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My Question is -

Each round Mike and Dean toss coin each. Mike tosses a not fair coin in which the probability to get heads is $0.6$. Dean tosses a not fair coin in which the probability to get heads is $0.1$. they toss the coins till they get the same results at the same time. What is the probability that there will be at most 5 rounds?

I started to calculate it as geometric distribution but something doesn't seem right in my calculations. I thought so since they are throwing till 'success" which defined Geometric probability.

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  • $\begingroup$ Geometric is fine here. Just compute the probability that the win on round $1$, round $2$, ..., round $5$, and sum. $\endgroup$
    – lulu
    May 17, 2020 at 20:27
  • $\begingroup$ First question: What is the probability of success in an individual round? $\endgroup$ May 17, 2020 at 20:27
  • $\begingroup$ It's a sequence of Bernoulli trials, all right. You'll have to show us your computations before we can tell if if they're wrong. $\endgroup$
    – saulspatz
    May 17, 2020 at 20:28
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    $\begingroup$ Alternatively, you could compute the complementary probability...the probability that they fail to match five times in a row. That is easier. $\endgroup$
    – lulu
    May 17, 2020 at 20:28
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    $\begingroup$ What probability of success did you calculate? It is helpful for you to show your work, even if it is wrong. Then ohers will be able to tell you what you did right, and where you went wrong. $\endgroup$ May 17, 2020 at 20:31

3 Answers 3

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The probability till Mike and Dean get the same results at most 5 rounds:

The final results can be... $${HH}\ or\ {TT}$$ The probability that Mike and Dean get the same results in a round $$={0.6}\times{0.1}+{(1-0.6)}\times{(1-0.1)}=0.42 $$ The probability that Mike and Dean do not get the same results in a round $$={1-0.42}=0.58$$

Let X be the number of rounds until Mike and Dean get the same results. $X\sim Geo(0.42)$

$$P(X\leq5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)$$ $$P(X\leq5)=0.42+0.58\times0.42+(0.58)^2\times0.42+(0.58)^3\times0.42+(0.58)^4\times0.42\approx0.9344\ (corr.to\ 4\ d.p.)$$

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Mark andDean have 5 rounds in which they need at least one successful toss. P(successful toss)=0.6*0.1=0.01

Next we use 1-(1-p)^n 1-(1-0.06)^5=0.2661

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The game ends when Mike and Dean toss different results.

  • Mike: Heads=0.6 + Tails=0.4
  • Dean: Heads=0.1 + Tails=0.9

The probability of round being:

  • Both Heads = 0.60 * 0.40 = 0.24
  • Both Tails = 0.10 * 0.90 = 0.09
  • Both Same = 0.24 + 0.09 = 0.33
  • Both Different = 1 - 0.33 = 0.67

What is the probability that of getting to round X?

  • 1 = 100% (first round is always played)
  • 2 = 0.33 (both same)
  • 3 = 0.33^2 = 0.1089 (both same, twice)
  • 4 = 0.33^3 = 0.0359 (both same, 3x)
  • 5 = 0.33^4 = 0.0118 (both same, 4x)
  • 6 = 0.33^5 = 0.0039  (both same, 5x)

What is the probability that there will be at most 5 rounds?

  • This is simply 1 minus the probability of getting to round 6
  • (1 - 0.33^5) = 0.99608
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