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Interview Question

Trader Interview(Student Candidate) New York, NY

A has 5 fair coins and B has 4 fair coins. A wins only if

  he flips more heads than B does. Probability of A winning?
Answer

Interview Answer

7 Answers

3

1/2

Interview Candidate on Mar 6, 2012
0

74.3489

ME on May 27, 2012
1

If A flips...
5 H's then A auto wins
4 H's then B must flip 0,1,2, or 3 H's
3 H's then B must flip 0,1, or 2 H's
2 H's then B must flip 0, or 1 H's
1 H then B must flip 0 H
0 H then A auto loses

So this is equal to...
(1/32) + (5/32)(15/16) + (10/32)(11/16) + (10/32)(5/16) + (5/32)(1/16) = .47

Anonymous on Jul 10, 2012
1

Anonymous' above analysis is correct, but the first term is 1 / 32 (times 16 / 16) or 16 / 512, so the answer comes out as 1/2 after all if you do the addition correctly.

M on Aug 6, 2012
11

Consider the game where each player has 4 coins. This game is symmetric and therefore p(win)=p(lose) and p(tie)=1-2p(win).

If I have 5 coins, flip one, and it is heads, I can view this game as the previous case where I "win" with probability p(win) + p(tie). If this coin is tails, there is no change. That is I win with probability p(win).

The probability I win in the 5 coins vs 4 coins case is therefore

1/2(p(win)) + 1/2(p(tie)+p(win))=p(win)+1/2p(tie) = p(win) + 1/2(1-2p(win)) = 1/2.

Observe that this argument was independent of the # of coins flipped so the probability of win if i flip n+1 coins and my opponent flips n coins is always 1/2.

Dan on Aug 19, 2012
0

draw a 4 * 5 matrix and tick the entries that A will win. very easy to see it is 10 (out of 20).

t on Nov 25, 2012
0

Consider A:
4T - 1
1H3T - 4
2H2T - 6
3H1T - 4
4H - 1

Consider B:
5T - 1
1H4T - 5
2H3T - 10
3H2T - 10
4H1T - 5
5H - 1

P(winning) = [(1/16)x(1/32)]x(1x31+4x26+6x16+4x6+1x1) = 1/2

Me on May 12, 2014

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