My initial approach is to not play paper (since you can only lose/draw so its not optimal) and play rock with probability $$a$$ and scissor with probability $$1-a$$. \mathbb{P}(\text{A wins} \mid \text{B leads}) = 1 - \mathbb{P}( \text{B wins} \mid \text{B leads}) = 1 - \mathbb{P}(\text{A wins} \mid \text{A leads}). Since the $6^{\text{th}}$ win must occur on Round 20, he/she must win $5$ and tie $1$ over the preceding six rounds.

Thanks for contributing an answer to Mathematics Stack Exchange! So the expected winning is $P(\text{Alice wins}) = 7p \cdot 44q + 3p \cdot 6q = \frac{7}{10}\cdot\frac{44}{50} + \frac{3}{10} \cdot {6}{50} = \frac{308}{500} + \frac{18}{500} = \frac{326}{500} = \frac{163}{250}$. To subscribe to this RSS feed, copy and paste this URL into your RSS reader.

I think you overlooked the fact that the five game lead can not happen before game 20. What are some familiar examples in our solar system, and can some still be closed? Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience.

If $X$ wins $5$, loses $1$, and ties $1$, then he/she must have won $4$, lost $1$, and tied $1$ over those six rounds. Rock, Paper, Scissors Probability For Multiple Players. Author has published a graph but won't share their results table. Thanks for contributing an answer to Mathematics Stack Exchange! Check out the resource. 20 ways. \begin{align} In a similiar way you can find $$\text{Alice wins 2, Bob wins 1, 7 draws}$$ Losie-pants had to win 6 out 7 games to come out 5 ahead. At round 20, one of them attains the 5 win lead and the game ends. Ask Question Asked 4 years, 6 months ago. It only takes a minute to sign up. Alice and Bob play Rock, Paper, Scissors until one or the other is 5 wins ahead. Can I afford to take this job's high-deductible health care plan? (You need to consider the number of orderings of these outcomes.) Why is the rate of return for website investments so high? They play several games against other students as well as a simulated game using cards, dice or a computer.

This means you should play the move that they just played. If your opponent plays paper w.p $$b$$ and scissor w.p $$1-b$$. Rock Paper Scissors is a new resource in Year 6 exploring concepts in Probability. Active 4 years, 6 months ago. $$\text{Alice wins 5, Bob wins 10, 5 draws} - Lose$$ Alice and Bob play Rock, Paper, Scissors until one or the other is 5 wins ahead. Probability Bob is ahead after 13 games: $3/10$. $P(\text{Alice wins}) = P(\text{Alice leads})P(\text{Alice wins } | \text{ Alice leads}) + P(\text{Bob leads})P(\text{Alice wins } | \text{ Bob leads})$. The simulated games in this lesson of Rock Paper Scissors are random. While it is marked as a Year 6 resource it could be used at many other year levels! After 10 rounds, Alice is 1 ahead. There are clearly $6$ ways for such a sequence of outcomes to happen.

How to Win at Rock, Paper, Scissors. However the losie pant's win most occur before the last winnie-pants win or else winnie-pants would have been up by 5. You write down Rock, Paper, and then Player 2 wins.

Finding zero cross of AC signal digitally. Are websites a good investment? $$\text{Alice wins 2, Bob wins 7, 11 draws} - Lose$$ $$\text{Alice wins 10, Bob wins 5, 5 draws} - Win$$ Viewed 2k times 1 $\begingroup$ I heard this recently and it got me thinking. I heard this recently and it got me thinking. p. Your strategy depends on your 'loss function'. How can I get readers to like a character they’ve never met? If winnie.-pants must be the winner, losie-pants cannot lose after the fourth win of winnie-pants, otherwise the game would have ended in round 18 or 19. I could have done a counting error. The main issue with your solution occurs when you assert that, all outcomes occur with equal probability. To learn more, see our tips on writing great answers. all of those are not possible. Doesn't matter. Statistical Variation: Foundation to Year 10. They play several games against other students as well as a simulated game using cards, dice or a computer. @fleablood, I take that in account by making sure the 19 round is won by Bob if Alice wins once in the last rounds, or round 19 is draw and round 18 is won by Bob, so that Alice wins before Bob has a 5 game lead. What are the chances that out of $n$ people playing rock, paper, scissors, only two choices are picked? The reSolve team is about to commence writing a series of new resources as part of our work on learning progressions. So there are 6 games Losie-pants could have drawn on. At round 20, one of them attains the 5 win lead and the game ends. This means you should play the move that was not played. Rock Paper Scissor - Probability Game. $$\text{Alice wins 1, Bob wins 0, 9 draws}$$ What is the optimal strategy for this game? Seems, scissors is the best choice directly. Similarly, we have $44q = P(\text{Alice wins } | \text{ Alice leads}) = P(\text{Bob wins } | \text{ Bob leads})$ and $6q = P(\text{Bob wins } | \text{ Alice leads}) = P(\text{Alice wins } | \text{ Bob leads})$, where $q = \frac{1}{50}$. For $X$ to win, you may expect that, similarly to above, we would have two distinct probabilities over the next seven rounds: This is nearly correct; however, we must account for the fact, that any win putting $X$ ahead by $5$ must not occur before Round 20. Probability Alice is ahead after 13 games: $7/10$. $$\text{Alice wins 4, Bob wins 5, 4 draws}$$ I assume Bob is winnie-pants, so Bob should win 18 if 19 is draw, so that Alice wins once before to prevent Bob from getting 5 ahead. Not every outcome is equaly likely at the end. The students use results from this game to determine how random the results really are in games against other people and in simulated games. $$\text{Alice wins 6, Bob wins 5, 2 draws}$$ First case can happen in 20 ways and the second case can happen in 4 ways. Rock Paper Scissors Type Games. Does this use of the perfect actually express something about the future? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. After 10 rounds, Alice is 1 ahead. Than you combine those probabilities to find that the probability Alice wins is $\frac{163}{250}$.

However, we also know that after 10 rounds, Alice has 1 more win than Bob, so the probability that Alice has the lead after 13 rounds is different than the probaliity Bob has the lead after 13 rounds. A Question Regarding the Meaning of a Question About the Probability of a R-P-S Game. Does it make any scientific sense that a comet coming to crush Earth would appear "sideways" from a telescope and on the sky (from Earth)? .. oh, scrod.

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# rock paper scissors probability calculator

And of the three remaining games: -- all were draws, so 3 of 6 where draws rest were winnie-pants wins: ${6 \choose 3} = 6*5*4/1*2*3 = 20$ ways.

First Polygon; SSA Counterexample: Don't Be an A** circumcenter "After 10 rounds, Alice is 1 ahead. Winnie pants won 3 other games in order to end up 5 ahead. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Why is Italiae used rather than Italis in the phrase "In hortis Italiae"? After 13 rounds, one or the other is 1 ahead. Please let me know if this is correct. site design / logo © 2020 Stack Exchange Inc; user contributions licensed under cc by-sa. We know that after 13 rounds, Alice or Bob has 1 win more than the other.

My initial approach is to not play paper (since you can only lose/draw so its not optimal) and play rock with probability $$a$$ and scissor with probability $$1-a$$. \mathbb{P}(\text{A wins} \mid \text{B leads}) = 1 - \mathbb{P}( \text{B wins} \mid \text{B leads}) = 1 - \mathbb{P}(\text{A wins} \mid \text{A leads}). Since the $6^{\text{th}}$ win must occur on Round 20, he/she must win $5$ and tie $1$ over the preceding six rounds.

Thanks for contributing an answer to Mathematics Stack Exchange! So the expected winning is $P(\text{Alice wins}) = 7p \cdot 44q + 3p \cdot 6q = \frac{7}{10}\cdot\frac{44}{50} + \frac{3}{10} \cdot {6}{50} = \frac{308}{500} + \frac{18}{500} = \frac{326}{500} = \frac{163}{250}$. To subscribe to this RSS feed, copy and paste this URL into your RSS reader.

I think you overlooked the fact that the five game lead can not happen before game 20. What are some familiar examples in our solar system, and can some still be closed? Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience.

If $X$ wins $5$, loses $1$, and ties $1$, then he/she must have won $4$, lost $1$, and tied $1$ over those six rounds. Rock, Paper, Scissors Probability For Multiple Players. Author has published a graph but won't share their results table. Thanks for contributing an answer to Mathematics Stack Exchange! Check out the resource. 20 ways. \begin{align} In a similiar way you can find $$\text{Alice wins 2, Bob wins 1, 7 draws}$$ Losie-pants had to win 6 out 7 games to come out 5 ahead. At round 20, one of them attains the 5 win lead and the game ends. Ask Question Asked 4 years, 6 months ago. It only takes a minute to sign up. Alice and Bob play Rock, Paper, Scissors until one or the other is 5 wins ahead. Can I afford to take this job's high-deductible health care plan? (You need to consider the number of orderings of these outcomes.) Why is the rate of return for website investments so high? They play several games against other students as well as a simulated game using cards, dice or a computer.

This means you should play the move that they just played. If your opponent plays paper w.p $$b$$ and scissor w.p $$1-b$$. Rock Paper Scissors is a new resource in Year 6 exploring concepts in Probability. Active 4 years, 6 months ago. $$\text{Alice wins 5, Bob wins 10, 5 draws} - Lose$$ Alice and Bob play Rock, Paper, Scissors until one or the other is 5 wins ahead. Probability Bob is ahead after 13 games: $3/10$. $P(\text{Alice wins}) = P(\text{Alice leads})P(\text{Alice wins } | \text{ Alice leads}) + P(\text{Bob leads})P(\text{Alice wins } | \text{ Bob leads})$. The simulated games in this lesson of Rock Paper Scissors are random. While it is marked as a Year 6 resource it could be used at many other year levels! After 10 rounds, Alice is 1 ahead. There are clearly $6$ ways for such a sequence of outcomes to happen.

How to Win at Rock, Paper, Scissors. However the losie pant's win most occur before the last winnie-pants win or else winnie-pants would have been up by 5. You write down Rock, Paper, and then Player 2 wins.

Finding zero cross of AC signal digitally. Are websites a good investment? $$\text{Alice wins 2, Bob wins 7, 11 draws} - Lose$$ $$\text{Alice wins 10, Bob wins 5, 5 draws} - Win$$ Viewed 2k times 1 $\begingroup$ I heard this recently and it got me thinking. I heard this recently and it got me thinking. p. Your strategy depends on your 'loss function'. How can I get readers to like a character they’ve never met? If winnie.-pants must be the winner, losie-pants cannot lose after the fourth win of winnie-pants, otherwise the game would have ended in round 18 or 19. I could have done a counting error. The main issue with your solution occurs when you assert that, all outcomes occur with equal probability. To learn more, see our tips on writing great answers. all of those are not possible. Doesn't matter. Statistical Variation: Foundation to Year 10. They play several games against other students as well as a simulated game using cards, dice or a computer. @fleablood, I take that in account by making sure the 19 round is won by Bob if Alice wins once in the last rounds, or round 19 is draw and round 18 is won by Bob, so that Alice wins before Bob has a 5 game lead. What are the chances that out of $n$ people playing rock, paper, scissors, only two choices are picked? The reSolve team is about to commence writing a series of new resources as part of our work on learning progressions. So there are 6 games Losie-pants could have drawn on. At round 20, one of them attains the 5 win lead and the game ends. This means you should play the move that was not played. Rock Paper Scissor - Probability Game. $$\text{Alice wins 1, Bob wins 0, 9 draws}$$ What is the optimal strategy for this game? Seems, scissors is the best choice directly. Similarly, we have $44q = P(\text{Alice wins } | \text{ Alice leads}) = P(\text{Bob wins } | \text{ Bob leads})$ and $6q = P(\text{Bob wins } | \text{ Alice leads}) = P(\text{Alice wins } | \text{ Bob leads})$, where $q = \frac{1}{50}$. For $X$ to win, you may expect that, similarly to above, we would have two distinct probabilities over the next seven rounds: This is nearly correct; however, we must account for the fact, that any win putting $X$ ahead by $5$ must not occur before Round 20. Probability Alice is ahead after 13 games: $7/10$. $$\text{Alice wins 4, Bob wins 5, 4 draws}$$ I assume Bob is winnie-pants, so Bob should win 18 if 19 is draw, so that Alice wins once before to prevent Bob from getting 5 ahead. Not every outcome is equaly likely at the end. The students use results from this game to determine how random the results really are in games against other people and in simulated games. $$\text{Alice wins 6, Bob wins 5, 2 draws}$$ First case can happen in 20 ways and the second case can happen in 4 ways. Rock Paper Scissors Type Games. Does this use of the perfect actually express something about the future? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. After 10 rounds, Alice is 1 ahead. Than you combine those probabilities to find that the probability Alice wins is $\frac{163}{250}$.

However, we also know that after 10 rounds, Alice has 1 more win than Bob, so the probability that Alice has the lead after 13 rounds is different than the probaliity Bob has the lead after 13 rounds. A Question Regarding the Meaning of a Question About the Probability of a R-P-S Game. Does it make any scientific sense that a comet coming to crush Earth would appear "sideways" from a telescope and on the sky (from Earth)? .. oh, scrod.