[Guest]

Although this seems to be a classic, I couldn't find it while searching?

Anyone who wants to be a member of our private tennis club, must play 3 games and get successive 2 games of these. He will meet two other member. One of them is a good player (G) while the other is the best player (B). He may choose the game order in two these: GBG or BGB. He will be admitted if he win at least two successive games. Which order would you choose? BGB or GBG?

Since there is only two alternatives, please don't post without explanation.

[Guest]

BGB. If I can beat "Best," I could probably beat good. Also, I would be "fresh" and have all of my energy for beating the better player. Most importantly, I would not go through the trouble of winning once and then getting the crap beat out of me. I could save time by losing on the first match!

[Guest]

i think this is the explanation:

i would go with bgb because the best player would be the hardest to beat obviously so i would have 2 tries to beat him. if i have enough skill to beat the best player even once i could beat the good player no problem so it wouldn't matter if i beat him my first or second try i would still have 2 consecutive wins assuming i beat the good player

[Guest]

Mickey and Jade12! You're right.

Though I missed to inform in OP that you're a very novice player. So don't be confident in easly beating the good player. But as a result, you're true. I would expect that playing twice with best player will come harder to you, but you, courageous players chose it, clappings.

Now I want you to prove this in math.

[Guest]

Although this seems to be a classic, I couldn't find it while searching?

Anyone who wants to be a member of our private tennis club, must play 3 games and get successive 2 games of these. He will meet two other member. One of them is a good player (G) while the other is the best player (B). He may choose the game order in two these: GBG or BGB. He will be admitted if he win at least two successive games. Which order would you choose? BGB or GBG?

Since there is only two alternatives, please don't post without explanation.

This is my first post, so excuse me if I'm not explaining it right.

We assume that there is a higher probability of beating the Good player rather than the Best one. For easier calculations, lets pretend that you have a 1/2 chance of beating the G player and a 1/4 of beating the B player.

We construct a probability tree (I cant attach an image now, as I am posting from work). The situations that would give you the right to become a member would be Win-Win-Win, W-W-L, or L-W-W.

If we play in the BGB order we get 1/32 + 3/32 + 3/32 = 7/32 chance of getting in the club.

If we play in the GBG order we get 1/16 + 1/16 + 1/16 = 3/16 = 6/32 chances of getting in the club.

So it's better to choose the BGB configuration.

Even by not introducing numeric probabilities: You would have higher chances of getting in if you play in the BGB order because even if you loose one game against the B player (highly likely), you still have one more shot if you win the other two games.

If you loose to the B player in the GBG configuration, you dont have any chance of getting in.

[Guest]

This is my first post, so excuse me if I'm not explaining it right.

We assume that there is a higher probability of beating the Good player rather than the Best one. For easier calculations, lets pretend that you have a 1/2 chance of beating the G player and a 1/4 of beating the B player.

We construct a probability tree (I cant attach an image now, as I am posting from work). The situations that would give you the right to become a member would be Win-Win-Win, W-W-L, or L-W-W.

If we play in the BGB order we get 1/32 + 3/32 + 3/32 = 7/32 chance of getting in the club.

If we play in the GBG order we get 1/16 + 1/16 + 1/16 = 3/16 = 6/32 chances of getting in the club.

So it's better to choose the BGB configuration.

Even by not introducing numeric probabilities: You would have higher chances of getting in if you play in the BGB order because even if you loose one game against the B player (highly likely), you still have one more shot if you win the other two games.

If you loose to the B player in the GBG configuration, you dont have any chance of getting in.

goob job

WWW, LWW or WLL will make you a member.

Let say b to the probability that you beat Best and g to the prob. that you beat Good.

And the probability that you can't beat B is of course (1-b)...

For BGB, your chance for WWW is = b.g.b, for WWL= b.g(1-b), for LWW=(1-b).g.b

Get sum: b.g.(2-b)

For GBG if we make same calculations --> sum of chance= b.g.(2-g)

Since g is greater (your chance to beat a good player), (2-g) will be smaller.

gb(2-g)<bg(2-b) ---> So p(GBG)<p(BGB) ---> BGB is a good deal.

[Guest]

I would pick GBG, I think that the best player will beat me every time. If I can win against the good player it will be WLW. If I play a second set of 3 games and win the first round, I would have 2 consecutive wins and satisfy the OP. It would be WLW WLW.

[Guest]

this club gets many new members. That's not a fiscally sound policy.

[Guest]

I would pick GBG, I think that the best player will beat me every time. If I can win against the good player it will be WLW. If I play a second set of 3 games and win the first round, I would have 2 consecutive wins and satisfy the OP. It would be WLW WLW.

If a second set is allowed, you're right. GBG GBG --> two successive games with good one will give you the premium chance. Of course one may say that this needs a further math proof, but I don't think it needs, it is obvious.

this club gets many new members.

That's not a fiscally sound policy.

Yes, it sounds a most cruel club in the universe.

Really, I should have modified the question as "best" to "good", and "good" to "bad".

[Guest]

Since it essential that the second game be won (otherwise it is impossible to get 2 consecutive wins) the best option would be BGB, to give the best chance at the second game. This leaves 2 chances at beating the BEST player. Although this player is harder, two chances are better than the one chance needed to win 2 consecutive games in the GBG scenario.

So, BGB is the safer option.

Edited March 4, 2009 by Wazzathebro

[Guest]

Answere is BGB

Explanation

To secure a entry one can loose only 1 match out of three.Since B is the best player probability of loosing against B is more that that against G.Now if one choose GBG combination ,he can not win even if he wins both matches against G but loose against B.So the appropriate combination is BGB,if one looses one match against B he can still win consecutive matches against G n b(WWL/LWW) and get the entry

[CaptainEd]

Here's the math:

G = probability of beating Good player in one game

B = probability of beating Best player in one game

G > B

Probability of getting in the club by playing the Good player first =

(prob of winning first two games) = G * B

+ (prob of losing first, then winning two games) = (1-G) * B

Total is G * B + (1 - G) * B * G = G * B * (2-G)

Similarly, Prob of getting by playing the Best player first = G * B * (2-B).

So, which is larger?

G*B*(2-G) or G*B*(2-B)

Since G > B, (2-G) < (2-B)

So the better chance is to play the Best player first.

[CaptainEd]

Here's the math:

G = probability of beating Good player in one game

B = probability of beating Best player in one game

G > B

Probability of getting in the club by playing the Good player first =

(prob of winning first two games) = G * B

+ (prob of losing first, then winning two games) = (1-G) * B * G

Total is G * B + (1 - G) * B * G = G * B * (2-G)

Similarly, Prob of getting by playing the Best player first = G * B * (2-B).

So, which is larger?

G*B*(2-G) or G*B*(2-B)

Since G > B, (2-G) < (2-B)

So the better chance is to play the Best player first.