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Say you need to win 4 times in a row at a 25% success chance for each event, how do I calculate the chance of winning? Overall, how does this calculation goes? given that 1st is a win, what is the chance that my second is a win, third, fourth etc. (Consider the events to be independent and can go for an infinity amount trials)

Pirategull
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  • That depends. Are the results of each attempt independent from one another? Consider the following scenario... I draw a card from a shuffled deck of four cards, one of which is a queen, hoping to get the queen. Whatever I draw, I put back on top of the deck without shuffling. My next try I just draw the top card which is the same card as what I drew last time... In such a scenario, given the 1st is a win the chance the second is a win is 100% and so on... This scenario also satisfies that it is a 25% chance of winning for a specific attempt (with no knowledge of earlier attempts) – JMoravitz Nov 12 '19 at 15:17
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    Compare this to if we shuffle the deck each time... As for calculating... if the outcomes are independent, we have that $Pr(A\cap B) = Pr(A)\times Pr(B)$, so winning the first and second time would have been $\frac{1}{4}\times \frac{1}{4} = \frac{1}{16}$, and winning three times out of three attempts would have been $\frac{1}{4}\times\frac{1}{4}\times\frac{1}{4}$, etc... Without knowledge of whether or not these are independent, there is no way of knowing. It could be anywhere from $0.25$ to $0$ – JMoravitz Nov 12 '19 at 15:20
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    Now... if your question isn't "what is the probability of winning four times in a row when trying four times" but rather "what is the probability of getting four wins in a row when trying $n$ times" with $n\geq 4$, that is a very different and much more complicated problem, but there should be duplicates elsewhere on this site that you can find. Try searching "heads in a row." You'll find questions like this one – JMoravitz Nov 12 '19 at 15:26

2 Answers2

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Given the first is won, the chance that the second is won is the same chance you had of winning the first, since the events are independent. To find the probability of getting 4 wins in a row, multiply the probability of a single win in each term by each other. So, if there is a 0.25 change of winning each round, the equation should appear as below: $$ 0.25 \times 0.25 \times 0.25 \times 0.25 $$ To make the multiplying a bit easier, let's use fractions (multiplying 4s is easier than multiplying 25s): $$ \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{256} $$ So, the probability of winning (by getting 4 wins in a row) is $ \frac{1}{256} $

suhbell
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    That's the probability of winning in 4 trials, which may be the most natural interpretation of the question, but not the only one. For example, when someone sees 4 consecutive home runs in a baseball game and says "What's the probability of that?" it's doubtful they mean "If I decide ahead of time to watch only 4 batters, what's the probability that they all hit home runs?" – Ned Nov 12 '19 at 17:50
  • I took the note in parentheses "Consider the events to be independent and can go for an infinity amount trials" to mean that the events are independent. – suhbell Nov 14 '19 at 21:32
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    It's not about independence -- that's fine -- it's about what you mean by "what's the probability of four wins in a row?" which is an incomplete question without specifying how many trials you will perform. In other words, do you mean "If I perform 4 trials, what's the probability that they are all successes?" That's $0.25^4$ like you found, but then why say anything about infinitely many trials? – Ned Nov 14 '19 at 22:09
  • Ah I see what you mean and yes, my answer assumed that the question was asking the probability of 4 wins in a row given a set of 4 games to play. – suhbell Nov 15 '19 at 17:40
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If the events are independent you get $$(1/4)^4 = \frac {1}{256}$$

Mohammad Riazi-Kermani
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