no. of heads $=0$

So, total no. of favourable outcomes $=3$ and total no. of possible outcomes $=8$

So, Probability $=\frac{\text{ No. of favourable outcomes }}{\text{ Total no. of outcomes }}=\boxed{\frac{3}{8}}$

as both can have 0 heads or one head as 2 heads is not possible in first case so probability of zero heads is 1/2 1/4=1/8 and probability of 1 head is 1/2 1/2=1/4 and total probability = 1/4+1/8=3/8

1st case: If Mark gets head (Probability = 1/2). Then obviously Dan have to get single head i.e. HT/ TH (Probability = 2/4). So probability in this case is 1/2 * 2/4 = 1/4. 2nd case: If Mark gets tail (Probability = 1/2). Then obviously Dan have to get both tails i.e. TT (Probability = 1/4). So probability in this case is 1/2 * 1/4 = 1/8. So total probability is 1/4 + 1/8 = 3/8.

There are 8 possible equi-probable outcomes 1. Mark gets H, Dan gets (H,H) 2. Mark gets T, Dan gets (H,H) 3. Mark gets H, Dan gets (H,T) 4. Mark gets T, Dan gets (H,T) 5. Mark gets H, Dan gets (T,H) 6. Mark gets T, Dan gets (T,H) 7. Mark gets H, Dan gets (T,T) 8. Mark gets T, Dan gets (T,T) Three of them, Mark gets same number of H as Dan gets. Hence, the answer is 3/8

Case 1: If Mark gets tail (probability=1/2), Dan should get both tails among 4 possibilities HH, HT,TH, TT (probability=1/4)

so probability that Dan gets same num of heads as Mark gets = (1/2) x (1/4)

Case 2: If Mark gets head (probability=1/2), Dan should get one head as HT or TH ( probability=2/4)

so probability that Dan gets same num of heads as Mark gets = (1/2) x (2/4)

              Net probability = (1/2 x 1/4) + (1/2 x 2/4) = 3/8

For Zero Heads 0.5 * 0.5 * 0.5 and For 1 head 0.5 0.5 0.5 2 .. multiplied by 2 as there are two possibilities so P = 0.5 0.5 0.5 3 = 3/8

Not a challenge at all for anyone who knows how to code. Just generate all the possibilities and ask for the ones in witch number of H's for player 0 is equal to number of H's for player 1.

To solve in the paper just make 2ˆn (n being the number of coins) and then figure out when they will both have the same number of heads.

The hard way: HHH HHN (1) HNH (2) HNN NHH NHN NNH NNN (3)