BDMO 2017 National round Secondary 1

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ahmedittihad
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BDMO 2017 National round Secondary 1

Unread post by ahmedittihad » Fri Feb 10, 2017 8:36 pm

Bangladesh plays India in a best of five game series. The team that wins $3$ games first wins the series. The series can end after $3$ games, or $4$ games, or $5$ games. If Bangladesh and India are equally strong, calculate

$(a)$ The probability that Bangladesh wins the series in $3$ games.
$(b)$ The probability that Bangladesh wins the series in $4$ games.
$(c)$ The probability that the series ends in exactly $5$ games.
Frankly, my dear, I don't give a damn.

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samiul_samin
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Re: BDMO 2017 National round Secondary 1

Unread post by samiul_samin » Fri Feb 02, 2018 2:11 am

(a)$1/8$☺☺☺

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Tasnood
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Re: BDMO 2017 National round Secondary 1

Unread post by Tasnood » Sat Feb 03, 2018 12:19 am

For $5$ match, result may be of $20$ ways in total. (Excluding unimportant results when series ends in 3/4 matches)

(a)If Bangladesh wins in 3 games, none but the sequence must be:$WWW$. So, the probability=$\frac{1}{20}$

(b)If Bangladesh wins in 4 matches, the sequence may be following: $WWLW, WLWW, LWWW$ (Excluding WWWL), in $3$ ways.
The probability=$\frac{3}{20}$

(c)When the series ends in the last match, the last one must be $W$. So, the sequence: $WWLLW,WLWLW,WLLWW,LWWLW,LWLWW,LLWWW$, in$6$ ways. (Rearranging the first four term with 2 identical W's and 2 identical L)
The probability=$\frac{6}{20}=\frac{3}{10}$
Last edited by Tasnood on Sat Feb 03, 2018 11:40 pm, edited 2 times in total.

Absur Khan Siam
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Re: BDMO 2017 National round Secondary 1

Unread post by Absur Khan Siam » Sat Feb 03, 2018 11:16 pm

Here are the list of all possibilities of the result of the game(Denoting $W$ and $L$ are the win and lose of Bangladesh):
  • $WWW$
  • $LLL$
  • $LWWW$
  • $WLWW$
  • $WWLW$
  • $WLLL$
  • $LWLL$
  • $LLWL$
  • $WWLLW$
  • $WLWLW$
  • $WLLWW$
  • $LWWLW$
  • $LWLWW$
  • $LLWWW$
  • $WWLLL$
  • $WLWLL$
  • $WLLWL$
  • $LWWLL$
  • $LWLWL$
  • $LLWWL$
Hence there are $20$ different possible results, not $2^5$
Last edited by Absur Khan Siam on Sun Feb 04, 2018 12:35 pm, edited 1 time in total.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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ahmedittihad
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Re: BDMO 2017 National round Secondary 1

Unread post by ahmedittihad » Sun Feb 04, 2018 2:47 am

Both the solutions are completely wrong. Because the 20 cases aren't equally likely.
Frankly, my dear, I don't give a damn.

Absur Khan Siam
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Re: BDMO 2017 National round Secondary 1

Unread post by Absur Khan Siam » Sun Feb 04, 2018 12:34 pm

My solution may be wrong.But the $20$ cases yield to a result that both team have a probability of $\frac{1}{2}$ to win the series.
"(To Ptolemy I) There is no 'royal road' to geometry." - Euclid

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Tasnood
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Re: BDMO 2017 National round Secondary 1

Unread post by Tasnood » Sun Feb 04, 2018 7:30 pm

Can anyone give the correct solution?

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ahmedittihad
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Re: BDMO 2017 National round Secondary 1

Unread post by ahmedittihad » Tue Feb 06, 2018 1:52 pm

(a) Bangladesh wins the series in $3$ matches iff they win all $3$ of the games. And the probability of that is $\dfrac{1}{8}$.

(b) Out of the $16$ cases of a $4$ length binary string, Only 3 of them satisfy (WWLW, WLWW,LWWW). So the probability here is $\dfrac{3}{16}.

(b) Probability of winning the series for Bangladesh is $\dfrac{1}{2}$. We can find the answer to this question by subtracting the answers of a and b from $\dfrac{1}{2}$. And, $\dfrac{1}{2}-\dfrac{3}{16}-\dfrac{1}{8}=\dfrac{3}{16}$
Frankly, my dear, I don't give a damn.

SMMamun
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Re: BDMO 2017 National round Secondary 1

Unread post by SMMamun » Thu Feb 08, 2018 1:35 am

The key to understanding this problem, as Ahmed pointed out, is that not all cases are equally likely to happen. The probabilities for the series to end in exactly 3 matches and to end in exactly 4 (or 5) matches are not equal, so the 20 cases listed above cannot be treated with equal weights. In general, extra care must be exercised when the no. of elements in probabilistic events changes.

Listing of the 20 cases by Siam and Tasnood are appreciable and are worthy of Secondary level at which students should elaborate and analyze problems as detailed as possible even if that means spending additional time. Now let's see how we can approach the problem.

(a) This part is easier: simple, independent probability.
Bangladesh must win the 1st match. What is the probability of this? It's $\frac{1}{2}$. End of the story? No, not yet!
Bangladesh must win the 2nd match. Probability of winning the 2nd match is again $\frac{1}{2}$. What is the total probability of winning the first 2 matches? $\frac{1}{2} × \frac{1}{2}$.
What is the probability of WWW then? It's $\frac{1}{2} × \frac{1}{2} × \frac{1}{2} = \frac{1}{8}$

Is any scenario / case other than WWW possible? NO. So the ultimate probability is $\frac{1}{8}$.

(b) Bangladesh must with the 4th match and lose exactly 1 in the first 3 matches. You have already listed all the three combinations:
LWWW
WLWW
WWLW

What is the probability in the first case? $\frac{1}{2} × \frac{1}{2} × \frac{1}{2} × \frac{1}{2} $= $\frac{1}{16}$
So the total probability in all three cases = $3 × \frac{1}{16} = \frac{3}{16}$

If Bangladesh and India have different probabilities, say 70% for Bangladesh winning and 30% for India winning, you would just apply those percentages appropriately in places of W and L.

(c) If the series does not end in the first 3 or 4 games, it must end in the 5th.
Probability of ending in 3 matches = Probability that Bangladesh wins or (i.e. +) Probability that India wins = $\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$
Probability of ending in 4 matches = Probability that Bangladesh wins or (i.e. +) Probability that India wins = $\frac{3}{16} + \frac{3}{16} = \frac{3}{8}$

Probability of ending in 5 matches = $1 - \frac{1}{4} - \frac{3}{8} = \frac{3}{8} = 0.375$

Another way to solve part (c): The series must have been tie after 4 matches: each team has won exactly two matches and lost exactly two matches, and the series goes to 5th match, 100% for sure. That can happen only in 6 possible cases ($4C2$ or $\frac{4!}{2!2!}$) if we consider 4 matches because in that case we do not need to know the outcome of the 5th match :
LLWW, WWLL
LWLW, WLWL
LWWL, WLLW

And the probability is $(6 × \frac{1}{16})$ or $0.375$.

Even another way to solve (c). There are 12 cases if we consider the outcome of the 5th match also.
WWLLW
WLWLW
WLLWW
LWWLW
LWLWW
LLWWW
WWLLL
WLWLL
WLLWL
LWWLL
LWLWL
LLWWL

And the probability is $(12 × \frac{1}{32})$ or $0.375$.

Hope it's clear. :)

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samiul_samin
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Re: BDMO 2017 National round Secondary 1

Unread post by samiul_samin » Tue Feb 20, 2018 1:07 am

SMMamun wrote:
Thu Feb 08, 2018 1:35 am
The key to understanding this problem, as Ahmed pointed out, is that not all cases are equally likely to happen. The probabilities for the series to end in exactly 3 matches and to end in exactly 4 (or 5) matches are not equal, so the 20 cases listed above cannot be treated with equal weights. In general, extra care must be exercised when the no. of elements in probabilistic events changes.

Listing of the 20 cases by Siam and Tasnood are appreciable and are worthy of Secondary level at which students should elaborate and analyze problems as detailed as possible even if that means spending additional time. Now let's see how we can approach the problem.

(a) This part is easier: simple, independent probability.
Bangladesh must win the 1st match. What is the probability of this? It's $\frac{1}{2}$. End of the story? No, not yet!
Bangladesh must win the 2nd match. Probability of winning the 2nd match is again $\frac{1}{2}$. What is the total probability of winning the first 2 matches? $\frac{1}{2} × \frac{1}{2}$.
What is the probability of WWW then? It's $\frac{1}{2} × \frac{1}{2} × \frac{1}{2} = \frac{1}{8}$

Is any scenario / case other than WWW possible? NO. So the ultimate probability is $\frac{1}{8}$.

(b) Bangladesh must with the 4th match and lose exactly 1 in the first 3 matches. You have already listed all the three combinations:
LWWW
WLWW
WWLW

What is the probability in the first case? $\frac{1}{2} × \frac{1}{2} × \frac{1}{2} × \frac{1}{2} $= $\frac{1}{16}$
So the total probability in all three cases = $3 × \frac{1}{16} = \frac{3}{16}$

If Bangladesh and India have different probabilities, say 70% for Bangladesh winning and 30% for India winning, you would just apply those percentages appropriately in places of W and L.

(c) If the series does not end in the first 3 or 4 games, it must end in the 5th.
Probability of ending in 3 matches = Probability that Bangladesh wins or (i.e. +) Probability that India wins = $\frac{1}{8} + \frac{1}{8} = \frac{1}{4}$
Probability of ending in 4 matches = Probability that Bangladesh wins or (i.e. +) Probability that India wins = $\frac{3}{16} + \frac{3}{16} = \frac{3}{8}$

Probability of ending in 5 matches = $1 - \frac{1}{4} - \frac{3}{8} = \frac{3}{8} = 0.375$

Another way to solve part (c): The series must have been tie after 4 matches: each team has won exactly two matches and lost exactly two matches, and the series goes to 5th match, 100% for sure. That can happen only in 6 possible cases ($4C2$ or $\frac{4!}{2!2!}$) if we consider 4 matches because in that case we do not need to know the outcome of the 5th match :
LLWW, WWLL
LWLW, WLWL
LWWL, WLLW

And the probability is $(6 × \frac{1}{16})$ or $0.375$.

Even another way to solve (c). There are 12 cases if we consider the outcome of the 5th match also.
WWLLW
WLWLW
WLLWW
LWWLW
LWLWW
LLWWW
WWLLL
WLWLL
WLLWL
LWWLL
LWLWL
LLWWL

And the probability is $(12 × \frac{1}{32})$ or $0.375$.

Hope it's clear. :)

No,It isn't clear to me because you counted twice in ($c$) and got the double of the real probablity.The answer of ($c$) is $\fbox {0.1875}$

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