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Tried to model a popular game I was playing, but the probabilities seemed off.

A game allows you to have up to 12 wins but only allows 3 losses. Each win/lose is independent from each other with a 50% probability and assuming we play until 12 wins are hit or 3 losses have occurred. What is the probability of having at least 7 wins?

Attempt: Let X be number of wins. I know that if X = 7, then there were 10 games total. X = 8, 11 games total. . . . For X = 12, there is a special case of 12, 13, 14 total games.

I assumed that X was a binomial random variable and summed up the possibilities for X = 7, 8, 9, 10, 11, 12 and it comes out to be around 30.6% which seems plausible, but if I up the win chance to 75%, then the sum of the probabilities becomes greater than one which implies my original logic is flawed.

Mid
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3 Answers3

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To have $7$ or more wins in total, you need to have won $7$, $8$ or $9$ of the first $9$ games, which (if the probability of winning each game is $p$) is

$${9 \choose 7}p^7(1-p)^2 + {9 \choose 8}p^8(1-p) + {9 \choose 9}p^9$$ $$= p^7 (28p^2-63p +36)$$ and this goes from $0$ to $1$ as $p$ goes from $0$ to $1$. For $p=0.5$, this gives about $0.09$.

Henry
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  • Thank-you for the reply, but my situation allows 3 losses at most and 12 would be the maximum cap for wins. Checked another answer and I am working out if following a negative binomial models it correctly. – Mid Sep 07 '15 at 06:57
  • It is your choice. You could try the infinite sum $${9 \choose 7}p^{7}(1-p)^3+{10 \choose 8}p^{8}(1-p)^3+{11 \choose 9}p^{10}(1-p)^3+{12 \choose 10}p^{10}(1-p)^3+{13 \choose 11}p^{11}(1-p)^3+{14 \choose 12}p^{12}(1-p)^3+{15 \choose 13}p^{13}(1-p)^2+\cdots$$ but you should get the same answer and mine is slightly quicker. – Henry Sep 07 '15 at 07:18
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    Sorry, misread! Your solution is really clever and simplified the problem tremendously. – Mid Sep 07 '15 at 07:30
  • @Henry: Why should there be an infinite sum ? – true blue anil Jun 30 '23 at 08:07
  • @Henry: Very elegant answer, as usual ! Cheers ! – true blue anil Jun 30 '23 at 16:38
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The general "adding up" strategy should work. But how did you calculate $\Pr(X=7)$?

We have $X=7$ if there were exactly $2$ losses in the first $9$ games, and then a loss on the $10$-th. The probability of this is $\binom{9}{2}/2^{10}$. Thus $X$ has a distribution that is more like the negative binomial than the binomial, apart from the special stopping condition at $12$ wins.

Calculate the probabilities of $X=8$, $9$, $10$, and $11$ in the same way. The case $X=12$ will have to be handled differently.

André Nicolas
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  • Thank-you, would this logic be right then: ${9 \choose 7}(0.5)^{10} + {10 \choose 8}(0.5)^{11} + {11 \choose 9}(0.5)^{12}+...+{13 \choose 11}(0.5)^{14}$, then X becomes a binomial for special case of X = 12 and lose no games but for lose 1 game or 2 games it would be negative binomial again? – Mid Sep 07 '15 at 06:51
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    For $12$, it would be $12$ wins in a row (easy) plus $1$ loss in first $12$ games, win on $12$-th, plus $2$ losses in first $13$, win on $14$. Both those have a negative binomial character. – André Nicolas Sep 07 '15 at 07:04
  • Using: ncr(9,7)(.5)^10 + ncr(10,8)(.5)^11 + ncr(11,9)(.5)^12 + ncr(12,10)(0.5)^13 + ncr(13,11)(0.5)^14 + ncr(12,12)(0.5)^12 + ncr(12,12)(0.5)^13 + ncr(13,12)(0.5)^14 gives only an 8.5% chance of winning at least 7 or more which seems plausible, I just want to check if my last line of logic for 12 wins is correct? – Mid Sep 07 '15 at 07:05
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    Let's look at last $3$ entries. The first is right. The next should be $\binom{12}{11}(0.5)^{13}$, or (I prefer, for no good reason) $\binom{12}{11}(0.5)^{13}$. The last should be $\binom{13}{11}(0.5)^{14}$. The correction will make no practical difference to the answer. I have not checked your $8.5%$ calculation. – André Nicolas Sep 07 '15 at 07:15
  • By the way, the solution of Henry is substantially more efficient. – André Nicolas Sep 07 '15 at 07:19
  • Why is it ncr(12,11) and ncr(13,11)? I thought it would be 12 wins so it would be ncr(12,12) and ncr(13,12). – Mid Sep 07 '15 at 07:30
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    I am finding the probability of exactly $12$ wins, to add to the probabilities of exactly $7$, exactly $8$, and so on. About $12$ wins in a row we agree. For $13$ games, we need to win $11$ of the first $12$, and then win. Probability $\binom{12}{11}(0.5)^{13}$. For $14$ games we need $11$ wins in the first $13$, then a win. Probability $\binom{13}{11}(0.5)^{14}$.. – André Nicolas Sep 07 '15 at 07:43
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I think the most difficult part is to visualize the sample space of the game. Let's denote with "W" the event that a game ended with a win (its probability $p$) and with "L" the event that the game ended with a loss (its probability $q,\ p+q=1$). Then we may have the following cases : $$ \eqalign{ 0. & LLL \ \ \ \ \ q^3 \cr 1. & \underbrace{W}_{LL}L \ \ \ \ \ {3 \choose 1}p^1 \cdot q^2 \cdot q \cr 2. & \underbrace{WW}_{LL}L \ \ \ \ \ {4 \choose 2}p^2 \cdot q^2 \cdot q \cr 3. & \underbrace{WWW}_{LL}L \ \ \ \ \ {5 \choose 3}p^3 \cdot q^2 \cdot q \cr 4. & \underbrace{WWWW}_{LL}L \ \ \ \ \ {6 \choose 4}p^4 \cdot q^2 \cdot q \cr 5. & \underbrace{WWWWW}_{LL}L \ \ \ \ \ {7 \choose 5}p^5 \cdot q^2 \cdot q \cr 6. & \underbrace{WWWWWW}_{LL}L \ \ \ \ \ {8 \choose 6}p^6 \cdot q^2 \cdot q \cr 7. & \underbrace{WWWWWWW}_{LL}L \ \ \ \ \ {9 \choose 7}p^7 \cdot q^2 \cdot q \cr 8. & \underbrace{WWWWWWWW}_{LL}L \ \ \ \ \ {10 \choose 8}p^8 \cdot q^2 \cdot q \cr 9. & \underbrace{WWWWWWWWW}_{LL}L \ \ \ \ \ {11 \choose 9}p^9 \cdot q^2 \cdot q \cr 10. & \underbrace{WWWWWWWWWW}_{LL}L \ \ \ \ \ {12 \choose 10}p^{10} \cdot q^2 \cdot q \cr 11. & \underbrace{WWWWWWWWWWW}_{LL}L \ \ \ \ \ {13 \choose 11}p^{11} \cdot q^2 \cdot q \cr 12. & WWWWWWWWWWWW \ \ \ \ \ p^{12} \cr } $$ From the above table we may compute the probability "of having at least 7 wins" by summing the probabilities of cases $7-12$.

nickchalkida
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