### Brain Teasers

# Two Games in a Row

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

A certain mathematician, his wife, and their teenage son all play a fair game of chess. One day when the son asked his father for 10 dollars for a Saturday night date, his father puffed his pipe for a moment and replied, "Let's do it this way. Today is Wednesday. You will play a game of chess tonight, tomorrow, and a third on Friday. If you win two games in a row, you get the money."

"Whom do I play first, you or mom?"

"You may have your choice," said the mathematician, his eyes twinkling.

The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?

"Whom do I play first, you or mom?"

"You may have your choice," said the mathematician, his eyes twinkling.

The son knew that his father played a stronger game than his mother. To maximize his chance of winning two games in succession, should he play father-mother-father or mother-father-mother?

### Hint

Who does he need to beat to win?### Answer

father-mother-fatherTo beat two games in a row, it is necessary to win the second game. This means that it would be to his advantage to play the second game against the weaker player. Though he plays his father twice, he has a higher chance of winning by playing his mother second.

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## Comments

I agree with the answer, but not your reasoning.

He has to beat each of his parents once. Thus he should take the F-M-F option - not because it makes Game #2 easier, but simply because it gives him two chances to defeat the stronger player, vs. just one chance.

He has to beat each of his parents once. Thus he should take the F-M-F option - not because it makes Game #2 easier, but simply because it gives him two chances to defeat the stronger player, vs. just one chance.

I like both of the conceptual explanations, but I confess that I needed to do the math, and I thought the answer was counter-intuitive. Here's how I solved it:

In either case, there are 8 possible outcomes- let a 1 be a win and a 0 be a loss: 000, 001, 010,011,100,101,110,111. The only outcomes that we care about are 011, 110, and 111. To figure the odds of these events under the two possible choices, let P be the probability of beating poppa, and M be the probability of beating momma. If dad goes first, the chances of winning are (1-P)*(M)(P) + P*M*(1-P) + P*M*P. Similarly, if mom goes first, probability is (1-M)*P*M + M*P*(1-M) + M * P * M. To figure out which chance is greater, subtract the second from the first and see whether the result if positive or negative.

Simplifying, we have

{2MP - MP^2 } - {2MP - M^2P}

Assuming P>0, divide by P and cancel the 2MP's:

-MP + M^2

Now assume M>0 as well, you get

M-P

Since we are given that M>P, then (M-P) is a positive number so the better chance is by playing Poppa first.

In either case, there are 8 possible outcomes- let a 1 be a win and a 0 be a loss: 000, 001, 010,011,100,101,110,111. The only outcomes that we care about are 011, 110, and 111. To figure the odds of these events under the two possible choices, let P be the probability of beating poppa, and M be the probability of beating momma. If dad goes first, the chances of winning are (1-P)*(M)(P) + P*M*(1-P) + P*M*P. Similarly, if mom goes first, probability is (1-M)*P*M + M*P*(1-M) + M * P * M. To figure out which chance is greater, subtract the second from the first and see whether the result if positive or negative.

Simplifying, we have

{2MP - MP^2 } - {2MP - M^2P}

Assuming P>0, divide by P and cancel the 2MP's:

-MP + M^2

Now assume M>0 as well, you get

M-P

Since we are given that M>P, then (M-P) is a positive number so the better chance is by playing Poppa first.

Bartman1 -- excellent explanation of a difficult concept. Well done!

And this is an excellent teaser, as well.

And this is an excellent teaser, as well.

Thanks BARTMAN. Great review, further enhances a good teaser.

But I will have wonder on the side if you are the cartoon character, or the guy that blew the playoffs for the Cubs (so they say) a few years back.

But I will have wonder on the side if you are the cartoon character, or the guy that blew the playoffs for the Cubs (so they say) a few years back.

Nice simple teaser, made me think. Excellent

That was easy with the hint

Great teaser, although WAY too easy.

This is what I thought:

He has to beat the middle person, so the middle should be the easiest. Also, if the Pappa is on the edges, he has two chances at beating him.

The best way would be - Mother-Mother-Mother (Although I know that is not a choice, that is what I would choose if I wanted the money.)

This is what I thought:

He has to beat the middle person, so the middle should be the easiest. Also, if the Pappa is on the edges, he has two chances at beating him.

The best way would be - Mother-Mother-Mother (Although I know that is not a choice, that is what I would choose if I wanted the money.)

Very nice teaser.

Let W = winning ; L=Losing. The sample space of possible outcomes are: { WWW, WWL, WLW, LWW, WLL, LWL, LLW, LLL}. Of these only WWW, WWL, and LWW results in the son winning the money.

Now let p = probability of beating dad; q = probability of beating mom.

Under the mom-dad-mom scenario, the probability of the son winning the money is:

q*p*q + q*p*(1-q) + (1-q)*p*q =

pq^2 + pq - pq^2 + pq -pq^2 =

2pq - pq^2 = pq(2 - q).

Under the dad-mom-dad scenario, the probability of winning the money is:

p*q*p + p*q*(1-p) + (1-p)*q*p =

qp^2+ pq -qp^2 +pq- qp^2 =

2pq - qp^2 = pq(2 - p).

Since p < q , 2 - p > 2 - q .

Hence, the probability of winning the money under the dad-mom-dad scenario is greater than under the mom-dad-mom alternative.

Specifically, if p= 0.2 and q= 0.5, then pq(2-p) = (0.2)(0.5)(1.= 0.18 or 18% chance of winning the money when playing dad-mom-dad.

Likewise, pq(2-q)= (0.2)(0.5)(1.5)= 0.15 = 15% of winning the money when playing mom-dad-mom.

Let W = winning ; L=Losing. The sample space of possible outcomes are: { WWW, WWL, WLW, LWW, WLL, LWL, LLW, LLL}. Of these only WWW, WWL, and LWW results in the son winning the money.

Now let p = probability of beating dad; q = probability of beating mom.

Under the mom-dad-mom scenario, the probability of the son winning the money is:

q*p*q + q*p*(1-q) + (1-q)*p*q =

pq^2 + pq - pq^2 + pq -pq^2 =

2pq - pq^2 = pq(2 - q).

Under the dad-mom-dad scenario, the probability of winning the money is:

p*q*p + p*q*(1-p) + (1-p)*q*p =

qp^2+ pq -qp^2 +pq- qp^2 =

2pq - qp^2 = pq(2 - p).

Since p < q , 2 - p > 2 - q .

Hence, the probability of winning the money under the dad-mom-dad scenario is greater than under the mom-dad-mom alternative.

Specifically, if p= 0.2 and q= 0.5, then pq(2-p) = (0.2)(0.5)(1.= 0.18 or 18% chance of winning the money when playing dad-mom-dad.

Likewise, pq(2-q)= (0.2)(0.5)(1.5)= 0.15 = 15% of winning the money when playing mom-dad-mom.

Note: = 8 in the previous post.

I agree with tpq76 that the solution's logic is not very good. It breaks down if say you add in a little brother that you can always beat. Would you want father-mother-father or brother-father-brother?

Using this solution's logic, father-mother-father is better because you must beat the middle person, and should hence choose the easiest opponent for game 2.

Using tpq76's logic (which I believe is correct), you must beat mother once and father once in father-mother-father, and beat brother once and father once in brother-father-brother. But this is the same as beating father once because you always beat brother. Clearly, beating father once is easier than beating mother once and beating father once.

Using this solution's logic, father-mother-father is better because you must beat the middle person, and should hence choose the easiest opponent for game 2.

Using tpq76's logic (which I believe is correct), you must beat mother once and father once in father-mother-father, and beat brother once and father once in brother-father-brother. But this is the same as beating father once because you always beat brother. Clearly, beating father once is easier than beating mother once and beating father once.

Sorry my previous comment is not quite correct. The answer will depend on the actual probabilities of beating mom and beating dad. Denote m the probability of beating mom and f the probability of beating dad. f < m.

Probability of getting $10 in father-mother-father is 2mf - fmf = mf (2-f).

Probability of getting $10 in brother-father-brother is f.

brother-father-brother is better if and only if m(2-f) < 1, which is indeed possible for say.. m = 0.3 f = 0.2.

But I believe this is still sufficient to show that you can't be certain fighting an easier opponent for game 2 will be the better option. Though, it is true for when there are only 2 opponents, but even then I can't seem to see why that is obvious without doing calculations.

Note I messed up in my previous post because brother-father-brother is not just beating father once, it's beating father once out of one game, whereas mother-father-mother is beating mother out of 1 game and father out of 2 games. So this does not necessary falsify tpq76's logic. It only shows it may not give you an answer with more than 2 opponents.

Probability of getting $10 in father-mother-father is 2mf - fmf = mf (2-f).

Probability of getting $10 in brother-father-brother is f.

brother-father-brother is better if and only if m(2-f) < 1, which is indeed possible for say.. m = 0.3 f = 0.2.

But I believe this is still sufficient to show that you can't be certain fighting an easier opponent for game 2 will be the better option. Though, it is true for when there are only 2 opponents, but even then I can't seem to see why that is obvious without doing calculations.

Note I messed up in my previous post because brother-father-brother is not just beating father once, it's beating father once out of one game, whereas mother-father-mother is beating mother out of 1 game and father out of 2 games. So this does not necessary falsify tpq76's logic. It only shows it may not give you an answer with more than 2 opponents.

As in a post above, let p = Prob(beat dad) and q = Prob(beat mom). Since the condition is winning twice in a row, then the kid's only two options are for dad-mom-dad:

- either wining the first two games (the 3rd one doesn't matter), this has probability pq

- lose the first and win the last two games, this has probability (1-p)pq

Hence the overall probability is pq + (1-p)pq = pq(2-p)

For mom-dad-mom, swap p and q to obtain pq(2-q).

Since p < q then dad-mom-dat is more likely to get him the money.

- either wining the first two games (the 3rd one doesn't matter), this has probability pq

- lose the first and win the last two games, this has probability (1-p)pq

Hence the overall probability is pq + (1-p)pq = pq(2-p)

For mom-dad-mom, swap p and q to obtain pq(2-q).

Since p < q then dad-mom-dat is more likely to get him the money.

The answer is wrong. It does not bmake any difference whether you play the mother or father first. you need to beat them both.

If you play MFM the you win with MF in the first 2 games or FM in games 2 and 3.

if you play FMF then you win with FM in the first 2 games or MF in games 2 and 3 so the chances are the same.

If you play MFM the you win with MF in the first 2 games or FM in games 2 and 3.

if you play FMF then you win with FM in the first 2 games or MF in games 2 and 3 so the chances are the same.

Multum, the answer is correct. To win the money, the son must either win the first two games (WWx), or lose the first and win the others (LWW).

If m and f are the probabilities of beating mother and father, respectively, then:

The probability of winning the first two games is mf, regardless of whom he plays first, and can be ignored.

If he plays dad first, the probability of losing the first game and winning the others is (1-f)mf. If he plays mum first it's (1-m)mf.

If m > f then (1-f)mf > (1-m)mf, so it's better to play dad first.

If m and f are the probabilities of beating mother and father, respectively, then:

The probability of winning the first two games is mf, regardless of whom he plays first, and can be ignored.

If he plays dad first, the probability of losing the first game and winning the others is (1-f)mf. If he plays mum first it's (1-m)mf.

If m > f then (1-f)mf > (1-m)mf, so it's better to play dad first.

Sorry, pretty much repeated what normadize wrote. Anyway, the key point, multum, is that you haven't taken into account the probability of losing the first game.

Spikethru4, I have looked at it again and must admit you are right. I overlooked that playing games 2 and 3 are conditional on losing game 1. So, the son has more chance of playing games 2 and 3 if he plays the father first. Therefore he should play father, mother, father.

Apologies for my earlier post, I should have looked at it closer.

Apologies for my earlier post, I should have looked at it closer.

the order of winning prob=a,b,a

Pr(win)=a*b + (1-a)*a*b=(2-a)*a*b

=> a is smaller : Pr(win) is bigger => hit daddy first

Pr(win)=a*b + (1-a)*a*b=(2-a)*a*b

=> a is smaller : Pr(win) is bigger => hit daddy first

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