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Suppose you are playing a game where you have 6 coins and your friend has 4.

You win in the event you toss more heads than your friend, otherwise you lose. Suppose that these are fair coins.

What is your probability of winning?

My thoughts:

Am I correct in thinking that the outcome of the first 4 coins don't matter ( i.e. ) symmetric and instead we should focus on what happens in the last two coins:
Out of those two coins, we must flip two heads - otherwise we lose.

Therefore the probability of winning is 1/4?

TYIA!

Sushiix
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  • A possible outcome : You - TTTTHH and Friend-HHHHTT ,Your friend wins in this case, thus the first 4 outcomes should also be considered –  Jun 01 '22 at 08:34
  • @RamanujanXXV This isn't actually a counter-example to the OP's claim. Consider, for example, the probability of getting an even number of heads in $n$ tosses: it's actually precisely $\tfrac12$ for all $n$. Indeed, regardless of the first $n-1$ tosses, the probability that the number after $n$ is even is $\tfrac12$. So, the first $n-1$ don't matter. The current question does not have this property, though, as I explain in my answer below – Sam OT Jun 01 '22 at 09:10

2 Answers2

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The correct answer can be found in a duplicate question from one of the SE-linked related questions: I flip M coins, my opponent flips N coins. Who has more heads wins. Is there a closed form for probability?. You should really check these. But still, it's not exactly a duplicate as you have asked for a proof verification. So, I'll just want to explain where you've gone wrong.

First, do you think that $1/4$ sounds reasonable? It seems a bit low to me given that you're expected to get $3$ and your friend $2$. More importantly, though, you have not justified why you can inore the first four—and you can't. Imagine you have 100002 coins and your friend 100000. Intuitively, these last two coins are going to make next to no difference, neither will the requirement to have more (instead of at least as many) heads. So, the probability will be very close to $1/2$.

In particular, your argument worked for $n+2$ vs $n$ coins, independently of $n$. But, what I've just shown, informally, is that the limit as $n \to \infty$ is $\tfrac12$.

Sam OT
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  • I suppose I'm trying to extrapolate from : https://math.stackexchange.com/questions/146802/probability-of-5-fair-coin-flips-having-strictly-more-heads-than-4-fair-coin?rq=1

    Where if n and n+1 coins are used, then we look only at the n+1 'th coin in order to determine the outcome. Hence I had thought that we now then look only at the last n+1th and n+2th coins, which have 4 total combinations ( HT, HH, TT, TH ), one which we win.

     – Sushiix Jun 01 '22 at 09:31
  • @Sushiix Yeah, it's not a bad idea for sure. You know the probability of winning is 1/2 when you only toss five, not six, coins by the linked question. That immediately tells you that the probability is at least—in fact, strictly larger than—1/2 when you toss the full six. So, your claim of 1/4 could not be right—maybe you meant 3/4 instead? I guess the difficulty in copying the linked question is that the claim "if friend is ahead after 4 tosses, then they win" no longer holds: you can be one behind, then toss two heads and win – Sam OT Jun 01 '22 at 10:28
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The answer is easy to calculate. 638/1024, roughly 62.3% probability. Say your outcomes are A, your friend's are B. There are 5 cases for B that you must beat. For each of B's outcomes, find the probability that you succeed. These are all specific cases, so they will have low-ish probabilities. Then add those together for the total probability of winning.

B = 0 Heads, A = 1+ (one or more) Heads: P(B zero) = (1/2)^4 (4C0) = 1/16 P(A not zero) = 1 - [(1/2)^6 (6C0)] = 1 - 1/64 = 63/64 Multiply, to get P(you win) for this case. (1/16)(63/64) = 63/1024

B = 1, A = 2+ : P(B=1) = (1/2)^4 (4C1) = 4/16 Four ways to get one Heads. P(A not 0 or 1)= 1 - [(1/2)^6 (6C1)] - 1/64 = 1 - 6/64 - 1/64 = 57/64 Six ways to get one Heads, still only one way to get zero Heads. Keep subtracting more and more from one, for each new case. Multiply again. (4/16)(57/64) = 228/1024

B = 2, A = 3+ : P(B=2) = 6/16 P(A not 0,1,2) = 1 - 15/64 - 6/64 - 1/64 = 42/64 Therefore, (6/16)(42/64) = 252/1024

B = 3, A = 4+ : P(B=3) = 4/16 P(A not 0,1,2,3) = 1 - 20/64 - 15/64 - 6/64 - 1/64 = 22/64 So, (4/16)(22/64) = 88/1024

B = 4, A = 5+ : P(B=4) = 1/16 P(A not 0,1,2,3,4) = 1 - 15/64 - 20/64 - 15/64 - 6/64 - 1/64 = 7/64 And, (1/16)(7/64) = 7/1024

Finally, the sum is 638/1024 ≈ .623

Bafs
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