Martin and Joey are playing a coin game in which each player tosses a

Question

Martin and Joey are playing a coin game in which each player tosses a fair coin alternately. The player who gets a ‘Heads’ first wins. The maximum number of tosses allowed in a single game for any player is 6. What is the probability that the person who tosses first will win the game?

dabaobao · Answer

Veritas Prep Official Solution

Probability of getting ‘Heads’ on a single toss = 1/2

Probability of getting ‘Tails’ on a single toss = 1/2

The person who starts the game can win the game if one of the following scenarios plays out:
The first person tosses the coin and gets a ‘Heads’ right away. The first person wins!
The first person tosses the coin and gets a ‘Tails’. The second person gets ‘Tails’ too. The first person tosses again and gets a ‘Heads’. The first person wins!
The first person tosses the coin and gets a ‘Tails’. The second person gets ‘Tails’ too. The first person tosses again and gets a ‘Tails’ again. The second person gets ‘Tails’ again too. Finally, the first person tosses and this time, gets a ‘Heads’. The first person wins!
and so on…

In the worst case, the first person will have to toss 6 times to get a ‘Heads’. He and the second person would end up getting ‘Tails’ on five previous tosses.

Probability that the first person tosses the coin and gets a ‘Heads’ right away = 1/2

Probability that the first person tosses the coin and gets a ‘Tails’ (1/2), the second person gets ‘Tails’ (1/2) and then the first person gets a ‘Heads’(1/2) = (1/2)*(1/2)*(1/2) = (1/2)^3

Probability that the first person tosses the coin and gets a ‘Tails’ (1/2), the second person gets ‘Tails’ (1/2) , the first person tosses again and gets a ‘Tails’ again (1/2), the second person gets ‘Tails’ again (1/2) and finally, the first person tosses and this time, gets a ‘Heads’ (1/2) = (1/2)^5

and so on…

Probability that the first person will have to toss 6 times to get a ‘Heads’ = (1/2)^11

To get the probability of the first person winning, we just need to add all these probabilities now.

Probability that the first person will win = (1/2) + (1/2)^3 + (1/2)^5 + (1/2)^7 + (1/2)^9 + (1/2)^11 => ANSWER

On the same lines, can you find the probability that the person who tosses second wins? I hope you understand that it is very similar to what we have already discussed. The person who tosses second will win if one of the following happens:
The person who tosses first gets ‘Tails’ and then the person who tosses second gets ‘Heads’.
The person who tosses first gets ‘Tails’, the person who tosses second gets ‘Tails’, the person who tosses first gets ‘Tails’ again and the second person then gets ‘Heads’.
and so on…

Probability that the second person will win = (1/2)^2 + (1/2)^4 + (1/2)^6 + (1/2)^8 + (1/2)^10 + (1/2)^12

Nalinikant123 · Answer

Martin wins in the following ways:

Martin tosses a head in the first toss, or
Martin tosses a tail in the first toss, Joey tosses a tail in the second toss, Martin tosses a head in the third toss, or
and on and on until Martin tosses a head
The probability of Martin tossing a head in the first toss = 1/2

The probability of Martin tossing a tail followed by Joey tossing a tail and then A tossing a head = (1/2)(1/2)(1/2)

So, the probability of Martin winning = 1/2 + (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2)(1/2)(1/2) + …

This is an infinite geometric series whose first term is 1/2 and common ratio is 1/4. Therefore, the sum is (1/2)/(1–1/4) = (1/2)/(3/4) = 2/3.

bumpbot · Answer

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Martin and Joey are playing a coin game in which each player tosses a

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