What is the Probability of Alvin Getting More Heads Than Andy in a Coin Toss?

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Homework Help Overview

The problem involves calculating the probability that Alvin, who tosses 4 coins, gets more heads than Andy, who tosses 3 coins, using fair coins. Participants are exploring various methods to approach this probability question.

Discussion Character

  • Exploratory, Assumption checking, Mixed

Approaches and Questions Raised

  • Some participants discuss the possible outcomes for Alvin and Andy's coin tosses and how to calculate the probabilities associated with each outcome. Others suggest considering the order of coin flips and the implications of ties between the two players. There is also mention of a potential shortcut involving symmetry and the outcomes of the first three flips.

Discussion Status

The discussion is ongoing, with various interpretations of the problem being explored. Some participants have proposed different methods and shortcuts, while others express uncertainty about the correctness of their calculations. There is no explicit consensus on the final probability, and multiple viewpoints are being considered.

Contextual Notes

Participants note the original phrasing of the problem and its implications, as well as the potential for misunderstanding the conditions under which Alvin wins. There is also mention of the generalization of the problem to different numbers of coins.

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Homework Statement



Alvin and Andy both have 4 coins and 3 coins respectively. The coins are fair. Both of them toss each of their coins once. What is the probability that Alvin gets more head than Andy?

Homework Equations





The Attempt at a Solution



(1,0),(2,0),(3,0),(4,0),(2,1),(3,1),(4,1),(3,2),(4,2),(4,3)

the probability is (1/2)^8 x 10=5/128
 
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So you have all the possible ways Alvin could have more heads than Andy (not gets more head than Andy O.o), but how can we work out the probability of each of them?

Eg (1,0) So Alvin must get exactly 1 head from 4 coins, then Andy must get exactly 0 from 3. Say it occurred in the order H T T T for Alvin, then T T T for Andy, then the probability of each of those events occurring is (1/2)^4 for Alvin, and (1/2)^3 for Andy, so 1/2^7 overall. But, Alvin could have gotten the heads in a different order, maybe T T H T, and that event also has the same chance of occurring. Since order doesn't matter, we have to multiply by the different orders the event could occur . For (1,0) case, there's only 4 different ways Alvin could get his H, and Andy can only get 0 in 1 way.

Once you consider the different orders each combination can happen and multiply, you can just add the probabilities of the cases.

PS If there's some big shortcut you used as there perhaps is, I'm sorry I don't see it.
 


Gib Z said:
So you have all the possible ways Alvin could have more heads than Andy (not gets more head than Andy O.o), but how can we work out the probability of each of them?

Eg (1,0) So Alvin must get exactly 1 head from 4 coins, then Andy must get exactly 0 from 3. Say it occurred in the order H T T T for Alvin, then T T T for Andy, then the probability of each of those events occurring is (1/2)^4 for Alvin, and (1/2)^3 for Andy, so 1/2^7 overall. But, Alvin could have gotten the heads in a different order, maybe T T H T, and that event also has the same chance of occurring. Since order doesn't matter, we have to multiply by the different orders the event could occur . For (1,0) case, there's only 4 different ways Alvin could get his H, and Andy can only get 0 in 1 way.

Once you consider the different orders each combination can happen and multiply, you can just add the probabilities of the cases.

PS If there's some big shortcut you used as there perhaps is, I'm sorry I don't see it.

Thanks. I tried for every cases and i got 87/128 which is not the correct answer.
 


A shortcut to this problem is this:

Compare Alvin's first 3 coinflips to Andy's first three coinflips. Sometimes they will be tied. If this occurs, then Alvin flips his 4th coin and wins (if its heads) or loses (if its tails). So he wins half of those. The rest of the time, someone will be ahead. If it's Alvin, he wins automatically. If it's Andy, Alvin can't catch up (the best he could do would be to tie with his 4th flip). Since they both flipped 3 coins, by symmetry, Alvin must win half of the cases when they are not tied.

So Alvin wins half the cases when they are tied, and half the cases when they are not tied, or half overall.
 


hgfalling said:
A shortcut to this problem is this:

Compare Alvin's first 3 coinflips to Andy's first three coinflips. Sometimes they will be tied. If this occurs, then Alvin flips his 4th coin and wins (if its heads) or loses (if its tails). So he wins half of those. The rest of the time, someone will be ahead. If it's Alvin, he wins automatically. If it's Andy, Alvin can't catch up (the best he could do would be to tie with his 4th flip). Since they both flipped 3 coins, by symmetry, Alvin must win half of the cases when they are not tied.

So Alvin wins half the cases when they are tied, and half the cases when they are not tied, or half overall.

I really can't agree with that. If Alvin is winning after 3, then Andy can't stop him. If Andy is winning after 3 then Alvin can stop him from winning by throwing a head on the last throw. Alvin wins more games.
 


No, Alvin only wins when he has more heads. When they tie Andy wins.

(The original problem asked "what is the probability that Alvin gets more heads than Andy," so when we turn it into a game, Alvin wins if he gets more heads.)
 


Actually it said "Alvin gets more head than Andy", which made me think it might be a joke. But you are right. The game is tllted in Andy's favor which compensates for Alvin's extra coin. Very perceptive.
 


Well, the cool thing is this obviously generalizes to any number of coins (n vs n+1); there's nothing special about 4 vs 3. If you multiply and sum out all the probabilities, you might be left with the idea that 4 vs 3 is like a special case that happens to end up as 1/2.
 


hgfalling said:
Well, the cool thing is this obviously generalizes to any number of coins (n vs n+1); there's nothing special about 4 vs 3. If you multiply and sum out all the probabilities, you might be left with the idea that 4 vs 3 is like a special case that happens to end up as 1/2.

thank you hgfalling!
 
  • #10


thereddevils said:
Thanks. I tried for every cases and i got 87/128 which is not the correct answer.

My method is not as nice as hgfalling's, but I just tried it myself and got 1/2 so you just did it wrong the first time.
 

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