The Futility Closet has this up today, and I don’t understand it.
The question asked is, if you flip 6 fair coins and I flip 5 fair coins, what is the probability that you’ll get more heads than I do?
Surely having more coins makes it more likely that you’ll get more heads? I mean, if you had 60 coins, and I had 1, you’re almost certain to get more heads than me.
It seems to matter, now that I think about it, that the balance of coins is such that you cannot get more heads and more tails, so there can only be one coin difference to make it work. But it makes my head head hurt, so can anyone explain it in an easy way?
Okay, let’s start with the case where we both flip 5 coins. There’s 3 possibilities there: you get more heads, I get more heads, or we tie.
The way the question is phrased above, you get an advantage in that you flip more coins. But since you only “win” if you get more heads than I do, essentially I win in case of a tie. (Or at least it’s counted the same as if I get more heads.)
It’s a tricky puzzle to actually work out all the probabilities, since your 6 coins has one probability distribution with 7 outcomes (from 0 heads up to 6 heads,) I have another distribution with 6 outcomes, leading to a grid with 42 different spots, each of which has a different chance of coming up. I’ll see what I can come up with quickly.
Sam flips 6 coins, Frank flips 5. As a result, one of the following three things must happen:
(1) Sam gets the same number of heads as Frank. Remove the coins that came up heads: since Sam had one more coin than Frank to begin with, he has one more coin (showing tails) remaining than Frank: Sam has more tails than Frank.
(2) Sam gets fewer heads than Frank. Remove the coins that came up heads, and Sam must have more coins remaining, so again, Sam has more tails than Frank.
(3) Sam gets more heads than Frank. This time, if you remove the coins that came up heads, you’ve removed more coins from Sam than from Frank, so now Frank has at least as many coins remaining as Sam does: Sam has the same or fewer tails as Frank.
Sam get more heads than Frank <–> Sam does not get more tails than Frank (#3)
Sam does not get more heads than Frank <–> Sam gets more tails than Frank. (#1 or 2)
So, one and only one of the following things must happen: “Sam gets more heads than Frank” or “Sam gets more tails than Frank.” There’s no reason for either of these things to be more likely than the other if the coin flips themselves are fair.
I assumed the first hand flips five coins and the second hand flips six.
First hand # of heads Probability % Second hand # of heads Probability %
0 0.03125 0 0.01563
1 0.15625 1 0.09375
2 0.3125 2 0.234375
3 0.3125 3 0.3125
4 0.15625 4 0.234375
5 0.03125 5 0.09375
6 0.01563
Here’s the probability distribution. If you run the numbers, it does in fact work out. The trick, though, is in how the ties are counted.
Only if somebody has a strange definition of “more”. Ties mean the condition of “more heads” is not achieved.
While true, that’s not an analogous situation. It’s more like if you had 60 coins, and I had 59. The key here is that one player has only 1 more coin than the other.
How it changes with the number of coins is an interesting thing to consider though. Say we have Players A and B.
Coins Player A = # Coins Player B
Then P(Player A has more heads) < 1/2
P(Player B has more heads) < 1/2
P(Player B has same or more heads) > 1/2
Say # Coins Player A = # Coins Player B + 1
Then P(Player A has more heads) = 1/2
P(Player B has same or more heads) = 1/2
Say # Coins Player A = # Coins Player B + n, n > 1
Then P(Player A has more heads) > 1/2
P(Player B has same or more heads) < 1/2
The way I think about it is if you had equal number of coins, then obviously you would have the same chance to get any number of heads. The difference when you have one more coin is the outcome of that one flip. So it becomes a 50/50 prospect for who has more heads.
Just came in to agree with Apex Rogers. Isn’t this one of those problems where you get hopelessly bogged down doing it the hard way, until you suddenly get the right perspective, and the answer is obvious? Is there any reason why this is wrong?
Like if you have 128 players in a singles tennis competition, how many matches will there be in total? You can faff about drawing up tables in a hierarchy to account for rounds quarter and semi finals, etc, or you can get that flash of insight that tells you that each game results in one person eliminated. That is true for all games except the last, where the winner is annointed. 128 players means there must be 127 games, and no messing about doing it the hard way required.
**Think of the 5-coin outcome probabilities as a pie chart (win, lose, tie). Converting half of the tie slice to wins means half the pie is wins (e.g. 50%), because the tie slice can be thought to eat into each player’s win percentage equally. ** The six coin provides additional wins in same size as half the tie slice.
E.g. If the probability of 5-coin Ties was 20%, then pre-6th coin each player would have 40% chance of winning (40%PlayerA win + 40%PlayerB win +20%Tie = 100%). The 6th coin let’s a player convert half the ties (10%) to wins, so their win rate jumps to 50%. The calculation of the actual Tie rate or any other probabilities is unnecessary.
The “answer explanation” on the website is ridiculously uninformative.