My girlfriend and I were having a debate about a magic trick.

I am a magician and I have a coin. I walk up to a complete stranger with the coin in my hand and know which side is up. I ask her to state heads or tails. If she picks the correct one I tell her, "I knew you would pick that one. You should marry me!". If however, she doesn't pick the one I know that I have in my hand facing up. I flip the coin, and if I get it correct, I instead say, "You should go and get a drink with me immediately!"

What are the chances that she will be told either that she should marry me or go out and get a drink with me? I have a 50% chance of her getting the first line 100% of the time. That leaves another 50% unaccounted for. I now have a 50% chance of her getting the second line, but only 25% of a chance over all.

Can these two numbers be added together to get 75% chance total? My girlfriend says it doesn't work like that: I am wrong.

  • 9
    $\begingroup$ The scenario sounds like a magic recipe for irritating your girlfriend. On the other hand you've made me curious what she thinks the correct answer is (if she has a specific answer in mind, it should be easy enough to conduct a tabletop experiment which should show one of the two to be wrong). $\endgroup$
    – Glen_b
    Commented Mar 8, 2014 at 22:09
  • 3
    $\begingroup$ She's right, it doesn't work like that. You don't talk to a girlfriend about picking other girls up. $\endgroup$ Commented Mar 8, 2014 at 23:16
  • $\begingroup$ @Glen_b, I have an interest in some of the concepts around mentalism. But, my girlfriend read this and said, "You just proved me right! The internet thinks you're a jerk!". Specifically, she said she can't really explain why this is wrong. $\endgroup$
    – Noah Clark
    Commented Mar 9, 2014 at 0:07
  • 3
    $\begingroup$ Some random person (i.e. me) thinking that behaviour (trying to get girls to go out with you in the guise of 'a magic trick' to resolve a probability argument with your girlfriend) might prove annoying to your girlfriend doesn't show you're a jerk; you might in general be a very nice person. But even if you were a jerk, it doesn't mean your probability calculations were wrong. (Some very good statisticians were also jerks. Indeed Fisher was kind of famous for being something of a jerk... and also for being right almost as often as he thought he was.) $\endgroup$
    – Glen_b
    Commented Mar 9, 2014 at 0:41
  • $\begingroup$ For good reason, some elements of the original post and its title have been flagged as rude or abusive (as well as being irrelevant to the question). Because attempts have been made to restore the original offending language, I have locked this thread. Should anyone feel discussion is warranted, please use our chat room or post a meta thread about your concern. Thank you. $\endgroup$
    – whuber
    Commented Nov 21, 2019 at 18:00

3 Answers 3


You can visualize and reason about problems like this using a probability tree diagram.

enter image description here

You multiply out probabilities horizontally, because probabilities of events happening multiply when the events are independent. So in your example the probability the girl doesn't pick the side of the coin facing up and then after flipping the coin the coin lands on side the that the girl chose is $0.5\times0.5=0.25$.

Can these two numbers be added together to get 75% chance total?

You add the probabilities that you are interested in vertically. This is because the different paths are mutually exclusive (this comment is dead right). So you add the probabilities resulting from the green paths on the diagram to get $0.5+0.25=0.75$. So yes, the probability is 75% total.

My girlfriend says it doesn't work like that I am wrong.

Perhaps she's not talking about the maths!

  • 2
    $\begingroup$ +1 for "Perhaps she's not talking about the maths!" How's my proof? $\endgroup$
    – Aaron Hall
    Commented Mar 8, 2014 at 22:18
  • $\begingroup$ @AaronHall I think your proof is sound (I edited it to romanize your text subscripts, hope you don't mind). What I like is the way that you explicitly use conditional probability; otoh I think at the start you could jump straight to using 0.5 for the probability of heads/tails for a fair coin just as you did for "Assuming a random choice by the one being approached, the probability of either side being chosen is .5.". $\endgroup$
    – TooTone
    Commented Mar 8, 2014 at 22:47

Imagine 1000 people trying out your approach. 500 of them get to ask the woman to marry them. (We are working with expected values here.) Out of the remaining 500 men, 250 of them get to offer the drink. So yes, your reasoning is correct.

  • 3
    $\begingroup$ P(A or B) = P(A)+P(B) if A and B are mutually exclusive. $\endgroup$
    – a11msp
    Commented Mar 8, 2014 at 20:07
  • $\begingroup$ The frequentist approach, right? :) $\endgroup$
    – Aaron Hall
    Commented Mar 9, 2014 at 0:21
  • $\begingroup$ Simplest to understand...and easy to write, too. $\endgroup$
    – soakley
    Commented Mar 9, 2014 at 5:41

For both actions, we can prove that there is a probability equal to .5 for either action, assuming a fair coin and a random choice by the one being asked:

\begin{eqnarray*} P_{H} & = & P_{T} \tag*{Fair Coin, Random Choice}\\ P_{H}+P_{T} & = & 1 \tag*{Axiom of Probability}\\ P_{H}+P_{H} & = & 1\tag*{Substition}\\ 2P_{H} & = & 1 \tag*{Combine} \end{eqnarray*}


\begin{eqnarray*} P_{H} & = & .5\\ P_{T} & = & .5 \end{eqnarray*}

Assuming a random choice by the one being approached, the probability of either side being chosen is .5.

\begin{eqnarray*} P_{\text{Correct Choice}} & = & .5\\ P_{\text{Incorrect Choice}} & = & .5 \end{eqnarray*}

Since this stops the procedure, there is no likelihood of a flip:

\begin{eqnarray*} P_{\text{Correct Flip|Correct Choice}} & = & 0\\ P_{\text{Incorrect Flip|Correct Choice}} & = & 0 \end{eqnarray*}

Assuming a fair coin, given that the wrong side was initially chosen, the probability of that a coin flip lands on the chosen side is .5.

\begin{eqnarray*} P_{\text{Correct Flip|Incorrect Choice}} & = & .5\\ P_{\text{Incorrect Flip|Incorrect Choice}} & = & .5 \end{eqnarray*}

And since $P(A\cap B) = P(A|B)\times P(B)$ the prior probabilities are

\begin{eqnarray*} P_{\text{Correct Flip}} & = & P_{\text{Incorrect Choice}} \times P_{\text{Correct Flip|Incorrect Choice}}\\ & = & .5\times.5\\ & = & .25 \end{eqnarray*}

And similarly:

\begin{eqnarray*} P_{\text{Incorrect Flip}} & = & .25 \end{eqnarray*}

The probabilities prior to the approach are therefore as follows:

\begin{eqnarray*} P_{\text{Correct Choice}} & = & .5\\ P_{\text{Correct Flip}} & = & .25\\ P_{\text{Incorrect Flip}} & = & .25 \end{eqnarray*}

And, since the union of disjoint probabilties is the sum of the probabilities, the probability of what the asker considers to be a successful outcome is therefore 75%.


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