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My teacher asked us this question in class but no one could explain why flipping a coin 2 times in a row is not the same as flipping 2 coins at the same time.

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closed as off-topic by Travis, Delta-u, B. Mehta, Namaste, JonMark Perry May 25 '18 at 0:14

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    $\begingroup$ Did your teacher specify whether or not the coins are fair? If they are not, you need to consider that the bias is different in each coin. $\endgroup$ – Avraham May 24 '18 at 14:43
  • $\begingroup$ It is the same, unless you plan to do something after the first flip that would be affected by the second one if you knew it. Related: math.stackexchange.com/questions/2755438/… (the other comments here are appropriate in other ways) $\endgroup$ – Ethan Bolker May 24 '18 at 14:44
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    $\begingroup$ This question is at best underspecified and at worst misleading. It's possible what your teacher is getting at is that two coins flipped at different times might be considered distinguishable (they are distinguished by the time they are flipped), whereas two coins flipped simultaneously might be considered indistinguishable. $\endgroup$ – Travis May 24 '18 at 14:44
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    $\begingroup$ Anyway, I've voted to close as unclear, because an effective answer depends on assumptions not given. $\endgroup$ – Travis May 24 '18 at 14:45
  • $\begingroup$ If you don't know whether the coin is fair, then knowing the outcome of the first toss says "something" about the outcome of the second toss. If you have two coins, then that is not the case (but not really since the two coins could have been manufactured under the exact same conditions etc.) $\endgroup$ – Calculon May 24 '18 at 14:48
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When we talk about probability, we need to specify what the sample space and probability measure are.

If we flip the coin two times, then the sample space is $\Omega=\{HH,HT,TH,TT\}$ and the probability is the same for each point in $\Omega$.

However, if we consider flip two coins at the same time, the sample space becomes $\Omega=\{\{H,H\},\{H,T\},\{T,T\}\}$. (unordered pairs) Then $P(\{H,H\})=P(\{T,T\})=0.25$ bu t $P(\{H,T\}) =0.5$.

Maybe that's why they are different.

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  • $\begingroup$ A probability space must be constructed, but fortunately we have a great freedom in doing so. IMHO the rules for modelling are not that strict. Further flipping two coins at the same time does not necessarily mean that the order gets lost. If it does then you can create a new order yourself (eg. which coin hits the floor as first?). $\endgroup$ – drhab May 24 '18 at 15:16
  • $\begingroup$ It all depends on whether you want to make the sample space coarser by gluing sample points. $\endgroup$ – Ben May 24 '18 at 15:51
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If the coins involved all have equal probability to produce e.g. a head then the probabilistic model of the first situation is also suitable for the second situation.

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When it comes to combinatorics, they are not the same.

If you consider the case where you flip both coins at once, the set of events is $\Omega_1=\{HH, HT,TT\}=\{HH,TH,TT\}$ as there is no order (although one could define the meaning of order differently. In this case, the order is determined by when you toss the coin(s)). So the cardinality is $\vert \Omega_1 \vert=3.$

But if you consider the case where you flip the coins consecutively, the set of events is $\Omega_1=\{HH, HT,TH,TT\}$, thus cardinality is $\vert \Omega_2 \vert=4.$

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